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Linear Equations in One Variable

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Question 1Question →

In the equation below, aa is a constant.

13(ax−2)−12(x−4a)=3x−5 \frac{1}{3}(ax - 2) - \frac{1}{2}(x - 4a) = 3x - 5

If the solution to the equation is x=2x = 2, what is the value of aa?

Show answer & explanation

Answer: 11

Answer

1
Substituting x=2x = 2 into the equation gives 13(2a−2)−12(2−4a)=3(2)−5\frac{1}{3}(2a - 2) - \frac{1}{2}(2 - 4a) = 3(2) - 5, which simplifies to 13(2a−2)−(1−2a)=1\frac{1}{3}(2a - 2) - (1 - 2a) = 1. Multiplying the entire equation by 3 to clear the fraction yields (2a−2)−3(1−2a)=3(2a - 2) - 3(1 - 2a) = 3. Distributing the negative 3 yields 2a−2−3+6a=32a - 2 - 3 + 6a = 3, which simplifies to 8a−5=38a - 5 = 3. Adding 5 to both sides results in 8a=88a = 8, which gives a=1a = 1.

Step-by-Step Solution

1
Substitute x=2x = 2 into the given equation.
13(2a−2)−12(2−4a)=3(2)−5\frac{1}{3}(2a - 2) - \frac{1}{2}(2 - 4a) = 3(2) - 5
Since x=2x = 2 is a solution to the equation, substituting it into the equation must yield a true mathematical statement.
2
Simplify the constant terms and distribute or simplify the expressions.
13(2a−2)−(1−2a)=1\frac{1}{3}(2a - 2) - (1 - 2a) = 1
Simplifying 3(2)−53(2) - 5 yields 11, and simplifying the second fraction term gives 12(2−4a)=1−2a\frac{1}{2}(2 - 4a) = 1 - 2a.
3
Multiply the entire equation by 3 to eliminate the fraction.
(2a−2)−3(1−2a)=3(2a - 2) - 3(1 - 2a) = 3
Multiplying all terms on both sides of the equation by the denominator 3 clears the fraction and simplifies the equation.
4
Distribute the negative 3 and combine like terms to solve for aa.
2a−2−3+6a=3  ⟹  8a−5=3  ⟹  8a=8  ⟹  a=12a - 2 - 3 + 6a = 3 \implies 8a - 5 = 3 \implies 8a = 8 \implies a = 1
Distributing −3(1−2a)-3(1 - 2a) yields −3+6a-3 + 6a. Combining like terms gives 8a−5=38a - 5 = 3, and adding 5 followed by dividing by 8 isolates the variable aa.

Key Concept

Solving linear equations in one variable containing fractions and constant parameters by substitution and term isolation.
Estimated Time:2m 0s
Question 2Question →

If 5−2(x+3)=−95 - 2(x + 3) = -9, what is the value of xx?

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Answer: 4

Answer

4
Distributing the −2-2 to both terms in the parentheses gives the equation 5−2x−6=−95 - 2x - 6 = -9. Combining the constants on the left side yields −2x−1=−9-2x - 1 = -9. Adding 11 to both sides gives −2x=−8-2x = -8. Finally, dividing by −2-2 yields the correct value of 44.

Step-by-Step Solution

1
Distribute the −2-2 to both terms inside the parentheses: −2-2 times xx and −2-2 times 33.
5−2x−6=−95 - 2x - 6 = -9
To remove the parentheses and simplify the equation.
2
Combine the constant terms on the left side of the equation: 5−65 - 6.
−2x−1=−9-2x - 1 = -9
To group like terms together before isolating the variable.
3
Add 11 to both sides of the equation.
−2x=−8-2x = -8
To isolate the variable term on the left side of the equation.
4
Divide both sides of the equation by −2-2.
x=4x = 4
To solve for xx.

Key Concept

Solving a linear equation in one variable by distributing, combining like terms, and isolating the variable.
Question 3Question →

If 23(3x−12)−34(2x−13)=16(x+4)−512\frac{2}{3}\left(3x - \frac{1}{2}\right) - \frac{3}{4}\left(2x - \frac{1}{3}\right) = \frac{1}{6}(x + 4) - \frac{5}{12}, what is the value of 12x−512x - 5?

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Answer: 7

Answer

7
The correct answer is 7. Expanding the left side of the equation yields 2x−13−32x+14=12x−1122x - \frac{1}{3} - \frac{3}{2}x + \frac{1}{4} = \frac{1}{2}x - \frac{1}{12}. Expanding the right side yields 16x+46−512=16x+14\frac{1}{6}x + \frac{4}{6} - \frac{5}{12} = \frac{1}{6}x + \frac{1}{4}. Setting the two sides equal gives 12x−112=16x+14\frac{1}{2}x - \frac{1}{12} = \frac{1}{6}x + \frac{1}{4}. Multiplying all terms by 12 clears the fractions, resulting in 6x−1=2x+36x - 1 = 2x + 3. Solving for xx gives 4x=44x = 4, which means x=1x = 1. Substituting x=1x = 1 into 12x−512x - 5 yields 12(1)−5=712(1) - 5 = 7.

