150 Best Algebra Problems With Solutions for Beginners

Algebra gets easier with practice, but only if you practice the right way. This page has 150 original problems that start with simplifying expressions and build up to quadratics, radicals and functions. Every problem has a full worked solution and a quick check, so you can confirm the answer yourself instead of trusting the answer key.

How to use this page:

Work each problem on paper before you open the solution. Getting stuck is part of learning, and it shows you exactly which step needs more practice.
Click Show solution to compare your steps with the worked solution, not just the final answer.
Read the Check line. Substituting your answer back into the original problem is the habit that catches most algebra mistakes.
If a section feels hard, go back to the previous one. The ten sections are graded from easier to harder, and each one uses skills from the earlier ones.

The topics follow the order used in free, openly licensed textbooks such as OpenStax's Elementary Algebra 2e and Intermediate Algebra 2e. Those books are good companions if you want a longer explanation of any rule used here. The problems on this page were written for this article and are not copied from any textbook.

Algebraic expressions (problems 1–15)

These warm-ups cover the order of operations, substitution, combining like terms and the distributive property. Most mistakes later in algebra trace back to a sign error made here, so take these slowly.

1. Order of operations

Evaluate $3 + 4 \cdot 2^2 - 6 \div 3$ .

Show solution

Work through the order of operations: exponents first, then multiplication and division from left to right, then addition and subtraction.

3 + 4 \cdot 2^2 - 6 \div 3 = 3 + 4 \cdot 4 - 6 \div 3 = 3 + 16 - 2 = 17

Answer: $17$

Check: Adding $3 + 4$ first is the classic mistake. It gives $7 \cdot 4 - 2 = 26$ instead of $17$ .

2. Substitute a negative value

Evaluate $2x^2 - 3x + 1$ when $x = -2$ .

Show solution

Replace every $x$ with $(-2)$ , keeping the parentheses so the signs stay correct.

2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 15

Answer: $15$

Check: $(-2)^2 = 4$ , not $-4$ , because the parentheses square the whole negative number. Then $8 + 6 + 1 = 15$ .

3. Combine like terms

Simplify $5a + 3b - 2a + 7b - b$ .

Show solution

Group the $a$ terms and the $b$ terms, then add their coefficients.

(5a - 2a) + (3b + 7b - b) = 3a + 9b

Answer: $3a + 9b$

Check: Let $a = 1$ and $b = 1$ . The original expression is $5 + 3 - 2 + 7 - 1 = 12$ , and $3(1) + 9(1) = 12$ .

4. Distribute a negative number

Simplify $-3(2x - 5)$ .

Show solution

Multiply $-3$ by each term inside the parentheses.

-3 \cdot 2x + (-3)(-5) = -6x + 15

Answer: $-6x + 15$

Check: With $x = 1$ : $-3(2 - 5) = -3(-3) = 9$ , and $-6 + 15 = 9$ .

5. Distribute twice and combine

Simplify $4(x + 2) - 2(3x - 1)$ .

Show solution

Distribute both factors, paying attention to the minus sign in front of the 2.

4x + 8 - 6x + 2 = -2x + 10

Answer: $-2x + 10$

Check: With $x = 2$ : $4(4) - 2(5) = 16 - 10 = 6$ , and $-2(2) + 10 = 6$ .

6. Evaluate a fraction expression

Evaluate $\dfrac{a + b}{a - b}$ when $a = 5$ and $b = -1$ .

Show solution

Substitute, then simplify the numerator and the denominator separately.

\frac{5 + (-1)}{5 - (-1)} = \frac{4}{6} = \frac{2}{3}

Answer: $\dfrac{2}{3}$

Check: Subtracting a negative adds: $5 - (-1) = 6$ . A common slip is writing $5 - 1 = 4$ in the denominator, which would give $1$ .

7. Translate words into algebra

Write an expression for "seven less than twice a number $n$ ."

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"Twice a number" is $2n$ . "Seven less than" something means you start with that something and subtract 7.

Answer: $2n - 7$

Check: If $n = 10$ , twice the number is 20 and seven less is 13. The expression gives $2(10) - 7 = 13$ . Writing $7 - 2n$ instead would give $-13$ .

8. Distribute a leading minus sign

Simplify $-(x^2 - 4x + 3)$ .

Show solution

A minus sign in front of parentheses means multiplying every term by $-1$ .

-(x^2 - 4x + 3) = -x^2 + 4x - 3

Answer: $-x^2 + 4x - 3$

Check: With $x = 1$ : $-(1 - 4 + 3) = 0$ , and $-1 + 4 - 3 = 0$ . With $x = 2$ : $-(4 - 8 + 3) = 1$ , and $-4 + 8 - 3 = 1$ .

9. Negative bases and exponents

Evaluate $(-3)^2 - 3^2$ .

Show solution

In $(-3)^2$ the base is $-3$ , so $(-3)^2 = (-3)(-3) = 9$ . In $3^2$ the base is $3$ , so $3^2 = 9$ .

(-3)^2 - 3^2 = 9 - 9 = 0

Answer: $0$

Check: Note that $-3^2$ (no parentheses) means $-(3^2) = -9$ . Only the parentheses make the negative sign part of the base.

10. Distribute fractions

Simplify $\frac{1}{2}(6y - 4) + \frac{1}{3}(9y + 3)$ .

Show solution

Distribute each fraction, then combine like terms.

3y - 2 + 3y + 1 = 6y - 1

Answer: $6y - 1$

Check: With $y = 1$ : $\frac{1}{2}(2) + \frac{1}{3}(12) = 1 + 4 = 5$ , and $6(1) - 1 = 5$ .

11. Perimeter expression

A rectangle has width $w$ and length $2w + 3$ . Write a simplified expression for its perimeter.

Show solution

Perimeter is twice the length plus twice the width.

P = 2(2w + 3) + 2w = 4w + 6 + 2w = 6w + 6

Answer: $P = 6w + 6$

Check: If $w = 2$ , the rectangle is 2 by 7, so the perimeter is $2 + 7 + 2 + 7 = 18$ . The formula gives $6(2) + 6 = 18$ .

12. Absolute value arithmetic

Evaluate $|4 - 9| - |-2|$ .

Show solution

Simplify inside each absolute value first, then take the distance from zero.

|4 - 9| - |-2| = |-5| - 2 = 5 - 2 = 3

Answer: $3$

Check: Absolute values are never negative, so both $|-5|$ and $|-2|$ become positive before you subtract.

13. Nested grouping symbols

Simplify $3x - [2 - (x - 4)]$ .

Show solution

Work from the innermost parentheses outward.

\begin{aligned} 3x - [2 - (x - 4)] &= 3x - [2 - x + 4] \\ &= 3x - [6 - x] \\ &= 3x - 6 + x \\ &= 4x - 6 \end{aligned}

Answer: $4x - 6$

Check: With $x = 5$ : $15 - [2 - 1] = 14$ , and $4(5) - 6 = 14$ .

14. Fractions and order of operations

Evaluate $\frac{2}{3} + \frac{3}{4} \cdot \frac{8}{9}$ .

Show solution

Multiply before adding. Cancel common factors to keep the numbers small.

\frac{3}{4} \cdot \frac{8}{9} = \frac{24}{36} = \frac{2}{3}, \qquad \frac{2}{3} + \frac{2}{3} = \frac{4}{3}

Answer: $\dfrac{4}{3}$

Check: In decimals: $0.667 + 0.75 \times 0.889 \approx 0.667 + 0.667 = 1.333$ , which matches $\frac{4}{3} \approx 1.333$ .

15. Simplify with two variables

Simplify $2(a - 3b) - 3(a + b) + 4b$ .

Show solution

Distribute, then collect the $a$ terms and the $b$ terms.

2a - 6b - 3a - 3b + 4b = -a - 5b

Answer: $-a - 5b$

Check: With $a = 1$ and $b = 1$ : $2(-2) - 3(2) + 4 = -4 - 6 + 4 = -6$ , and $-1 - 5 = -6$ .

Linear equations (problems 16–30)

A linear equation is solved by doing the same operation to both sides until the variable stands alone. The last few problems include an identity, a contradiction and two short word problems.

16. One-step addition equation

Solve $x + 7 = 12$ .

Show solution

Subtract 7 from both sides to undo the addition.

x + 7 - 7 = 12 - 7 \quad\Rightarrow\quad x = 5

Answer: $x = 5$

Check: $5 + 7 = 12$ . ✓

17. One-step multiplication equation

Solve $4x = -28$ .

Show solution

Divide both sides by 4.

x = \frac{-28}{4} = -7

Answer: $x = -7$

Check: $4(-7) = -28$ . ✓

18. Two-step equation

Solve $3x - 5 = 16$ .

Show solution

Undo the subtraction first, then the multiplication.

\begin{aligned} 3x - 5 &= 16 \\ 3x &= 21 \\ x &= 7 \end{aligned}

Answer: $x = 7$

Check: $3(7) - 5 = 21 - 5 = 16$ . ✓

19. Equation with a fraction

Solve $\frac{x}{4} + 3 = 9$ .

Show solution

Subtract 3, then multiply both sides by 4.

\frac{x}{4} = 6 \quad\Rightarrow\quad x = 24

Answer: $x = 24$

Check: $\frac{24}{4} + 3 = 6 + 3 = 9$ . ✓

20. Variables on both sides

Solve $5x + 2 = 3x + 14$ .

Show solution

Move the variable terms to one side and the constants to the other.

\begin{aligned} 5x - 3x &= 14 - 2 \\ 2x &= 12 \\ x &= 6 \end{aligned}

Answer: $x = 6$

Check: Left: $5(6) + 2 = 32$ . Right: $3(6) + 14 = 32$ . ✓

21. Parentheses on both sides

Solve $2(x - 3) = 3(x + 1)$ .