Step-by-Step Solution

1
Distribute the factors on the left side of the equation: 23(3x−12)\frac{2}{3}\left(3x - \frac{1}{2}\right) and −34(2x−13)-\frac{3}{4}\left(2x - \frac{1}{3}\right).
2x−13−32x+142x - \frac{1}{3} - \frac{3}{2}x + \frac{1}{4}
To eliminate the parentheses and prepare the left side of the equation for combining like terms.
2
Combine the variable terms and the constant terms on the left side: (2x−32x)+(−13+14)\left(2x - \frac{3}{2}x\right) + \left(-\frac{1}{3} + \frac{1}{4}\right).
12x−112\frac{1}{2}x - \frac{1}{12}
To simplify the left side into a single linear expression with a common denominator for the constants.
3
Expand and simplify the right side of the equation: 16(x+4)−512\frac{1}{6}(x + 4) - \frac{5}{12}.
16x+14\frac{1}{6}x + \frac{1}{4}
By distributing 16\frac{1}{6}, we get 16x+46−512\frac{1}{6}x + \frac{4}{6} - \frac{5}{12}. Finding a common denominator of 12 for the constant terms yields 812−512=312=14\frac{8}{12} - \frac{5}{12} = \frac{3}{12} = \frac{1}{4}.
4
Equate the simplified left and right sides: 12x−112=16x+14\frac{1}{2}x - \frac{1}{12} = \frac{1}{6}x + \frac{1}{4}. Multiply both sides of the equation by 12 to clear the fractions.
6x−1=2x+36x - 1 = 2x + 3, which simplifies to 4x=44x = 4, and thus x=1x = 1.
To solve for the variable xx in a simplified integer form.
5
Substitute x=1x = 1 into the requested expression 12x−512x - 5.
12(1)−5=712(1) - 5 = 7
To calculate the final value asked by the question.

Key Concept

Solving multi-step linear equations in one variable with fractional coefficients, distributing negative signs, and evaluating expressions.
Question 4Question →

If 12(4x−8)−23(3−9x)=5(x−1)+7\frac{1}{2}(4x - 8) - \frac{2}{3}(3 - 9x) = 5(x - 1) + 7, what is the value of 3x−23x - 2?

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Answer: 6

Answer

6
The correct value is 6. After correctly distributing the coefficients on both sides, the equation becomes 8x−6=5x+28x - 6 = 5x + 2. Isolating the variable yields 3x=83x = 8, which gives x=83x = \frac{8}{3}. Substituting this back into the target expression 3x−23x - 2 results in 3(83)−2=63(\frac{8}{3}) - 2 = 6.

Step-by-Step Solution

1
Distribute the fraction coefficients on the left side of the equation.
12(4x−8)−23(3−9x)=2x−4−2+6x=8x−6\frac{1}{2}(4x - 8) - \frac{2}{3}(3 - 9x) = 2x - 4 - 2 + 6x = 8x - 6
To simplify the expression by removing the parentheses on the left side.
2
Distribute and simplify the right side of the equation.
5(x−1)+7=5x−5+7=5x+25(x - 1) + 7 = 5x - 5 + 7 = 5x + 2
To simplify the expression by removing the parentheses on the right side.
3
Equate the simplified left and right sides, then solve for xx.
8x−6=5x+2  ⟹  3x=8  ⟹  x=838x - 6 = 5x + 2 \implies 3x = 8 \implies x = \frac{8}{3}
To isolate the variable xx.
4
Substitute the value of xx into the expression 3x−23x - 2.
3(83)−2=8−2=63\left(\frac{8}{3}\right) - 2 = 8 - 2 = 6
To find the final value requested by the question.

Key Concept

Solving linear equations in one variable by distributing coefficients, combining like terms, and isolating the variable.
Estimated Time:2m 0s
Question 5Question →
In the equation below, xx is a real number.
56(3x−4)−38(4x−12)=14(2x+6)\frac{5}{6}(3x - 4) - \frac{3}{8}(4x - 12) = \frac{1}{4}(2x + 6)
What is the value of 6x−56x - 5?
Show answer & explanation

Answer: −1-1

Answer

−1-1
The correct answer is −1-1. First, distribute the fractions across the parentheses on both sides of the equation:
56(3x−4)−38(4x−12)=14(2x+6)\frac{5}{6}(3x - 4) - \frac{3}{8}(4x - 12) = \frac{1}{4}(2x + 6)
52x−103−32x+92=12x+32\frac{5}{2}x - \frac{10}{3} - \frac{3}{2}x + \frac{9}{2} = \frac{1}{2}x + \frac{3}{2}
Combine the like terms on the left side:
(52x−32x)+(−103+92)=12x+32\left(\frac{5}{2}x - \frac{3}{2}x\right) + \left(-\frac{10}{3} + \frac{9}{2}\right) = \frac{1}{2}x + \frac{3}{2}
x+76=12x+32x + \frac{7}{6} = \frac{1}{2}x + \frac{3}{2}
Subtract 12x\frac{1}{2}x from both sides:
12x+76=32\frac{1}{2}x + \frac{7}{6} = \frac{3}{2}
Subtract 76\frac{7}{6} from both sides:
12x=96−76=26=13\frac{1}{2}x = \frac{9}{6} - \frac{7}{6} = \frac{2}{6} = \frac{1}{3}
Multiply by 22 to solve for xx:
x=23x = \frac{2}{3}
Finally, substitute x=23x = \frac{2}{3} into the expression 6x−56x - 5:
6(23)−5=4−5=−16\left(\frac{2}{3}\right) - 5 = 4 - 5 = -1

Step-by-Step Solution

1
Distribute the coefficients to the terms within the parentheses on both sides of the equation.
52x−103−32x+92=12x+32\frac{5}{2}x - \frac{10}{3} - \frac{3}{2}x + \frac{9}{2} = \frac{1}{2}x + \frac{3}{2}
To simplify the linear equation, parenthetical expressions must be expanded.
2
Combine like terms on the left side of the equation.
x+76=12x+32x + \frac{7}{6} = \frac{1}{2}x + \frac{3}{2}
Grouping the variable terms (xx) and constant terms simplifies the equation prior to isolation.
3
Isolate the variable term on one side of the equation by subtracting 12x\frac{1}{2}x and 76\frac{7}{6} from both sides.
12x=13\frac{1}{2}x = \frac{1}{3}
This isolates the variable xx on the left side and the constants on the right side.
4
Solve for xx by multiplying both sides by 22.
x=23x = \frac{2}{3}
Multiplying by the reciprocal of the coefficient of xx gives the value of xx.
5
Substitute the value of xx into the expression 6x−56x - 5.
−1-1
The question asks for the value of the expression 6x−56x - 5, not the variable xx.

Key Concept

Solving multi-step linear equations in one variable containing fractions and parentheses.
Question 6Question →

If 5(x−2)=3x+85(x - 2) = 3x + 8, what is the value of xx?

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Answer: 9

Answer

The value of xx is 9.
Distributing the 5 to both terms inside the parentheses yields 5x−10=3x+85x - 10 = 3x + 8. Subtracting 3x3x from both sides gives 2x−10=82x - 10 = 8. Adding 10 to both sides yields 2x=182x = 18. Dividing both sides by 2 gives the correct value of xx, which is 9.

Step-by-Step Solution

1
Distribute the 5 to the terms inside the parentheses on the left side of the equation
5x−10=3x+85x - 10 = 3x + 8
To simplify the left side of the equation by removing the parentheses.
2
Subtract 3x3x from both sides of the equation to collect the variable terms on one side
2x−10=82x - 10 = 8
To isolate the variable term on the left side of the equation.
3
Add 10 to both sides of the equation to collect the constant terms on the other side
2x=182x = 18
To further isolate the variable term.
4
Divide both sides of the equation by 2
x=9x = 9
To solve for the variable xx.

Key Concept

Solving linear equations in one variable using distributive property and isolation of the variable.
Question 7Question →

If xx is the solution to the equation 14(23x−8)−56(12−35x)=13(x−9)−1\frac{1}{4}\left(\frac{2}{3}x - 8\right) - \frac{5}{6}\left(12 - \frac{3}{5}x\right) = \frac{1}{3}(x - 9) - 1, what is the value of 12x−5\frac{1}{2}x - 5?

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Answer: 7

Answer

7
The correct answer is obtained by first distributing the fractions across the terms in the parentheses, which yields 16x−2−10+12x=13x−4\frac{1}{6}x - 2 - 10 + \frac{1}{2}x = \frac{1}{3}x - 4. Combining the like terms on the left side gives 23x−12=13x−4\frac{2}{3}x - 12 = \frac{1}{3}x - 4. Subtracting 13x\frac{1}{3}x from both sides and adding 1212 to both sides yields 13x=8\frac{1}{3}x = 8, which means x=24x = 24. Substituting 2424 for xx in the expression 12x−5\frac{1}{2}x - 5 gives 12(24)−5=12−5=7\frac{1}{2}(24) - 5 = 12 - 5 = 7.

Step-by-Step Solution

1
Distribute the coefficients outside the parentheses on both sides of the equation.
16x−2−10+12x=13x−3−1\frac{1}{6}x - 2 - 10 + \frac{1}{2}x = \frac{1}{3}x - 3 - 1
To eliminate parentheses and simplify the terms.
2
Combine the like terms on the left and right sides of the equation.
23x−12=13x−4\frac{2}{3}x - 12 = \frac{1}{3}x - 4
To group the coefficients of xx and the constant values.
3
Isolate the variable xx by subtracting 13x\frac{1}{3}x and adding 1212 to both sides of the equation.
13x=8  ⟹  x=24\frac{1}{3}x = 8 \implies x = 24
To solve for xx.
4
Substitute the value of xx into the expression 12x−5\frac{1}{2}x - 5 to find the final value.
12(24)−5=12−5=7\frac{1}{2}(24) - 5 = 12 - 5 = 7
To evaluate the requested expression.

Key Concept

Solving multi-step linear equations in one variable involving distribution, fractions, and grouping like terms.