Show solution

Distribute on both sides, then collect terms.

\begin{aligned} 2x - 6 &= 3x + 3 \\ -6 - 3 &= 3x - 2x \\ x &= -9 \end{aligned}

Answer: $x = -9$

Check: Left: $2(-12) = -24$ . Right: $3(-8) = -24$ . ✓

22. Clear fractions with the LCD

Solve $\frac{x}{2} - \frac{x}{3} = 4$ .

Show solution

Multiply every term by the least common denominator, 6.

\begin{aligned} 6 \cdot \frac{x}{2} - 6 \cdot \frac{x}{3} &= 6 \cdot 4 \\ 3x - 2x &= 24 \\ x &= 24 \end{aligned}

Answer: $x = 24$

Check: $\frac{24}{2} - \frac{24}{3} = 12 - 8 = 4$ . ✓

23. Equation with decimals

Solve $0.2x + 1.5 = 0.5x - 0.3$ .

Show solution

Multiply every term by 10 to clear the decimals.

\begin{aligned} 2x + 15 &= 5x - 3 \\ 18 &= 3x \\ x &= 6 \end{aligned}

Answer: $x = 6$

Check: Left: $0.2(6) + 1.5 = 2.7$ . Right: $0.5(6) - 0.3 = 2.7$ . ✓

24. Watch the subtraction sign

Solve $7 - 2(x + 4) = 3x + 9$ .

Show solution

Distribute $-2$ (not $2$ ) across the parentheses.

\begin{aligned} 7 - 2x - 8 &= 3x + 9 \\ -2x - 1 &= 3x + 9 \\ -10 &= 5x \\ x &= -2 \end{aligned}

Answer: $x = -2$

Check: Left: $7 - 2(2) = 3$ . Right: $3(-2) + 9 = 3$ . ✓

25. Proportion with binomials

Solve $\frac{2x + 1}{3} = \frac{x - 2}{2}$ .

Show solution

Cross-multiply (or multiply both sides by 6).

\begin{aligned} 2(2x + 1) &= 3(x - 2) \\ 4x + 2 &= 3x - 6 \\ x &= -8 \end{aligned}

Answer: $x = -8$

Check: Left: $\frac{-15}{3} = -5$ . Right: $\frac{-10}{2} = -5$ . ✓

26. An identity

Solve $3(x + 2) - x = 2(x + 3)$ .

Show solution

Simplify both sides.

3x + 6 - x = 2x + 6 \quad\Rightarrow\quad 2x + 6 = 2x + 6

Both sides are identical, so the equation is true for every value of $x$ .

Answer: all real numbers

Check: Try any two values. $x = 0$ gives $6 = 6$ and $x = 10$ gives $26 = 26$ .

27. A contradiction

Solve $4x - 3 = 2(2x + 1)$ .

Show solution

Distribute and simplify.

4x - 3 = 4x + 2 \quad\Rightarrow\quad -3 = 2

The variable disappears and leaves a false statement, so no value of $x$ works.

Answer: no solution

Check: For any $x$ , the left side is always 5 less than the right side. For example, $x = 0$ gives $-3$ and $2$ .

28. Solve a formula for a variable

Solve the perimeter formula $P = 2l + 2w$ for $w$ .

Show solution

Treat every letter except $w$ as a known number.

\begin{aligned} P - 2l &= 2w \\ w &= \frac{P - 2l}{2} \end{aligned}

Answer: $w = \dfrac{P - 2l}{2}$ (equivalently $w = \dfrac{P}{2} - l$ )

Check: A 10 by 4 rectangle has $P = 28$ . The formula gives $w = \frac{28 - 20}{2} = 4$ . ✓

29. Consecutive integers

The sum of three consecutive integers is 72. Find the integers.

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Call the integers $n$ , $n + 1$ and $n + 2$ .

\begin{aligned} n + (n + 1) + (n + 2) &= 72 \\ 3n + 3 &= 72 \\ n &= 23 \end{aligned}

Answer: 23, 24 and 25

Check: $23 + 24 + 25 = 72$ , and the numbers are consecutive. ✓

30. Phone bill word problem

A phone plan costs $20 per month plus $0.10 per minute. One month's bill was $32.50. How many minutes were used?

Show solution

Let $m$ be the number of minutes.

\begin{aligned} 20 + 0.10m &= 32.50 \\ 0.10m &= 12.50 \\ m &= 125 \end{aligned}

Answer: 125 minutes

Check: $20 + 0.10(125) = 20 + 12.50 = 32.50$ . ✓

Linear inequalities and absolute value (problems 31–45)

Inequalities are solved almost like equations, with one key exception: when you multiply or divide both sides by a negative number, the inequality sign reverses (OpenStax, Elementary Algebra 2e, section 2.7).

31. One-step inequality

Solve $x - 4 > 3$ .

Show solution

Add 4 to both sides. Adding or subtracting never changes the direction of an inequality.

x > 7

Answer: $x > 7$ , or $(7, \infty)$ in interval notation

Check: Test $x = 8$ : $8 - 4 = 4 > 3$ is true. Test $x = 7$ : $3 > 3$ is false, so 7 itself is excluded.

32. Divide by a negative

Solve $-3x \le 12$ .

Show solution

Divide both sides by $-3$ . Dividing by a negative number reverses the inequality sign.

x \ge \frac{12}{-3} \quad\Rightarrow\quad x \ge -4

Answer: $x \ge -4$ , or $[-4, \infty)$

Check: Test $x = 0$ : $0 \le 12$ is true. Test $x = -5$ : $15 \le 12$ is false, which confirms that numbers below $-4$ fail.

33. Two-step inequality

Solve $2x + 5 < 17$ .

Show solution

Subtract 5, then divide by 2. The divisor is positive, so the sign stays the same.

2x < 12 \quad\Rightarrow\quad x < 6

Answer: $x < 6$ , or $(-\infty, 6)$

Check: Test $x = 5$ : $15 < 17$ is true. Test $x = 6$ : $17 < 17$ is false.

34. Negative coefficient on the variable

Solve $5 - 2x \ge 11$ .

Show solution

Subtract 5, then divide by $-2$ and flip the sign.

-2x \ge 6 \quad\Rightarrow\quad x \le -3

Answer: $x \le -3$ , or $(-\infty, -3]$

Check: Test $x = -3$ : $5 + 6 = 11 \ge 11$ is true. Test $x = 0$ : $5 \ge 11$ is false.

35. Inequality with parentheses

Solve $3(x - 2) > 2x + 1$ .

Show solution

Distribute, then collect terms.

\begin{aligned} 3x - 6 &> 2x + 1 \\ x &> 7 \end{aligned}

Answer: $x > 7$

Check: Test $x = 8$ : $3(6) = 18 > 17$ is true. Test $x = 7$ : $12 > 15$ is false.

36. Multiply by a negative

Solve $\frac{x}{-2} + 1 < 4$ .

Show solution

Subtract 1, then multiply both sides by $-2$ and reverse the sign.

\frac{x}{-2} < 3 \quad\Rightarrow\quad x > -6

Answer: $x > -6$

Check: Test $x = 0$ : $0 + 1 = 1 < 4$ is true. Test $x = -8$ : $4 + 1 = 5 < 4$ is false.

37. Compound "and" inequality

Solve $-3 < 2x + 1 \le 9$ .

Show solution

Do the same operation to all three parts.

\begin{aligned} -4 &< 2x \le 8 \\ -2 &< x \le 4 \end{aligned}

Answer: $-2 < x \le 4$ , or $(-2, 4]$

Check: $x = 4$ gives $2(4) + 1 = 9$ , which is allowed. $x = -2$ gives $-3$ , which is not, because the left inequality is strict.

38. Compound "or" inequality

Solve $x + 2 < -1$ or $3x \ge 12$ .

Show solution

Solve each part separately. Then keep every number that satisfies either part.

x < -3 \quad\text{or}\quad x \ge 4

Answer: $(-\infty, -3) \cup [4, \infty)$

Check: $x = -4$ satisfies the first part, $x = 5$ satisfies the second, and $x = 0$ satisfies neither.

39. Absolute value less than

Solve $|x| < 5$ .

Show solution

$|x| < 5$ means $x$ is less than 5 units from zero, so $x$ lies between $-5$ and $5$ .

Answer: $-5 < x < 5$ , or $(-5, 5)$

Check: $|4| = 4 < 5$ is true, and $|-6| = 6 < 5$ is false.

40. Shifted absolute value

Solve $|x - 3| \le 2$ .

Show solution

Rewrite as a compound inequality, then add 3 to all three parts.

-2 \le x - 3 \le 2 \quad\Rightarrow\quad 1 \le x \le 5

Answer: $[1, 5]$

Check: Both endpoints work: $|1 - 3| = 2$ and $|5 - 3| = 2$ . $x = 6$ fails, since $|3| = 3 > 2$ .

41. Absolute value greater than

Solve $|2x + 1| > 7$ .

Show solution

A "greater than" absolute value splits into two separate inequalities.

2x + 1 > 7 \quad\text{or}\quad 2x + 1 < -7

x > 3 \quad\text{or}\quad x < -4

Answer: $(-\infty, -4) \cup (3, \infty)$

Check: $x = 4$ gives $|9| > 7$ ✓. $x = -5$ gives $|-9| > 7$ ✓. $x = 0$ gives $|1| > 7$ , which is false, so 0 is correctly excluded.

42. Inequality with a fraction

Solve $\frac{2x - 1}{3} \ge x - 2$ .

Show solution

Multiply both sides by 3 (positive, so no flip), then collect terms.

\begin{aligned} 2x - 1 &\ge 3x - 6 \\ 5 &\ge x \end{aligned}

Answer: $x \le 5$

Check: At $x = 5$ both sides equal 3. At $x = 2$ : $1 \ge 0$ is true. At $x = 8$ : $5 \ge 6$ is false.