Alternative Method

Instead of distributing fractions first, we can multiply the entire equation by the least common multiple of all denominators, which is 1212. This eliminates all fractions immediately: 3(23x−8)−10(12−35x)=4(x−9)−123\left(\frac{2}{3}x - 8\right) - 10\left(12 - \frac{3}{5}x\right) = 4(x - 9) - 12. Expanding this gives 2x−24−120+6x=4x−36−12  ⟹  8x−144=4x−48  ⟹  4x=96  ⟹  x=242x - 24 - 120 + 6x = 4x - 36 - 12 \implies 8x - 144 = 4x - 48 \implies 4x = 96 \implies x = 24. Substituting x=24x = 24 into the expression gives 77.
Estimated Time:3m 0s
Question 8Question →

If 2(3x−4)−5(x−1)=3(x+2)−12(3x - 4) - 5(x - 1) = 3(x + 2) - 1, what is the value of 2x+32x + 3?

Show answer & explanation

Answer: -5

Answer

The correct answer is −5-5.
Solving the given equation step-by-step yields x=−4x = -4. Substituting x=−4x = -4 into the expression 2x+32x + 3 results in 2(−4)+3=−52(-4) + 3 = -5.

Step-by-Step Solution

1
Distribute the values outside the parentheses on both sides of the equation.
6x−8−5x+5=3x+6−16x - 8 - 5x + 5 = 3x + 6 - 1
This step removes parentheses so that like terms can be combined.
2
Combine like terms on each side of the equation.
x−3=3x+5x - 3 = 3x + 5
Simplifying both sides makes it easier to isolate the variable.
3
Isolate the variable xx by subtracting xx and 55 from both sides.
x=−4x = -4
This determines the value of xx which is required to evaluate the final expression.
4
Substitute the value of xx into the expression 2x+32x + 3.
2(−4)+3=−52(-4) + 3 = -5
This calculates the final value requested by the question.

Key Concept

Solving linear equations in one variable involving parentheses and distributing terms.
Question 9Question →

If 3(k+4)=5k−83(k + 4) = 5k - 8, what is the value of kk?

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Answer: 10

Answer

10
Distributing the 33 on the left side yields 3k+12=5k−83k + 12 = 5k - 8. Subtracting 3k3k from both sides yields 12=2k−812 = 2k - 8. Adding 88 to both sides yields 20=2k20 = 2k. Dividing by 22 gives the solution k=10k = 10.

Step-by-Step Solution

1
Distribute 33 to the terms inside the parentheses
3k+12=5k−83k + 12 = 5k - 8
To simplify the left side of the equation
2
Subtract 3k3k from both sides of the equation
12=2k−812 = 2k - 8
To group the variable terms on one side
3
Add 88 to both sides of the equation
20=2k20 = 2k
To isolate the variable term
4
Divide both sides by 22
k=10k = 10
To solve for kk

Key Concept

Solving linear equations with variables on both sides using distributive property
Question 10Question →
In the equation below, kk is a constant.
5x−2(3−k)x=9x−15x - 2(3 - k)x = 9x - 1
If the equation has no solution, what is the value of kk?
Show answer & explanation

Answer: 5

Answer

5
To find the value of kk for which the equation has no solution, we simplify the left side of the equation: 5x−2(3−k)x=5x−6x+2kx=−x+2kx=(2k−1)x5x - 2(3 - k)x = 5x - 6x + 2kx = -x + 2kx = (2k - 1)x. This yields the simplified equation (2k−1)x=9x−1(2k - 1)x = 9x - 1. A linear equation in one variable has no solution when the variable terms on both sides cancel each other out (meaning the coefficients of xx are equal) but the constant terms are different. Setting the coefficients of xx equal to each other gives 2k−1=92k - 1 = 9. Adding 11 to both sides gives 2k=102k = 10, and dividing by 22 yields k=5k = 5. Since the constant term on the left side is 00 and the constant term on the right side is −1-1, the constants are different, confirming that the equation has no solution when kk equals 55.

Step-by-Step Solution

1
Distribute the term −2-2 to the expressions inside the parentheses on the left side of the equation.
5x−6x+2kx=9x−15x - 6x + 2kx = 9x - 1
To eliminate the parentheses so that all like terms can be grouped.
2
Combine the xx terms on the left side of the equation.
−x+2kx=9x−1-x + 2kx = 9x - 1, which can be factored as (2k−1)x=9x−1(2k - 1)x = 9x - 1
Grouping the coefficients of the variable xx allows direct comparison of both sides of the linear equation.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
2k−1=92k - 1 = 9
For a linear equation in one variable to have no solution, the variable terms on both sides must cancel out (meaning their coefficients must be equal), while the constant terms must remain unequal.
4
Solve the resulting equation for kk.
2k=10  ⟹  k=52k = 10 \implies k = 5
Isolating kk by adding 11 to both sides and then dividing by 22 determines the specific constant value.
5
Verify that the constant terms are different when k=5k = 5.
Substitute k=5k = 5 back into the original equation to get 9x=9x−19x = 9x - 1, which simplifies to 0=−10 = -1.
Since 0=−10 = -1 is a false statement, the equation has no solution, confirming that k=5k = 5 is correct.

Key Concept

Linear Equations in One Variable

Alternative Method

Instead of algebraic simplification, the value of kk can be found by substituting the answer choices into the equation to see which value eliminates the variable xx while leaving an untrue statement. Plugging in 55 for kk gives 5x−2(3−5)x=9x−1  ⟹  5x−2(−2)x=9x−1  ⟹  9x=9x−1  ⟹  0=−15x - 2(3 - 5)x = 9x - 1 \implies 5x - 2(-2)x = 9x - 1 \implies 9x = 9x - 1 \implies 0 = -1. Because this statement is false, the equation has no solution, verifying that 55 is the correct answer.
Estimated Time:2m 0s
Question 11Question →

In the equation 34(8x−12)+kx=10x−9\frac{3}{4}(8x - 12) + kx = 10x - 9, kk is a constant. If the equation has infinitely many solutions, what is the value of kk?

Show answer & explanation

Answer: 4

Answer

The value of kk is 44.
Distributing the fraction on the left side of the equation yields 34(8x)−34(12)=6x−9\frac{3}{4}(8x) - \frac{3}{4}(12) = 6x - 9. Substituting this back gives 6x−9+kx=10x−96x - 9 + kx = 10x - 9. Factoring out xx on the left side gives (6+k)x−9=10x−9(6 + k)x - 9 = 10x - 9. For a linear equation to have infinitely many solutions, the coefficients of xx on both sides must be identical, and the constants must be identical. Since the constants on both sides are already −9-9, we set the coefficients equal: 6+k=106 + k = 10. Subtracting 6 from both sides gives the correct value k=4k = 4.

Step-by-Step Solution

1
Distribute the fraction 34\frac{3}{4} to the terms inside the parentheses.
6x−9+kx=10x−96x - 9 + kx = 10x - 9
To simplify the expression and eliminate the parentheses.
2
Factor out xx from the terms on the left side of the equation.
(6+k)x−9=10x−9(6 + k)x - 9 = 10x - 9
To group the xx terms together to easily compare coefficients.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
6+k=106 + k = 10
For the equation to have infinitely many solutions, the coefficients of the variable on both sides must be equal when the constant terms are equal.
4
Solve for kk by subtracting 6 from both sides of the equation.
k=4k = 4
To isolate the constant kk.

Key Concept

A linear equation in one variable has infinitely many solutions when it can be simplified to an identity of the form Ax+B=Ax+BAx + B = Ax + B, meaning both the coefficients of xx and the constant terms on both sides of the equation are equal.
Question 12Question →

In the equation below, aa and bb are constants.

13(2a−5x)−34(x−b)=−2912x+5\frac{1}{3}(2a - 5x) - \frac{3}{4}(x - b) = -\frac{29}{12}x + 5

If the equation has infinitely many solutions for xx, what is the value of 8a+9b8a + 9b?

Show answer & explanation

Answer: 60

Answer

The value of 8a+9b8a + 9b is 6060.
To find the value of 8a+9b8a + 9b that makes the equation have infinitely many solutions, we first expand and simplify the left side of the equation: 23a−53x−34x+34b=−2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5. Combining the xx terms gives -\frac{29}{12}x + \(\frac{2}{3}a + \frac{3}{4}b\) = -\frac{29}{12}x + 5. Since the coefficients of xx on both sides are equal (−2912-\frac{29}{12}), the equation will have infinitely many solutions if the constant terms on both sides are also equal. This requires 23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5. Multiplying this entire equation by the least common multiple of the denominators, which is 12, yields 8a+9b=608a + 9b = 60.

Step-by-Step Solution

1
Distribute the constants through the parentheses on the left side of the equation.
23a−53x−34x+34b=−2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5
This separates the variable terms from the constant terms so the equation can be simplified.
2
Combine the coefficients of the xx terms on the left side using 12 as the common denominator.
−2912x+23a+34b=−2912x+5-\frac{29}{12}x + \frac{2}{3}a + \frac{3}{4}b = -\frac{29}{12}x + 5
Simplifying the variable terms allows us to compare the coefficients on both sides of the equation.
3
Equate the constant terms from the left and right sides of the equation.
23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5
A linear equation has infinitely many solutions when the coefficients of the variable on both sides are equal and the constant terms on both sides are also equal.
4
Multiply both sides of the equation by 12 to eliminate the fractional denominators.
8a+9b=608a + 9b = 60
Multiplying the equation by the common denominator directly evaluates the target expression 8a+9b8a + 9b.

Key Concept

For a linear equation in one variable to have infinitely many solutions, it must be reducible to an identity of the form cx+d=cx+dcx + d = cx + d, where both the variable coefficients and the constant terms on both sides of the equation are equal.
Question 13Question →

If 20−2(x+3)=820 - 2(x + 3) = 8, what is the value of xx?

Show answer & explanation

Answer: 3

Answer

3
The correct answer is 3. Distributing −2-2 to the terms inside the parentheses gives 20−2x−6=820 - 2x - 6 = 8. Simplifying the left side yields 14−2x=814 - 2x = 8. Subtracting 14 from both sides results in −2x=−6-2x = -6. Dividing both sides by −2-2 gives x=3x = 3.

Step-by-Step Solution

1
Distribute the coefficient −2-2 to all terms inside the parentheses (x+3)(x + 3).
20−2x−6=820 - 2x - 6 = 8
Applying the distributive property requires multiplying both xx and 33 by −2-2.
2
Combine the constant terms 2020 and −6-6 on the left side of the equation.
14−2x=814 - 2x = 8
Simplifying the expression makes it easier to isolate the variable term.
3
Subtract 1414 from both sides of the equation to isolate the variable term −2x-2x.
−2x=−6-2x = -6
This moves the constant term to the opposite side of the equation.
4
Divide both sides of the equation by −2-2 to solve for xx.
x=3x = 3
Dividing by the coefficient of the variable isolates the variable completely.