43. Answer in interval notation

Solve $4x - 7 \le 6x + 3$ and write the answer in interval notation.

Show solution

Move the $x$ terms to the right so the coefficient stays positive.

\begin{aligned} -7 - 3 &\le 6x - 4x \\ -10 &\le 2x \\ -5 &\le x \end{aligned}

Answer: $[-5, \infty)$

Check: At $x = -5$ both sides equal $-27$ . At $x = -6$ : $-31 \le -33$ is false, as expected.

44. Test average word problem

A student scored 78, 85 and 90 on three tests. What score on the fourth test gives an average of at least 85?

Show solution

Let $s$ be the fourth score. The average of four tests must be at least 85.

\begin{aligned} \frac{78 + 85 + 90 + s}{4} &\ge 85 \\ 253 + s &\ge 340 \\ s &\ge 87 \end{aligned}

Answer: at least 87

Check: With $s = 87$ the average is $\frac{340}{4} = 85$ exactly. With $s = 86$ it is $84.75$ , which falls short.

45. An absolute value that can't be negative

Solve $|x + 4| < -2$ .

Show solution

An absolute value is never negative, so it can never be less than $-2$ .

Answer: no solution

Check: Any test value fails. For example, $x = -4$ gives $0 < -2$ , which is false.

Systems of equations (problems 46–60)

A system's solution must satisfy every equation at once. These problems use substitution and elimination, include one system with no solution and one with infinitely many, and finish with word problems.

46. Add the equations

Solve the system $x + y = 10$ and $x - y = 4$ .

Show solution

Add the two equations. The $y$ terms cancel.

2x = 14 \quad\Rightarrow\quad x = 7

Substitute into $x + y = 10$ : $7 + y = 10$ , so $y = 3$ .

Answer: $(7, 3)$

Check: $7 + 3 = 10$ ✓ and $7 - 3 = 4$ ✓.

47. Substitution with y = 2x

Solve the system $y = 2x$ and $x + y = 12$ .

Show solution

Replace $y$ with $2x$ in the second equation.

x + 2x = 12 \quad\Rightarrow\quad 3x = 12 \quad\Rightarrow\quad x = 4

Then $y = 2(4) = 8$ .

Answer: $(4, 8)$

Check: $8 = 2(4)$ ✓ and $4 + 8 = 12$ ✓.

48. Solve one equation for x first

Solve the system $2x + 3y = 12$ and $x - y = 1$ .

Show solution

The second equation gives $x = y + 1$ . Substitute it into the first.

\begin{aligned} 2(y + 1) + 3y &= 12 \\ 5y + 2 &= 12 \\ y &= 2 \end{aligned}

Then $x = 2 + 1 = 3$ .

Answer: $(3, 2)$

Check: $2(3) + 3(2) = 12$ ✓ and $3 - 2 = 1$ ✓.

49. Opposite coefficients

Solve the system $3x + 2y = 16$ and $5x - 2y = 0$ .

Show solution

The $y$ coefficients are already opposites, so add the equations.

8x = 16 \quad\Rightarrow\quad x = 2

Substitute into the first equation: $6 + 2y = 16$ , so $y = 5$ .

Answer: $(2, 5)$

Check: $3(2) + 2(5) = 16$ ✓ and $5(2) - 2(5) = 0$ ✓.

50. Multiply before eliminating

Solve the system $4x - 3y = -1$ and $2x + 5y = 19$ .

Show solution

Multiply the second equation by 2 so that both equations have $4x$ .

4x + 10y = 38

Subtract the first equation from this one.

(4x + 10y) - (4x - 3y) = 38 - (-1) \quad\Rightarrow\quad 13y = 39 \quad\Rightarrow\quad y = 3

Then $2x + 15 = 19$ , so $x = 2$ .

Answer: $(2, 3)$

Check: $4(2) - 3(3) = -1$ ✓ and $2(2) + 5(3) = 19$ ✓.

51. Two lines in slope-intercept form

Solve the system $y = 3x - 4$ and $y = -x + 8$ .

Show solution

Both expressions equal $y$ , so set them equal to each other.

3x - 4 = -x + 8 \quad\Rightarrow\quad 4x = 12 \quad\Rightarrow\quad x = 3

Then $y = 3(3) - 4 = 5$ .

Answer: $(3, 5)$

Check: The second line gives $y = -3 + 8 = 5$ too, so the lines cross at $(3, 5)$ . ✓

52. A system with no solution

Solve the system $2x + y = 5$ and $4x + 2y = 7$ .

Show solution

Multiply the first equation by 2: $4x + 2y = 10$ . Subtract the second equation.

0 = 3

That statement is false, so the lines are parallel and never meet.

Answer: no solution (an inconsistent system)

Check: In slope-intercept form the lines are $y = -2x + 5$ and $y = -2x + 3.5$ . They have the same slope and different intercepts, so they are parallel.

53. Infinitely many solutions

Solve the system $x - 2y = 3$ and $3x - 6y = 9$ .

Show solution

The second equation is exactly 3 times the first. Subtracting 3 times the first from it gives $0 = 0$ , which is always true.

Answer: infinitely many solutions, all points on the line $x - 2y = 3$ (for example $(3, 0)$ and $(5, 1)$ )

Check: $(5, 1)$ satisfies both equations: $5 - 2 = 3$ and $15 - 6 = 9$ .

54. System with fractions

Solve the system $\frac{x}{2} + \frac{y}{3} = 4$ and $x - y = 3$ .

Show solution

Multiply the first equation by 6 to clear fractions: $3x + 2y = 24$ . From the second equation, $x = y + 3$ .

\begin{aligned} 3(y + 3) + 2y &= 24 \\ 5y + 9 &= 24 \\ y &= 3 \end{aligned}

Then $x = 6$ .

Answer: $(6, 3)$

Check: $\frac{6}{2} + \frac{3}{3} = 3 + 1 = 4$ ✓ and $6 - 3 = 3$ ✓.

55. Ticket sales

A school play sold 50 tickets. Adult tickets cost $12, child tickets cost $7, and total sales were $470. How many of each type were sold?

Show solution

Let $a$ be adult tickets and $c$ be child tickets.

a + c = 50, \qquad 12a + 7c = 470

Substitute $c = 50 - a$ .

12a + 7(50 - a) = 470 \quad\Rightarrow\quad 5a + 350 = 470 \quad\Rightarrow\quad a = 24

So $c = 26$ .

Answer: 24 adult tickets and 26 child tickets

Check: $24 + 26 = 50$ tickets, and $12(24) + 7(26) = 288 + 182 = 470$ dollars. ✓

56. Mixture problem

Coffee that costs $8 per pound is mixed with coffee that costs $12 per pound to make 20 pounds of a blend worth $9.50 per pound. How many pounds of each are used?

Show solution

Let $x$ be pounds of the $8 coffee and $y$ be pounds of the $12 coffee. The weights add to 20, and the costs add to $20 \times 9.50 = 190$ dollars.

x + y = 20, \qquad 8x + 12y = 190

Substitute $y = 20 - x$ .

8x + 240 - 12x = 190 \quad\Rightarrow\quad -4x = -50 \quad\Rightarrow\quad x = 12.5

So $y = 7.5$ .

Answer: 12.5 lb of the $8 coffee and 7.5 lb of the $12 coffee

Check: $8(12.5) + 12(7.5) = 100 + 90 = 190$ dollars for 20 lb, which is $9.50 per pound. ✓

57. Sum and difference

Two numbers add to 45 and differ by 9. Find the numbers.

Show solution

Let the larger number be $x$ and the smaller be $y$ : $x + y = 45$ and $x - y = 9$ . Add the equations.

2x = 54 \quad\Rightarrow\quad x = 27, \qquad y = 45 - 27 = 18

Answer: 27 and 18

Check: $27 + 18 = 45$ ✓ and $27 - 18 = 9$ ✓.

58. Coin problem

A jar holds 30 coins, all nickels and dimes, worth $2.15 in total. How many of each coin are there?

Show solution

Let $n$ be nickels and $d$ be dimes. Working in cents avoids decimals.

n + d = 30, \qquad 5n + 10d = 215

Substitute $d = 30 - n$ .

5n + 300 - 10n = 215 \quad\Rightarrow\quad -5n = -85 \quad\Rightarrow\quad n = 17

So $d = 13$ .

Answer: 17 nickels and 13 dimes

Check: $17 + 13 = 30$ coins, and $85 + 130 = 215$ cents. ✓

59. Three equations, three unknowns

Solve the system $x + y + z = 6$ , $x - y + z = 2$ and $2x + y - z = 1$ .

Show solution

Subtract the second equation from the first to eliminate $x$ and $z$ .

2y = 4 \quad\Rightarrow\quad y = 2

Substitute $y = 2$ into the first and third equations.

x + z = 4, \qquad 2x - z = -1

Add these two equations: $3x = 3$ , so $x = 1$ and $z = 3$ .

Answer: $(x, y, z) = (1, 2, 3)$

Check: $1 + 2 + 3 = 6$ ✓, $1 - 2 + 3 = 2$ ✓ and $2 + 2 - 3 = 1$ ✓.

60. Boat and current

A boat travels 24 miles downstream in 2 hours and returns the same 24 miles upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

Show solution

Let $b$ be the boat's speed and $c$ the current's speed, in miles per hour. Downstream the speeds add, and upstream they subtract.

b + c = \frac{24}{2} = 12, \qquad b - c = \frac{24}{3} = 8

Add the equations: $2b = 20$ , so $b = 10$ and $c = 2$ .

Answer: boat 10 mph, current 2 mph

Check: Downstream: $12 \times 2 = 24$ miles ✓. Upstream: $8 \times 3 = 24$ miles ✓.