Key Concept

Solving linear equations in one variable with parentheses by distributing coefficients, combining like terms, and isolating the variable.
Question 14Question →

In the equation 12(kx−4)−23(x−k)=56x−1\frac{1}{2}(kx - 4) - \frac{2}{3}(x - k) = \frac{5}{6}x - 1, where kk is a constant, the equation has no solution for xx. What is the value of kk?

Show answer & explanation

Answer: 3

Answer

The value of the constant kk for which the equation has no solution is 33.
For a linear equation of the form Ax+B=Cx+DAx + B = Cx + D to have no solution, the coefficients of the variable terms must be equal (A=CA = C) but the constant terms must be unequal (B≠DB \neq D). Distributing the terms on the left side of the given equation and grouping them yields (12k−23)x+(23k−2)=56x−1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1. Setting the coefficients of xx equal to each other gives 12k−23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}. Adding 23\frac{2}{3} to both sides yields 12k=96\frac{1}{2}k = \frac{9}{6}, or 12k=32\frac{1}{2}k = \frac{3}{2}, which simplifies to k=3k = 3. Evaluating the constant terms when k=3k = 3 gives 23(3)−2=0\frac{2}{3}(3) - 2 = 0 on the left side and −1-1 on the right side. Since 0≠−10 \neq -1, the equation has no solution when k=3k = 3.

Step-by-Step Solution

1
Distribute the fractional coefficients on the left side of the equation.
12kx−2−23x+23k=56x−1\frac{1}{2}kx - 2 - \frac{2}{3}x + \frac{2}{3}k = \frac{5}{6}x - 1
To separate the variable terms from the constants for further algebraic manipulation.
2
Group the terms on the left side into a coefficient for xx and a single constant term.
(12k−23)x+(23k−2)=56x−1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1
To represent the equation in the standard linear format Ax+B=Cx+DAx + B = Cx + D.
3
Set the coefficients of xx from both sides equal to each other.
12k−23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}
A linear equation has no solution when the variable terms on both sides cancel out, requiring their coefficients to be identical.
4
Solve the linear equation to isolate the constant kk.
k=3k = 3
Add 23\frac{2}{3} to both sides to get 12k=96\frac{1}{2}k = \frac{9}{6}, which simplifies to 12k=32\frac{1}{2}k = \frac{3}{2}. Multiplying both sides by 22 yields k=3k = 3.
5
Verify that the constant terms are not equal when substituting k=3k = 3.
The left-side constant is 00 and the right-side constant is −1-1. Since 0≠−10 \neq -1, the condition is satisfied.
To ensure the equation does not simplify to an identity with infinitely many solutions (which occurs when both the variable coefficients and the constant terms are equal).

Key Concept

Determining the conditions under which a linear equation in one variable has no solution.

Alternative Method

To avoid working with fractions, multiply every term in the equation by 66 (the least common multiple of 2,3,62, 3, 6) at the start: 3(kx−4)−4(x−k)=5x−63(kx - 4) - 4(x - k) = 5x - 6. Expand the parentheses to get 3kx−12−4x+4k=5x−63kx - 12 - 4x + 4k = 5x - 6, and group the terms: (3k−4)x+(4k−12)=5x−6(3k - 4)x + (4k - 12) = 5x - 6. For there to be no solution, set the coefficients of xx equal to each other: 3k−4=5  ⟹  3k=9  ⟹  k=33k - 4 = 5 \implies 3k = 9 \implies k = 3. Check the constant terms with k=3k = 3: 4(3)−12=04(3) - 12 = 0, which is unequal to −6-6. This confirms k=3k = 3 is the correct answer.
Estimated Time:2m 0s
Question 15Question →

If 23(x−6)−12(x−4)=4\frac{2}{3}(x - 6) - \frac{1}{2}(x - 4) = 4, what is the value of 3x−123x - 12?

Show answer & explanation

Answer: 96

Answer

96
The correct answer is 96. Multiplying the entire equation by the common denominator 6 clears the fractions to yield 4(x−6)−3(x−4)=244(x - 6) - 3(x - 4) = 24. Distributing the factors gives 4x−24−3x+12=244x - 24 - 3x + 12 = 24, which simplifies to x−12=24x - 12 = 24. Adding 12 to both sides determines that x=36x = 36. Evaluating the expression 3x−123x - 12 with x=36x = 36 yields 3(36)−12=963(36) - 12 = 96.