Exponents and scientific notation (problems 61–75)

Any nonzero number raised to the power 0 equals 1, and a negative exponent means a reciprocal: $a^{-n} = \frac{1}{a^n}$ (OpenStax, sections 6.5 and 6.7). Unless a problem says otherwise, assume variables in denominators are not zero.

61. Product rule

Simplify $x^3 \cdot x^5$ .

Show solution

When you multiply powers with the same base, add the exponents.

x^3 \cdot x^5 = x^{3 + 5} = x^8

Answer: $x^8$

Check: With $x = 2$ : $8 \cdot 32 = 256$ , and $2^8 = 256$ . ✓

62. Power of a power

Simplify $(y^4)^3$ .

Show solution

When you raise a power to a power, multiply the exponents.

(y^4)^3 = y^{4 \cdot 3} = y^{12}

Answer: $y^{12}$

Check: $(y^4)^3 = y^4 \cdot y^4 \cdot y^4$ , and adding the exponents gives $4 + 4 + 4 = 12$ . ✓

63. Power of a product

Simplify $(2a^3)^4$ .

Show solution

Raise each factor to the fourth power.

(2a^3)^4 = 2^4 \cdot (a^3)^4 = 16a^{12}

Answer: $16a^{12}$

Check: With $a = 1$ : $2^4 = 16$ , and $16(1) = 16$ . Forgetting to raise the 2 to the fourth power would give $2a^{12}$ .

64. Quotient rule

Simplify $\dfrac{m^9}{m^4}$ for $m \neq 0$ .

Show solution

When you divide powers with the same base, subtract the exponents.

\frac{m^9}{m^4} = m^{9 - 4} = m^5

Answer: $m^5$

Check: With $m = 2$ : $\frac{512}{16} = 32 = 2^5$ . ✓

65. Zero and negative exponents

Evaluate $5^0 + 3^{-2}$ .

Show solution

Any nonzero number to the power 0 is 1, and a negative exponent means a reciprocal.

5^0 + 3^{-2} = 1 + \frac{1}{3^2} = 1 + \frac{1}{9} = \frac{10}{9}

Answer: $\dfrac{10}{9}$

Check: $3^{-2}$ is a small positive number, about $0.111$ , not $-9$ or $-6$ . So the total is about $1.111 = \frac{10}{9}$ .

66. Divide monomials

Simplify $\dfrac{12x^5y^2}{4x^2y^6}$ for $x, y \neq 0$ .

Show solution

Divide the coefficients, then subtract exponents for each variable.

\frac{12}{4} \cdot x^{5 - 2} \cdot y^{2 - 6} = 3x^3y^{-4} = \frac{3x^3}{y^4}

Answer: $\dfrac{3x^3}{y^4}$

Check: With $x = 1$ and $y = 2$ : $\frac{12 \cdot 4}{4 \cdot 64} = \frac{48}{256} = \frac{3}{16}$ , and $\frac{3}{2^4} = \frac{3}{16}$ . ✓

67. Negative base raised to a power

Simplify $(-2x^2y)^3$ .

Show solution

Cube each factor. An odd power of a negative number is negative.

(-2)^3 (x^2)^3 y^3 = -8x^6y^3

Answer: $-8x^6y^3$

Check: With $x = 1$ and $y = 1$ : $(-2)^3 = -8$ . ✓

68. Negative exponents in a product

Simplify $x^{-3}y^{2} \cdot x^{5}y^{-4}$ and write the answer with positive exponents.

Show solution

Add the exponents for each base.

x^{-3 + 5} \, y^{2 + (-4)} = x^2 y^{-2} = \frac{x^2}{y^2}

Answer: $\dfrac{x^2}{y^2}$

Check: With $x = 2$ and $y = 1$ : $\frac{1}{8} \cdot 1 \cdot 32 \cdot 1 = 4$ , and $\frac{4}{1} = 4$ . ✓

69. Power of a quotient

Simplify $\left(\dfrac{3a^2}{b^3}\right)^2$ .

Show solution

Square the numerator and the denominator separately.

\frac{3^2 (a^2)^2}{(b^3)^2} = \frac{9a^4}{b^6}

Answer: $\dfrac{9a^4}{b^6}$

Check: With $a = 1$ and $b = 1$ : $3^2 = 9$ . ✓

70. Negative exponent on a fraction

Evaluate $\left(\dfrac{2}{5}\right)^{-2}$ .

Show solution

A negative exponent flips the fraction. Then square it.

\left(\frac{2}{5}\right)^{-2} = \left(\frac{5}{2}\right)^{2} = \frac{25}{4}

Answer: $\dfrac{25}{4}$

Check: $\left(\frac{2}{5}\right)^2 = \frac{4}{25}$ , and its reciprocal is $\frac{25}{4}$ . ✓

71. Multiply in scientific notation

Compute $(3 \times 10^4)(2.5 \times 10^{-6})$ and write the answer in scientific notation.

Show solution

Multiply the decimal parts, then add the exponents of 10.

(3 \times 2.5) \times 10^{4 + (-6)} = 7.5 \times 10^{-2}

Answer: $7.5 \times 10^{-2}$ (that is, $0.075$ )

Check: $30\,000 \times 0.0000025 = 0.075$ . ✓

72. Divide in scientific notation

Compute $\dfrac{6 \times 10^{8}}{1.5 \times 10^{3}}$ .

Show solution

Divide the decimal parts, then subtract the exponents of 10.

\frac{6}{1.5} \times 10^{8 - 3} = 4 \times 10^5

Answer: $4 \times 10^5$

Check: $600\,000\,000 \div 1500 = 400\,000$ . ✓

73. Rational exponent

Evaluate $16^{3/4}$ .

Show solution

The denominator 4 means a fourth root, and the numerator 3 means cube it. Taking the root first keeps the numbers small.

16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8

Answer: $8$

Check: $8^4 = 4096$ and $16^3 = 4096$ , so $8$ is the fourth root of $16^3$ . ✓

74. Negative rational exponent

Evaluate $27^{-2/3}$ .

Show solution

Take the reciprocal for the negative sign, the cube root for the 3 and the square for the 2.

27^{-2/3} = \frac{1}{27^{2/3}} = \frac{1}{\left(\sqrt[3]{27}\right)^2} = \frac{1}{3^2} = \frac{1}{9}

Answer: $\dfrac{1}{9}$

Check: $\left(\frac{1}{9}\right)^{-3/2} = 9^{3/2} = 3^3 = 27$ , so working backward recovers 27. ✓

75. Combine several exponent rules

Simplify $\dfrac{(2x^{3})^{2} \cdot x^{-1}}{8x^{4}}$ for $x \neq 0$ .

Show solution

Simplify the numerator first, then divide.

\frac{4x^6 \cdot x^{-1}}{8x^4} = \frac{4x^5}{8x^4} = \frac{x}{2}

Answer: $\dfrac{x}{2}$

Check: With $x = 2$ : $\frac{(16)^2 \cdot \frac{1}{2}}{128} = \frac{128}{128} = 1$ , and $\frac{2}{2} = 1$ . ✓

Polynomials and factoring (problems 76–90)

The first half of this section multiplies polynomials, and the second half reverses the process by factoring. Always look for a greatest common factor (GCF) first. You can check any factoring answer by multiplying it back out.

76. Add polynomials

Simplify $(3x^2 - 2x + 5) + (x^2 + 6x - 8)$ .

Show solution

Combine like terms: $x^2$ with $x^2$ , $x$ with $x$ , and constants with constants.

(3 + 1)x^2 + (-2 + 6)x + (5 - 8) = 4x^2 + 4x - 3

Answer: $4x^2 + 4x - 3$

Check: With $x = 1$ : $6 + (-1) = 5$ , and $4 + 4 - 3 = 5$ . ✓

77. Subtract polynomials

Simplify $(5x^2 + 3x - 1) - (2x^2 - 4x + 6)$ .

Show solution

Subtracting a polynomial means changing the sign of every term in it.

5x^2 + 3x - 1 - 2x^2 + 4x - 6 = 3x^2 + 7x - 7

Answer: $3x^2 + 7x - 7$

Check: With $x = 1$ : $7 - 4 = 3$ , and $3 + 7 - 7 = 3$ . ✓

78. Multiply two binomials

Multiply $(x + 4)(x - 7)$ .

Show solution

Multiply each term in the first binomial by each term in the second.

x^2 - 7x + 4x - 28 = x^2 - 3x - 28

Answer: $x^2 - 3x - 28$

Check: With $x = 1$ : $(5)(-6) = -30$ , and $1 - 3 - 28 = -30$ . ✓

79. Square a binomial

Expand $(2x - 3)^2$ .

Show solution

Use $(a - b)^2 = a^2 - 2ab + b^2$ , or multiply $(2x - 3)(2x - 3)$ out in full.

(2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9

Answer: $4x^2 - 12x + 9$

Check: With $x = 1$ : $(-1)^2 = 1$ , and $4 - 12 + 9 = 1$ . Writing $4x^2 + 9$ (no middle term) would give 13.

80. Difference of squares product

Multiply $(3x + 5)(3x - 5)$ .

Show solution

The middle terms cancel: $(a + b)(a - b) = a^2 - b^2$ .

(3x)^2 - 5^2 = 9x^2 - 25

Answer: $9x^2 - 25$

Check: With $x = 2$ : $(11)(1) = 11$ , and $36 - 25 = 11$ . ✓

81. Binomial times trinomial

Multiply $(x + 2)(x^2 - 3x + 4)$ .

Show solution

Distribute $x$ and then $2$ across the trinomial, and combine like terms.

\begin{aligned} &x^3 - 3x^2 + 4x + 2x^2 - 6x + 8 \\ &= x^3 - x^2 - 2x + 8 \end{aligned}

Answer: $x^3 - x^2 - 2x + 8$

Check: With $x = 1$ : $(3)(2) = 6$ , and $1 - 1 - 2 + 8 = 6$ . ✓

82. Factor out the GCF

Factor $6x^3 - 9x^2 + 15x$ .

Show solution

The greatest common factor of the coefficients is 3, and every term contains at least one $x$ . So factor out $3x$ .

6x^3 - 9x^2 + 15x = 3x(2x^2 - 3x + 5)

Answer: $3x(2x^2 - 3x + 5)$

Check: Distributing $3x$ gives back $6x^3 - 9x^2 + 15x$ . The trinomial does not factor further over the integers, because its discriminant is $9 - 40 < 0$ .

83. Factor a simple trinomial

Factor $x^2 + 9x + 20$ .

Show solution

Find two numbers that multiply to 20 and add to 9. Those numbers are 4 and 5.

x^2 + 9x + 20 = (x + 4)(x + 5)

Answer: $(x + 4)(x + 5)$

Check: $(x + 4)(x + 5) = x^2 + 5x + 4x + 20 = x^2 + 9x + 20$ . ✓

84. Trinomial with a negative constant

Factor $x^2 - 2x - 35$ .

Show solution

You need two numbers that multiply to $-35$ and add to $-2$ . Those numbers are $-7$ and $5$ .

x^2 - 2x - 35 = (x - 7)(x + 5)

Answer: $(x - 7)(x + 5)$

Check: $(x - 7)(x + 5) = x^2 + 5x - 7x - 35 = x^2 - 2x - 35$ . ✓

85. Leading coefficient other than 1

Factor $2x^2 + 7x + 3$ .

Show solution

Use the AC method. Here $a \cdot c = 6$ , and you need two numbers that multiply to 6 and add to 7: 1 and 6. Split the middle term and factor by grouping.

\begin{aligned} 2x^2 + x + 6x + 3 &= x(2x + 1) + 3(2x + 1) \\ &= (2x + 1)(x + 3) \end{aligned}

Answer: $(2x + 1)(x + 3)$

Check: $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$ . ✓

86. AC method with a negative

Factor $6x^2 - x - 2$ .

Show solution

Here $a \cdot c = -12$ , and you need two numbers that multiply to $-12$ and add to $-1$ : $-4$ and $3$ .

\begin{aligned} 6x^2 - 4x + 3x - 2 &= 2x(3x - 2) + 1(3x - 2) \\ &= (3x - 2)(2x + 1) \end{aligned}

Answer: $(3x - 2)(2x + 1)$

Check: $(3x - 2)(2x + 1) = 6x^2 + 3x - 4x - 2 = 6x^2 - x - 2$ . ✓

87. Difference of squares

Factor $49y^2 - 16$ .

Show solution

Both terms are perfect squares: $49y^2 = (7y)^2$ and $16 = 4^2$ . Use $a^2 - b^2 = (a - b)(a + b)$ .

Answer: $(7y - 4)(7y + 4)$

Check: $(7y - 4)(7y + 4) = 49y^2 + 28y - 28y - 16 = 49y^2 - 16$ . ✓

88. Perfect square trinomial

Factor $x^2 - 12x + 36$ .

Show solution

The first and last terms are squares ( $x^2$ and $6^2$ ), and the middle term is $-2 \cdot x \cdot 6$ . So the trinomial is a perfect square.

Answer: $(x - 6)^2$

Check: $(x - 6)^2 = x^2 - 12x + 36$ . ✓

89. Factor by grouping

Factor $x^3 + 3x^2 - 4x - 12$ completely.

Show solution

Group the first two terms and the last two terms.

\begin{aligned} x^2(x + 3) - 4(x + 3) &= (x + 3)(x^2 - 4) \\ &= (x + 3)(x - 2)(x + 2) \end{aligned}

Answer: $(x + 3)(x - 2)(x + 2)$

Check: With $x = 1$ : $1 + 3 - 4 - 12 = -12$ , and $(4)(-1)(3) = -12$ . ✓

90. Factor completely

Factor $3x^3 - 75x$ completely.

Show solution

Take out the GCF $3x$ first, then factor the difference of squares that remains.

3x^3 - 75x = 3x(x^2 - 25) = 3x(x - 5)(x + 5)

Answer: $3x(x - 5)(x + 5)$

Check: With $x = 1$ : $3 - 75 = -72$ , and $3(1)(-4)(6) = -72$ . ✓

Quadratic equations (problems 91–105)

To solve by factoring, set one side equal to zero and use the zero product property: if $a \cdot b = 0$ , then $a = 0$ or $b = 0$ (OpenStax, section 7.6). When factoring fails, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Its discriminant, $b^2 - 4ac$ , tells you whether there are two, one or no real solutions (section 10.3).

91. Square root property

Solve $x^2 = 49$ .

Show solution

Take the square root of both sides, and remember that there are two square roots: a positive one and a negative one.

x = \pm\sqrt{49} = \pm 7

Answer: $x = 7$ or $x = -7$

Check: $7^2 = 49$ and $(-7)^2 = 49$ . ✓

92. Square root of a squared binomial

Solve $(x - 3)^2 = 16$ .

Show solution

Take square roots, then solve the two linear equations.

x - 3 = \pm 4 \quad\Rightarrow\quad x = 3 + 4 = 7 \quad\text{or}\quad x = 3 - 4 = -1

Answer: $x = 7$ or $x = -1$

Check: $(7 - 3)^2 = 16$ ✓ and $(-1 - 3)^2 = 16$ ✓.

93. Solve by factoring

Solve $x^2 - 5x + 6 = 0$ .

Show solution

Factor, then use the zero product property: if a product is 0, at least one factor is 0.

(x - 2)(x - 3) = 0 \quad\Rightarrow\quad x = 2 \quad\text{or}\quad x = 3

Answer: $x = 2$ or $x = 3$

Check: $4 - 10 + 6 = 0$ ✓ and $9 - 15 + 6 = 0$ ✓.

94. Set the equation equal to zero first

Solve $x^2 + 4x = 21$ .

Show solution

The zero product property only works when one side is 0, so subtract 21 first.

x^2 + 4x - 21 = 0 \quad\Rightarrow\quad (x + 7)(x - 3) = 0

Answer: $x = -7$ or $x = 3$

Check: $49 - 28 = 21$ ✓ and $9 + 12 = 21$ ✓.

95. Don't divide by x

Solve $2x^2 - 8x = 0$ .

Show solution

Factor out $2x$ . Dividing both sides by $x$ would lose the solution $x = 0$ .

2x(x - 4) = 0 \quad\Rightarrow\quad x = 0 \quad\text{or}\quad x = 4

Answer: $x = 0$ or $x = 4$

Check: $0 - 0 = 0$ ✓ and $2(16) - 32 = 0$ ✓.

96. Factor with a leading coefficient

Solve $3x^2 + 10x - 8 = 0$ .

Show solution

Here $a \cdot c = -24$ , and you need two numbers that multiply to $-24$ and add to 10: 12 and $-2$ .

\begin{aligned} 3x^2 + 12x - 2x - 8 &= 3x(x + 4) - 2(x + 4) \\ &= (3x - 2)(x + 4) = 0 \end{aligned}

Answer: $x = \dfrac{2}{3}$ or $x = -4$

Check: $x = -4$ : $48 - 40 - 8 = 0$ ✓. $x = \frac{2}{3}$ : $\frac{4}{3} + \frac{20}{3} - 8 = 8 - 8 = 0$ ✓.

97. Complete the square

Solve $x^2 + 6x - 7 = 0$ by completing the square.

Show solution

Move the constant to the right. Then add $\left(\frac{6}{2}\right)^2 = 9$ to both sides.

\begin{aligned} x^2 + 6x + 9 &= 7 + 9 \\ (x + 3)^2 &= 16 \\ x + 3 &= \pm 4 \end{aligned}

Answer: $x = 1$ or $x = -7$

Check: $1 + 6 - 7 = 0$ ✓ and $49 - 42 - 7 = 0$ ✓.

98. Complete the square with an irrational answer

Solve $x^2 - 4x - 1 = 0$ by completing the square.

Show solution

Add 1 to both sides, then add $\left(\frac{-4}{2}\right)^2 = 4$ .

\begin{aligned} x^2 - 4x + 4 &= 1 + 4 \\ (x - 2)^2 &= 5 \\ x &= 2 \pm \sqrt{5} \end{aligned}

Answer: $x = 2 + \sqrt{5}$ or $x = 2 - \sqrt{5}$

Check: The two roots add to $4$ and multiply to $4 - 5 = -1$ . Those match $-b/a = 4$ and $c/a = -1$ for this equation. ✓

99. Quadratic formula

Solve $x^2 + 2x - 4 = 0$ with the quadratic formula.

Show solution

Here $a = 1$ , $b = 2$ and $c = -4$ .

x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2(1)} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}

Answer: $x = -1 + \sqrt{5}$ or $x = -1 - \sqrt{5}$

Check: $-1 + \sqrt{5} \approx 1.236$ , and $1.236^2 + 2(1.236) - 4 \approx 1.528 + 2.472 - 4 = 0$ . ✓

100. Quadratic formula with a ≠ 1

Solve $2x^2 - 3x - 4 = 0$ .

Show solution

The trinomial does not factor nicely, so use the formula with $a = 2$ , $b = -3$ and $c = -4$ .

x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-4)}}{2(2)} = \frac{3 \pm \sqrt{9 + 32}}{4} = \frac{3 \pm \sqrt{41}}{4}

Answer: $x = \dfrac{3 + \sqrt{41}}{4}$ or $x = \dfrac{3 - \sqrt{41}}{4}$

Check: $\frac{3 + \sqrt{41}}{4} \approx 2.351$ , and $2(2.351)^2 - 3(2.351) - 4 \approx 11.054 - 7.053 - 4 \approx 0$ . ✓

101. Negative discriminant

How many real solutions does $x^2 + 2x + 5 = 0$ have?

Show solution

Compute the discriminant $b^2 - 4ac$ .