Step-by-Step Solution

1
Multiply the entire equation by 6 (the least common multiple of the denominators 3 and 2) to eliminate the fractions.
4(x−6)−3(x−4)=244(x - 6) - 3(x - 4) = 24
Clearing the denominators simplifies the algebraic manipulation and reduces the risk of fractional arithmetic errors.
2
Distribute the coefficients to the terms inside the parentheses, taking care to distribute the negative sign for the second term.
4x−24−3x+12=244x - 24 - 3x + 12 = 24
Applying the distributive property expands the expression so that like terms can be grouped.
3
Combine like terms on the left side of the equation.
x−12=24x - 12 = 24
Grouping 4x−3x4x - 3x to get xx and −24+12-24 + 12 to get −12-12 simplifies the equation to a single variable and constant.
4
Isolate the variable xx by adding 12 to both sides of the equation.
x=36x = 36
Adding the opposite of the constant term isolates the variable on one side.
5
Substitute x=36x = 36 into the requested expression 3x−123x - 12 to find its value.
3(36)−12=963(36) - 12 = 96
The question asks for the value of the expression 3x−123x - 12, not just the value of xx.

Key Concept

Solving multi-step linear equations in one variable and evaluating algebraic expressions.
Estimated Time:1m 30s
Question 16Question →

A rental company charges a flat fee of 3535 dollars plus 1212 dollars per hour to rent a bicycle. If a customer was charged a total of 9595 dollars, for how many hours did the customer rent the bicycle?

Show answer & explanation

Answer: 5

Answer

5
To find the number of hours the bicycle was rented, we can set up the equation 12h+35=9512h + 35 = 95, where hh is the number of hours. Subtracting 3535 from both sides gives 12h=6012h = 60. Dividing both sides by 1212 gives h=5h = 5. Thus, the customer rented the bicycle for 55 hours.

Step-by-Step Solution

1
Set up a linear equation representing the total cost.
12h+35=9512h + 35 = 95, where hh represents the number of hours rented.
The total cost consists of a flat fee of 3535 dollars plus an hourly charge of 1212 dollars multiplied by the number of hours hh.
2
Subtract the flat fee from both sides of the equation to isolate the variable term.
12h=6012h = 60
Subtracting 3535 from both sides simplifies the equation to find the total hourly charge portion of the cost.
3
Divide by the hourly rate to solve for the number of hours hh.
h=5h = 5
Dividing both sides by 1212 isolates hh to find the number of rental hours.

Key Concept

Linear Equations in One Variable
Question 17Question →

In the equation below, mm is a constant.

35(5x−10)−2(x−m)=18\frac{3}{5}(5x - 10) - 2(x - m) = 18

If the solution to the equation is x=12x = 12, what is the value of mm?

Show answer & explanation

Answer: 6

Answer

The correct value of mm is 6.
Substituting x=12x = 12 into the equation yields 35(5(12)−10)−2(12−m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18, which simplifies to 30−24+2m=1830 - 24 + 2m = 18. Combining constant terms gives 6+2m=186 + 2m = 18. Subtracting 6 from both sides gives 2m=122m = 12, and dividing by 2 yields m=6m = 6.

Step-by-Step Solution

1
Substitute x=12x = 12 into the given equation.
35(5(12)−10)−2(12−m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18
Since x=12x = 12 is the solution, it must satisfy the equation.
2
Simplify the expression inside the first set of parentheses.
5(12)−10=60−10=505(12) - 10 = 60 - 10 = 50
Evaluate the terms inside the parentheses first.
3
Multiply the simplified term by the fraction 35\frac{3}{5}.
35(50)=30\frac{3}{5}(50) = 30
Multiply 50 by 3 and divide by 5.
4
Substitute 30 back into the equation and distribute −2-2 to the terms inside the second set of parentheses.
30−24+2m=1830 - 24 + 2m = 18
Distributing −2-2 to 12−m12 - m yields −24+2m-24 + 2m due to sign rules.
5
Combine the constant terms on the left side of the equation.
6+2m=186 + 2m = 18
30−24=630 - 24 = 6.
6
Subtract 6 from both sides to isolate the term with mm.
2m=122m = 12
Isolate the variable term on one side of the equation.
7
Divide both sides by 2 to solve for mm.
m=6m = 6
Divide 12 by 2 to find the final value.

Key Concept

Solving a linear equation in one variable by substitution and isolation.
Question 18Question →
In the equation below, aa and bb are constants.
a(x−1)−2(3x+b)3=4(x−5)−a(b−x)6\frac{a(x - 1) - 2(3x + b)}{3} = \frac{4(x - 5) - a(b - x)}{6}
If the equation has infinitely many solutions for xx, what is the value of a−ba - b?
Show answer & explanation

Answer: 15

Answer

The value of a−ba - b is 15.
The correct answer is 15. Multiplying the entire equation by 6 to clear the denominators, distributing the terms, and grouping them yields (2a−12)x−(2a+4b)=(a+4)x−(ab+20)(2a - 12)x - (2a + 4b) = (a + 4)x - (ab + 20). For the equation to have infinitely many solutions, the coefficients of xx on both sides must be equal (2a−12=a+42a - 12 = a + 4), and the constant terms must also be equal (2a+4b=ab+202a + 4b = ab + 20). Solving this system of equations yields a=16a = 16 and b=1b = 1. The difference a−ba - b is 16−1=1516 - 1 = 15.