2^2 - 4(1)(5) = 4 - 20 = -16

The discriminant is negative, so the equation has no real solutions.

Answer: no real solutions

Check: Completing the square gives $(x + 1)^2 + 4 = 0$ . A square plus 4 is always at least 4, so it can never equal 0.

102. Zero discriminant

Solve $4x^2 - 12x + 9 = 0$ .

Show solution

The discriminant is $(-12)^2 - 4(4)(9) = 144 - 144 = 0$ , so there is exactly one real solution. The trinomial is a perfect square.

(2x - 3)^2 = 0 \quad\Rightarrow\quad x = \frac{3}{2}

Answer: $x = \dfrac{3}{2}$

Check: $4\left(\frac{9}{4}\right) - 12\left(\frac{3}{2}\right) + 9 = 9 - 18 + 9 = 0$ . ✓

103. Vertex of a parabola

Find the vertex of $y = x^2 - 6x + 5$ .

Show solution

The vertex's $x$ -coordinate is $-\dfrac{b}{2a}$ .

x = -\frac{-6}{2(1)} = 3, \qquad y = 3^2 - 6(3) + 5 = -4

Answer: $(3, -4)$

Check: Completing the square gives $y = (x - 3)^2 - 4$ , whose lowest point is at $(3, -4)$ . ✓

104. Consecutive integers with a product

The product of two consecutive positive integers is 132. Find them.

Show solution

Let the integers be $n$ and $n + 1$ .

n(n + 1) = 132 \quad\Rightarrow\quad n^2 + n - 132 = 0 \quad\Rightarrow\quad (n + 12)(n - 11) = 0

The problem asks for positive integers, so reject $n = -12$ .

Answer: 11 and 12

Check: $11 \times 12 = 132$ . ✓

105. Projectile height

A ball's height in feet after $t$ seconds is modeled by $h = -16t^2 + 48t + 64$ . When does the ball hit the ground?

Show solution

The ball hits the ground when $h = 0$ . Divide by $-16$ to simplify, then factor.

\begin{aligned} -16t^2 + 48t + 64 &= 0 \\ t^2 - 3t - 4 &= 0 \\ (t - 4)(t + 1) &= 0 \end{aligned}

Time cannot be negative, so reject $t = -1$ .

Answer: after 4 seconds

Check: $-16(16) + 48(4) + 64 = -256 + 192 + 64 = 0$ . ✓

Rational expressions and equations (problems 106–120)

A rational expression is a fraction with polynomials in it. Before solving a rational equation, note the values that make a denominator zero. Any "solution" equal to one of them is extraneous and must be discarded (OpenStax, section 8.6).

106. Excluded values

For which values of $x$ is $\dfrac{x + 1}{x^2 - 9}$ undefined?

Show solution

A rational expression is undefined where its denominator is 0.

x^2 - 9 = 0 \quad\Rightarrow\quad (x - 3)(x + 3) = 0 \quad\Rightarrow\quad x = \pm 3

Answer: $x = 3$ and $x = -3$

Check: Substituting either value makes the denominator $9 - 9 = 0$ .

107. Simplify by canceling a factor

Simplify $\dfrac{x^2 - 4}{x + 2}$ .

Show solution

Factor the numerator, then cancel the common factor. You can only cancel factors, never individual terms.

\frac{(x - 2)(x + 2)}{x + 2} = x - 2, \qquad x \neq -2

Answer: $x - 2$ (for $x \neq -2$ )

Check: With $x = 3$ : $\frac{5}{5} = 1$ , and $3 - 2 = 1$ . ✓

108. Simplify a ratio of trinomials

Simplify $\dfrac{x^2 + 5x + 6}{x^2 - x - 6}$ .

Show solution

Factor the numerator and the denominator.

\frac{(x + 2)(x + 3)}{(x - 3)(x + 2)} = \frac{x + 3}{x - 3}, \qquad x \neq -2,\ 3

Answer: $\dfrac{x + 3}{x - 3}$ (for $x \neq -2, 3$ )

Check: With $x = 4$ : $\frac{16 + 20 + 6}{16 - 4 - 6} = \frac{42}{6} = 7$ , and $\frac{7}{1} = 7$ . ✓

109. Multiply rational expressions

Multiply $\dfrac{3x}{x + 1} \cdot \dfrac{x^2 - 1}{6x^2}$ .

Show solution

Factor everything, then cancel common factors.

\frac{3x}{x + 1} \cdot \frac{(x - 1)(x + 1)}{6x^2} = \frac{x - 1}{2x}, \qquad x \neq 0,\ -1

Answer: $\dfrac{x - 1}{2x}$

Check: With $x = 2$ : $\frac{6}{3} \cdot \frac{3}{24} = \frac{1}{4}$ , and $\frac{1}{4} = \frac{1}{4}$ . ✓

110. Divide rational expressions

Divide $\dfrac{x^2 - 9}{4x} \div \dfrac{x + 3}{8x^2}$ .

Show solution

Multiply by the reciprocal of the second fraction, then cancel.

\frac{(x - 3)(x + 3)}{4x} \cdot \frac{8x^2}{x + 3} = 2x(x - 3), \qquad x \neq 0,\ -3

Answer: $2x(x - 3)$ , or $2x^2 - 6x$

Check: With $x = 1$ : $\frac{-8}{4} \div \frac{4}{8} = -2 \cdot 2 = -4$ , and $2(1)(-2) = -4$ . ✓

111. Same denominator

Simplify $\dfrac{2x}{x - 5} - \dfrac{10}{x - 5}$ .

Show solution

Subtract the numerators and keep the denominator. Then factor and cancel.

\frac{2x - 10}{x - 5} = \frac{2(x - 5)}{x - 5} = 2, \qquad x \neq 5

Answer: $2$ (for $x \neq 5$ )

Check: With $x = 7$ : $\frac{14}{2} - \frac{10}{2} = 7 - 5 = 2$ . ✓

112. Add with different denominators

Add $\dfrac{3}{x} + \dfrac{2}{x + 1}$ .

Show solution

The LCD is $x(x + 1)$ . Rewrite each fraction over the LCD, then add the numerators.

\frac{3(x + 1)}{x(x + 1)} + \frac{2x}{x(x + 1)} = \frac{5x + 3}{x(x + 1)}

Answer: $\dfrac{5x + 3}{x(x + 1)}$

Check: With $x = 1$ : $3 + 1 = 4$ , and $\frac{8}{2} = 4$ . ✓

113. Subtract with different denominators

Simplify $\dfrac{5}{x - 2} - \dfrac{3}{x + 2}$ .

Show solution

Use the LCD $(x - 2)(x + 2)$ , and distribute the minus sign over the whole second numerator.

\frac{5(x + 2) - 3(x - 2)}{(x - 2)(x + 2)} = \frac{5x + 10 - 3x + 6}{x^2 - 4} = \frac{2x + 16}{x^2 - 4}

Answer: $\dfrac{2x + 16}{x^2 - 4}$ , or $\dfrac{2(x + 8)}{(x - 2)(x + 2)}$

Check: With $x = 3$ : $5 - \frac{3}{5} = \frac{22}{5}$ , and $\frac{22}{5}$ from the formula. ✓

114. Factor the denominator first

Simplify $\dfrac{x}{x^2 - 1} + \dfrac{1}{x + 1}$ .

Show solution

Since $x^2 - 1 = (x - 1)(x + 1)$ , the LCD is $(x - 1)(x + 1)$ .

\frac{x}{(x - 1)(x + 1)} + \frac{x - 1}{(x - 1)(x + 1)} = \frac{2x - 1}{(x - 1)(x + 1)}

Answer: $\dfrac{2x - 1}{(x - 1)(x + 1)}$

Check: With $x = 2$ : $\frac{2}{3} + \frac{1}{3} = 1$ , and $\frac{3}{(1)(3)} = 1$ . ✓

115. Complex fraction

Simplify $\dfrac{1 + \frac{1}{x}}{1 - \frac{1}{x^2}}$ .

Show solution

Multiply the numerator and the denominator by $x^2$ , the LCD of the small fractions.

\frac{x^2 + x}{x^2 - 1} = \frac{x(x + 1)}{(x - 1)(x + 1)} = \frac{x}{x - 1}

Answer: $\dfrac{x}{x - 1}$ (for $x \neq 0, \pm 1$ )

Check: With $x = 2$ : $\frac{1.5}{0.75} = 2$ , and $\frac{2}{1} = 2$ . ✓

116. Rational equation

Solve $\dfrac{4}{x} + \dfrac{1}{2} = \dfrac{3}{x}$ .

Show solution

Note that $x \neq 0$ . Multiply every term by the LCD $2x$ .

\begin{aligned} 8 + x &= 6 \\ x &= -2 \end{aligned}

Answer: $x = -2$

Check: $\frac{4}{-2} + \frac{1}{2} = -2 + 0.5 = -1.5$ , and $\frac{3}{-2} = -1.5$ . ✓

117. Cross-multiplication

Solve $\dfrac{3}{x - 1} = \dfrac{5}{x + 3}$ .

Show solution

Note that $x \neq 1$ and $x \neq -3$ . Cross-multiply.

\begin{aligned} 3(x + 3) &= 5(x - 1) \\ 3x + 9 &= 5x - 5 \\ 14 &= 2x \\ x &= 7 \end{aligned}

Answer: $x = 7$

Check: $\frac{3}{6} = \frac{1}{2}$ and $\frac{5}{10} = \frac{1}{2}$ . ✓

118. Extraneous solution

Solve $\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 3$ .

Show solution

Note that $x \neq 2$ . Multiply every term by $x - 2$ .

\begin{aligned} x &= 2 + 3(x - 2) \\ x &= 3x - 4 \\ x &= 2 \end{aligned}

But $x = 2$ makes the denominators zero, so it is an extraneous solution and must be rejected.

Answer: no solution

Check: Substituting $x = 2$ into the original equation gives $\frac{2}{0}$ , which is undefined, so the candidate fails.

119. Rational equation that becomes quadratic

Solve $\dfrac{1}{x} + \dfrac{1}{x + 2} = \dfrac{4}{3}$ .

Show solution

Note that $x \neq 0$ and $x \neq -2$ . Multiply every term by $3x(x + 2)$ .

\begin{aligned} 3(x + 2) + 3x &= 4x(x + 2) \\ 6x + 6 &= 4x^2 + 8x \\ 0 &= 4x^2 + 2x - 6 \\ 0 &= 2x^2 + x - 3 \\ 0 &= (2x + 3)(x - 1) \end{aligned}

Neither candidate is an excluded value, so both are solutions.

Answer: $x = 1$ or $x = -\dfrac{3}{2}$

Check: $x = 1$ : $1 + \frac{1}{3} = \frac{4}{3}$ ✓. $x = -\frac{3}{2}$ : $-\frac{2}{3} + 2 = \frac{4}{3}$ ✓.

120. Work-rate problem

One pipe fills a tank in 6 hours and a second pipe fills it in 3 hours. How long does it take with both pipes open?

Show solution

The first pipe fills $\frac{1}{6}$ of the tank per hour and the second fills $\frac{1}{3}$ . Let $t$ be the time together.

\frac{t}{6} + \frac{t}{3} = 1 \quad\Rightarrow\quad t + 2t = 6 \quad\Rightarrow\quad t = 2

Answer: 2 hours

Check: In 2 hours the first pipe fills $\frac{2}{6} = \frac{1}{3}$ and the second fills $\frac{2}{3}$ . Together that is one full tank. ✓

Radicals and radical equations (problems 121–135)

Simplify radicals by pulling out perfect-square (or perfect-cube) factors. When you square both sides of a radical equation, you can create a value that does not satisfy the original equation, so checking is a required step, not an optional one (OpenStax, section 9.6).

121. Simplify a square root

Simplify $\sqrt{72}$ .

Show solution

Find the largest perfect-square factor of 72, which is 36.

\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}

Answer: $6\sqrt{2}$

Check: $(6\sqrt{2})^2 = 36 \cdot 2 = 72$ . ✓

122. Square root with a variable

Simplify $\sqrt{50x^3}$ for $x \ge 0$ .

Show solution

Pull out perfect squares from both the number and the variable: $50 = 25 \cdot 2$ and $x^3 = x^2 \cdot x$ .

\sqrt{25x^2 \cdot 2x} = 5x\sqrt{2x}

Answer: $5x\sqrt{2x}$

Check: $(5x\sqrt{2x})^2 = 25x^2 \cdot 2x = 50x^3$ . ✓

123. Cube root of a negative

Simplify $\sqrt[3]{-54}$ .

Show solution

Cube roots of negative numbers are real. The largest perfect-cube factor is $-27$ .

\sqrt[3]{-54} = \sqrt[3]{-27 \cdot 2} = -3\sqrt[3]{2}

Answer: $-3\sqrt[3]{2}$

Check: $(-3\sqrt[3]{2})^3 = -27 \cdot 2 = -54$ . ✓

124. Add like radicals

Simplify $3\sqrt{5} + 4\sqrt{5} - \sqrt{5}$ .

Show solution

Like radicals combine the same way as like terms.

(3 + 4 - 1)\sqrt{5} = 6\sqrt{5}

Answer: $6\sqrt{5}$

Check: With $\sqrt{5} \approx 2.236$ : $6.708 + 8.944 - 2.236 = 13.416 \approx 6(2.236)$ . ✓

125. Simplify before adding

Simplify $\sqrt{12} + \sqrt{27}$ .

Show solution

Simplify each radical first, then combine.

\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}

Answer: $5\sqrt{3}$

Check: $\sqrt{12} \approx 3.464$ and $\sqrt{27} \approx 5.196$ , which sum to $8.660$ . Also $5\sqrt{3} \approx 8.660$ . Note that $\sqrt{12 + 27} = \sqrt{39} \approx 6.245$ is different.

126. Multiply radicals

Simplify $\sqrt{6} \cdot \sqrt{15}$ .

Show solution

Multiply under one radical, then simplify.

\sqrt{6 \cdot 15} = \sqrt{90} = \sqrt{9 \cdot 10} = 3\sqrt{10}

Answer: $3\sqrt{10}$

Check: $(3\sqrt{10})^2 = 90 = 6 \cdot 15$ . ✓

127. Multiply binomials with radicals

Multiply $(2 + \sqrt{3})(4 - \sqrt{3})$ .

Show solution

Multiply each term in the first factor by each term in the second, remembering that $\sqrt{3} \cdot \sqrt{3} = 3$ .

8 - 2\sqrt{3} + 4\sqrt{3} - 3 = 5 + 2\sqrt{3}

Answer: $5 + 2\sqrt{3}$

Check: Numerically, $(3.732)(2.268) \approx 8.464$ , and $5 + 2(1.732) = 8.464$ . ✓

128. Square a radical binomial

Expand $(\sqrt{7} - 2)^2$ .

Show solution

Use $(a - b)^2 = a^2 - 2ab + b^2$ .

(\sqrt{7})^2 - 2 \cdot 2\sqrt{7} + 4 = 11 - 4\sqrt{7}

Answer: $11 - 4\sqrt{7}$

Check: $\sqrt{7} - 2 \approx 0.6458$ , and its square is about $0.417$ . Also $11 - 4(2.6458) \approx 0.417$ . ✓

129. Rationalize a denominator

Rationalize $\dfrac{6}{\sqrt{3}}$ .

Show solution

Multiply the top and the bottom by $\sqrt{3}$ .

\frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}

Answer: $2\sqrt{3}$

Check: $\frac{6}{1.732} \approx 3.464$ and $2(1.732) = 3.464$ . ✓

130. Rationalize with a conjugate

Rationalize $\dfrac{4}{3 - \sqrt{5}}$ .

Show solution

Multiply the top and the bottom by the conjugate $3 + \sqrt{5}$ .

\frac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{4(3 + \sqrt{5})}{9 - 5} = 3 + \sqrt{5}

Answer: $3 + \sqrt{5}$

Check: $(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$ , so $3 + \sqrt{5}$ really is $4 \div (3 - \sqrt{5})$ . ✓

131. Basic radical equation

Solve $\sqrt{x + 3} = 5$ .

Show solution

Square both sides to remove the radical.

x + 3 = 25 \quad\Rightarrow\quad x = 22

Answer: $x = 22$

Check: $\sqrt{22 + 3} = \sqrt{25} = 5$ . ✓

132. Isolate the radical first

Solve $\sqrt{2x - 1} + 3 = 6$ .

Show solution

Get the radical alone before squaring.

\begin{aligned} \sqrt{2x - 1} &= 3 \\ 2x - 1 &= 9 \\ x &= 5 \end{aligned}

Answer: $x = 5$

Check: $\sqrt{9} + 3 = 6$ . ✓

133. Radical equation with an extraneous root

Solve $\sqrt{x + 6} = x$ .

Show solution

Square both sides and solve the resulting quadratic.

\begin{aligned} x + 6 &= x^2 \\ 0 &= x^2 - x - 6 \\ 0 &= (x - 3)(x + 2) \end{aligned}

Check both candidates in the original equation. $x = 3$ : $\sqrt{9} = 3$ ✓. $x = -2$ : $\sqrt{4} = 2 \neq -2$ ✗, so $-2$ is extraneous.

Answer: $x = 3$

Check: Squaring can create extra solutions, because $2^2$ and $(-2)^2$ are both 4. That is why the check in the original equation is required.

134. Another extraneous root

Solve $\sqrt{3x + 1} = x - 3$ .

Show solution

Square both sides, remembering that $(x - 3)^2 = x^2 - 6x + 9$ .

\begin{aligned} 3x + 1 &= x^2 - 6x + 9 \\ 0 &= x^2 - 9x + 8 \\ 0 &= (x - 1)(x - 8) \end{aligned}

$x = 1$ : $\sqrt{4} = 2$ but $1 - 3 = -2$ ✗. $x = 8$ : $\sqrt{25} = 5$ and $8 - 3 = 5$ ✓.

Answer: $x = 8$

Check: Only $x = 8$ makes both sides equal. $x = 1$ is extraneous.

135. Pythagorean theorem with a radical

A right triangle has legs of length 4 and 6. Find the exact length of the hypotenuse.

Show solution

Use $a^2 + b^2 = c^2$ .

c^2 = 4^2 + 6^2 = 16 + 36 = 52 \quad\Rightarrow\quad c = \sqrt{52} = 2\sqrt{13}

Lengths are positive, so take only the positive root.

Answer: $2\sqrt{13} \approx 7.21$

Check: $(2\sqrt{13})^2 = 4 \cdot 13 = 52 = 16 + 36$ . ✓

Functions and word problems (problems 136–150)

A function assigns exactly one output to each input (OpenStax, Intermediate Algebra 2e, section 3.5). The section ends with word problems that turn a situation into a function and then solve it.

136. Evaluate a function

If $f(x) = 3x - 5$ , find $f(4)$ and $f(-2)$ .

Show solution

Replace $x$ with each input.

f(4) = 3(4) - 5 = 7, \qquad f(-2) = 3(-2) - 5 = -11

Answer: $f(4) = 7$ and $f(-2) = -11$

Check: $f(x)$ is a line with slope 3. Moving from $x = -2$ to $x = 4$ is a change of 6 in $x$ , so the output changes by $3 \times 6 = 18$ , and $-11 + 18 = 7$ . ✓

137. Evaluate a quadratic function

If $g(x) = x^2 - 2x$ , find $g(-3)$ .