Step-by-Step Solution

1
Eliminate the denominators by multiplying both sides of the equation by 6.
2[a(x−1)−2(3x+b)]=4(x−5)−a(b−x)2[a(x - 1) - 2(3x + b)] = 4(x - 5) - a(b - x)
Multiplying both sides by the least common multiple of 3 and 6 simplifies the equation by removing the fractions.
2
Expand both sides of the equation using the distributive property.
2[ax−a−6x−2b]=4x−20−ab+ax2[ax - a - 6x - 2b] = 4x - 20 - ab + ax which simplifies to 2ax−2a−12x−4b=ax+4x−ab−202ax - 2a - 12x - 4b = ax + 4x - ab - 20
Distributing the constants outside the parentheses allows us to group like terms.
3
Group the xx-terms and constant terms on each side of the equation.
(2a−12)x−(2a+4b)=(a+4)x−(ab+20)(2a - 12)x - (2a + 4b) = (a + 4)x - (ab + 20)
Structuring the equation in the form Ax+B=Cx+DAx + B = Cx + D makes it easier to compare the coefficients.
4
Set up a system of equations by equating the coefficients of xx and the constant terms from both sides.
2a−12=a+42a - 12 = a + 4 and 2a+4b=ab+202a + 4b = ab + 20
For a linear equation to have infinitely many solutions, the coefficient of xx on both sides must be equal, and the constant terms on both sides must also be equal.
5
Solve the first equation for aa.
a=16a = 16
Subtracting aa and adding 12 to both sides isolates the variable aa.
6
Substitute a=16a = 16 into the second equation and solve for bb.
2(16)+4b=16b+20  ⟹  32+4b=16b+20  ⟹  12=12b  ⟹  b=12(16) + 4b = 16b + 20 \implies 32 + 4b = 16b + 20 \implies 12 = 12b \implies b = 1
Substituting the known value of aa leaves a single linear equation in terms of bb.
7
Calculate the value of a−ba - b.
16−1=1516 - 1 = 15
This is the final value requested by the question.

Key Concept

Solving linear equations in one variable with infinitely many solutions by equating coefficients on both sides of the equation.
Question 19Question →

If 5(x−3)−2x=95(x - 3) - 2x = 9, what is the value of xx?

Show answer & explanation

Answer: 8

Answer

The value of xx is 88.
To solve the equation, distribute the 5 to the terms inside the parentheses to get 5x−15−2x=95x - 15 - 2x = 9. Combine the like terms 5x5x and −2x-2x to simplify the equation to 3x−15=93x - 15 = 9. Add 15 to both sides to isolate the variable term, which yields 3x=243x = 24. Finally, divide both sides by 3 to find x=8x = 8.

Step-by-Step Solution

1
Distribute the 55 to the terms inside the parentheses.
5x−15−2x=95x - 15 - 2x = 9
To remove the parentheses and begin simplifying the equation.
2
Combine the like variable terms 5x5x and −2x-2x.
3x−15=93x - 15 = 9
To group the x terms together.
3
Add 1515 to both sides of the equation.
3x=243x = 24
To isolate the term with the variable on one side.
4
Divide both sides of the equation by 33.
x=8x = 8
To solve for xx.

Key Concept

Solving linear equations in one variable using the distributive property, combining like terms, and applying inverse operations.
Estimated Time:45s
Question 20Question →

If 25(3x−4)−13(2x+5)=15x+1115\frac{2}{5}(3x - 4) - \frac{1}{3}(2x + 5) = \frac{1}{5}x + \frac{11}{15}, what is the value of 2x−72x - 7?

Show answer & explanation

Answer: 17

Answer

The correct answer is 17.
To find the value of 2x−72x - 7, first solve the linear equation for xx. Distributing the coefficients on the left side of the equation gives 65x−85−23x−53=15x+1115\frac{6}{5}x - \frac{8}{5} - \frac{2}{3}x - \frac{5}{3} = \frac{1}{5}x + \frac{11}{15}. Combining the variable terms and constants on the left side results in 815x−4915=315x+1115\frac{8}{15}x - \frac{49}{15} = \frac{3}{15}x + \frac{11}{15}. Subtracting 315x\frac{3}{15}x and adding 4915\frac{49}{15} to both sides yields 515x=6015\frac{5}{15}x = \frac{60}{15}, which simplifies to 13x=4\frac{1}{3}x = 4, or x=12x = 12. Finally, substituting 12 into the expression 2x−72x - 7 gives 2(12)−7=172(12) - 7 = 17.

Step-by-Step Solution

1
Distribute the coefficients to the terms inside the parentheses.
65x−85−23x−53=15x+1115\frac{6}{5}x - \frac{8}{5} - \frac{2}{3}x - \frac{5}{3} = \frac{1}{5}x + \frac{11}{15}
To eliminate parentheses and allow grouping of like terms.
2
Combine the variable terms and the constant terms on the left side using a common denominator of 15.
815x−4915=315x+1115\frac{8}{15}x - \frac{49}{15} = \frac{3}{15}x + \frac{11}{15}
To simplify the linear equation into a standard two-sided form.
3
Subtract the variable term from the right side and add the constant term from the left side.
515x=6015\frac{5}{15}x = \frac{60}{15}, which simplifies to x=12x = 12
To isolate the variable xx on one side of the equation.
4
Evaluate the expression 2x−72x - 7 using the value of xx.
2(12)−7=172(12) - 7 = 17
To solve for the final requested quantity.

Key Concept

Linear Equations in One Variable
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