Show solution

Keep the input in parentheses.

g(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

Answer: $15$

Check: Factoring gives $g(x) = x(x - 2)$ , so $g(-3) = (-3)(-5) = 15$ . ✓

138. Is it a function?

Is the relation $\{(1, 2), (2, 4), (1, 5)\}$ a function?

Show solution

A function assigns exactly one output to each input. Here the input 1 is paired with both 2 and 5.

Answer: no, it is not a function

Check: List the inputs: 1, 2, 1. A repeated input with different outputs breaks the definition.

139. Domain of a rational function

Find the domain of $f(x) = \dfrac{1}{x - 4}$ .

Show solution

Division by zero is undefined, so exclude any $x$ that makes $x - 4 = 0$ .

Answer: all real numbers except $4$ , or $(-\infty, 4) \cup (4, \infty)$

Check: $f(4) = \frac{1}{0}$ is undefined. Every other input, such as $f(5) = 1$ or $f(0) = -\frac{1}{4}$ , gives a real number.

140. Domain of a square root function

Find the domain of $f(x) = \sqrt{x + 2}$ .

Show solution

For a real square root, the expression inside must be at least 0.

x + 2 \ge 0 \quad\Rightarrow\quad x \ge -2

Answer: $[-2, \infty)$

Check: $f(-2) = \sqrt{0} = 0$ is defined. $f(-3) = \sqrt{-1}$ is not a real number.

141. Slope between two points

Find the slope of the line through $(2, 3)$ and $(6, 11)$ .

Show solution

Slope is the change in $y$ divided by the change in $x$ .

m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2

Answer: $m = 2$

Check: Starting at $(2, 3)$ and moving 4 units right at slope 2 raises $y$ by 8, which lands on $(6, 11)$ . ✓

142. Point-slope form

Write the equation of the line through $(1, 4)$ with slope $-3$ in slope-intercept form.

Show solution

Start from point-slope form, $y - y_1 = m(x - x_1)$ .

\begin{aligned} y - 4 &= -3(x - 1) \\ y - 4 &= -3x + 3 \\ y &= -3x + 7 \end{aligned}

Answer: $y = -3x + 7$

Check: At $x = 1$ : $y = -3 + 7 = 4$ , so the line passes through $(1, 4)$ . ✓

143. Line through two points

Find the equation of the line through $(-2, 1)$ and $(4, 13)$ .

Show solution

First find the slope, then the $y$ -intercept.

m = \frac{13 - 1}{4 - (-2)} = \frac{12}{6} = 2

Substitute $(4, 13)$ into $y = 2x + b$ : $13 = 8 + b$ , so $b = 5$ .

Answer: $y = 2x + 5$

Check: $2(-2) + 5 = 1$ ✓ and $2(4) + 5 = 13$ ✓.

144. Solve for an input

If $f(x) = 2x + 1$ , find $x$ so that $f(x) = 13$ .

Show solution

Set the function equal to 13 and solve.

2x + 1 = 13 \quad\Rightarrow\quad 2x = 12 \quad\Rightarrow\quad x = 6

Answer: $x = 6$

Check: $f(6) = 12 + 1 = 13$ . ✓

145. Composition order matters

Let $f(x) = x + 3$ and $g(x) = 2x$ . Find $f(g(5))$ and $g(f(5))$ .

Show solution

Always work from the inside out.

f(g(5)) = f(10) = 13, \qquad g(f(5)) = g(8) = 16

Answer: $f(g(5)) = 13$ and $g(f(5)) = 16$

Check: The two results differ, which shows that composition is generally not commutative. In general $f(g(x)) = 2x + 3$ and $g(f(x)) = 2x + 6$ .

146. Linear cost model

Suppose a taxi charges $3 to start plus $2.50 per mile. Write the cost $C$ as a function of miles $m$ , and find the cost of an 8-mile ride.

Show solution

The fixed charge is the intercept and the per-mile rate is the slope.

C(m) = 3 + 2.5m, \qquad C(8) = 3 + 2.5(8) = 23

Answer: $C(m) = 3 + 2.5m$ , and an 8-mile ride costs $23

Check: 8 miles at $2.50 is $20, and adding the $3 start fee gives $23. ✓

147. Growth word problem

A plant is 4 cm tall and grows 1.5 cm per week. After how many weeks will it be 13 cm tall?

Show solution

Let $t$ be weeks. Height is $h(t) = 4 + 1.5t$ .

4 + 1.5t = 13 \quad\Rightarrow\quad 1.5t = 9 \quad\Rightarrow\quad t = 6

Answer: 6 weeks

Check: $4 + 1.5(6) = 4 + 9 = 13$ cm. ✓

148. Distance-rate-time

Two cars leave the same point at the same time and drive in opposite directions, one at 55 mph and the other at 65 mph. When will they be 300 miles apart?

Show solution

The distance between them grows at $55 + 65 = 120$ miles per hour.

120t = 300 \quad\Rightarrow\quad t = 2.5

Answer: after 2.5 hours (2 hours 30 minutes)

Check: In 2.5 hours the cars travel $137.5$ and $162.5$ miles, which add to 300. ✓

149. Percent discount

After a 20% discount, a jacket costs $48. What was the original price?

Show solution

A 20% discount leaves 80% of the original price $p$ .

0.8p = 48 \quad\Rightarrow\quad p = \frac{48}{0.8} = 60

Answer: $60

Check: 20% of $60 is $12, and $60 - 12 = 48$ . Adding 20% to $48 instead gives $57.60, which is the common mistake.

150. Maximize an area

A rectangle is enclosed with 40 m of fencing. Write its area as a function of the width $w$ and find the largest possible area.

Show solution

The perimeter gives $2l + 2w = 40$ , so $l = 20 - w$ . The area is then

A(w) = w(20 - w) = -w^2 + 20w

This is a downward parabola, and its vertex is at $w = -\frac{20}{2(-1)} = 10$ .

A(10) = 10(20 - 10) = 100

Answer: $A(w) = -w^2 + 20w$ , with a maximum area of $100\ \text{m}^2$ (a 10 m by 10 m square)

Check: Nearby widths give less area: $A(9) = 9 \times 11 = 99$ and $A(11) = 11 \times 9 = 99$ . ✓

Common mistakes to watch for

These are the errors the problems above are built to expose. Each one has a quick test you can run on your own work.

Dropping the middle term when squaring. $(a + b)^2$ is $a^2 + 2ab + b^2$ , not $a^2 + b^2$ . A university course handout on serious algebra errors lists this one alongside $\sqrt{x + y} = \sqrt{x} + \sqrt{y}$ and $\frac{1}{x + y} = \frac{1}{x} + \frac{1}{y}$ , which are also false (UIC Math 220 handout). Test with $a = b = 1$ : $(1 + 1)^2 = 4$ , but $1^2 + 1^2 = 2$ . See problems 79 and 128.
Distributing a minus sign to only the first term. $-(x - 4)$ is $-x + 4$ . The minus sign multiplies every term in the parentheses. See problems 8, 24 and 77.
Forgetting to flip the inequality. Multiplying or dividing by a negative number reverses the sign (OpenStax, section 2.7). Test a number from your answer in the original inequality. See problems 32, 34 and 36.
Treating $-3^2$ like $(-3)^2$ . Without parentheses, the exponent applies only to the 3, so $-3^2 = -9$ while $(-3)^2 = 9$ . See problems 2 and 9.
Canceling terms instead of factors. In $\frac{x^2 - 4}{x + 2}$ you cannot cancel the $x$ 's or the 2s. Factor first, then cancel whole factors. See problems 107 and 108.
Dividing both sides by a variable. In $2x^2 = 8x$ , dividing by $x$ loses the solution $x = 0$ . Move everything to one side and factor instead. See problem 95.
Skipping the check on rational and radical equations. Multiplying by a variable denominator or squaring both sides can introduce extraneous solutions (OpenStax, sections 8.6 and 9.6). See problems 118, 133 and 134.
Forgetting the $\pm$ . $x^2 = 49$ has two solutions, $7$ and $-7$ . See problems 91 and 92.
Keeping answers that don't fit the situation. A negative time or length can solve the equation and still make no sense for the problem. See problems 104, 105 and 135.

Next steps

Keep the key rules from this page in one place with Math Formulas Every Student Should Know.
Once your answers are reliably correct, work on speed with Math Tricks to Solve Problems Faster.
See where functions and equations show up in data with Probability and Statistics: Real-World Examples.
For an idea of where systems of equations lead next, read Linear Algebra Concepts for Machine Learning.
Browse more articles and blogs in the Mathematics category.
For more exercises with answers, the free OpenStax books in the references below include end-of-section practice sets. The Algebra and Trigonometry 2e book picks up where this page ends, with functions, polynomials and logarithms.

References

Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. Elementary Algebra 2e. OpenStax, Rice University, 2020. CC BY-NC-SA 4.0. https://openstax.org/details/books/elementary-algebra-2e. Sections cited: 2.7 Solve Linear Inequalities, 6.5 Divide Monomials, 6.7 Integer Exponents and Scientific Notation, 7.6 Quadratic Equations, 8.6 Solve Rational Equations, 9.6 Solve Equations with Square Roots, 10.3 Solve Quadratic Equations Using the Quadratic Formula.
Marecek, L., and Honeycutt Mathis, A. Intermediate Algebra 2e. OpenStax, Rice University, 2020. CC BY-NC-SA 4.0. https://openstax.org/details/books/intermediate-algebra-2e. Section cited: 3.5 Relations and Functions.
Abramson, J., et al. Algebra and Trigonometry 2e. OpenStax, Rice University, 2021. CC BY-NC-SA 4.0. https://openstax.org/details/books/algebra-and-trigonometry-2e.
"Serious Algebra and Calculus Errors." Math 220 course handout, University of Illinois Chicago, 2006. https://homepages.math.uic.edu/~tier/M220/Errors/errors.pdf.