TACHS Quiz: One Variable Linear Equations

13 questions · exam conditions

0:00

One Variable Linear EquationsQuestion 1 of 13

Solve for $z$ : $7 - 2(3z - 4) = 5z + 1$

\dfrac{14}{11}

\dfrac{11}{14}

-\dfrac{14}{11}

-\dfrac{11}{14}

← Back to quizzes

TACHS Quiz

TACHS Quiz: One Variable Linear Equations

Practice One Variable Linear Equations in TACHS with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One Variable Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for TACHS.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve for $z$ : $7 - 2(3z - 4) = 5z + 1$

$\dfrac{14}{11}$ (correct answer)
$\dfrac{11}{14}$
$-\dfrac{14}{11}$
$-\dfrac{11}{14}$

Explanation: This equation tests your ability to solve linear equations with parentheses and variables on both sides. When you see parentheses with a negative sign in front, that's your cue to carefully distribute the negative through all terms inside. Start by distributing the

-2

through the parentheses:

7 - 2(3z - 4) = 7 - 6z + 8 = 15 - 6z

. The equation becomes

15 - 6z = 5z + 1

. Next, collect all variables on one side by adding

6z

to both sides:

15 = 11z + 1

. Subtract 1 from both sides:

14 = 11z

. Therefore,

z = \frac{14}{11}

. Looking at the wrong answers: Choice B gives

\frac{11}{14}

, which happens when students flip the fraction by mistake—they might solve correctly to get

14 = 11z

but then write

z = \frac{11}{14}

. Choice C gives

-\frac{14}{11}

, which results from sign errors during distribution or when moving terms between sides. Choice D gives

-\frac{11}{14}

, combining both the fraction flip error and a sign mistake. You can verify the answer by substituting

z = \frac{14}{11}

back into the original equation—both sides should equal the same value. Strategy tip: Always double-check your distribution step, especially with negative signs. When you get your final answer, substitute it back into the original equation as a quick verification—this catches most algebraic errors before you submit your answer.

Question 2

Solve for $k$ : $12 - (4k + 8) = 6k - 4$

$-\dfrac{5}{4}$
$\dfrac{5}{4}$
$-\dfrac{4}{5}$
$\dfrac{4}{5}$ (correct answer)

Explanation: When you encounter an equation with parentheses and variables on both sides, your goal is to isolate the variable through systematic algebraic steps. Start by distributing the negative sign through the parentheses:

12 - (4k + 8) = 12 - 4k - 8 = 4 - 4k

. Now your equation becomes

4 - 4k = 6k - 4

. Next, collect all terms with

k

on one side and constants on the other. Add

4k

to both sides:

4 = 10k - 4

. Then add

4

to both sides:

8 = 10k

. Finally, divide by

10

k = \frac{8}{10} = \frac{4}{5}

. Let's verify: substituting

k = \frac{4}{5}

back into the original equation gives us

12 - (4 \cdot \frac{4}{5} + 8) = 12 - (\frac{16}{5} + 8) = 12 - \frac{56}{5} = \frac{4}{5}

, and

6 \cdot \frac{4}{5} - 4 = \frac{24}{5} - 4 = \frac{4}{5}

. Both sides equal

\frac{4}{5}

, confirming our answer. Choice A gives

-\frac{5}{4}

, which would result from incorrectly combining terms or sign errors. Choice B gives

\frac{5}{4}

, which might come from flipping the correct fraction. Choice C gives

-\frac{4}{5}

, likely from a sign error when distributing or moving terms. Remember: when distributing a negative sign, every term inside the parentheses changes sign. Always check your work by substituting back into the original equation—this catches most algebraic mistakes.

Question 3

Solve for $t$ : $5(t - 0.6) + 1.8 = 3(2t - 0.3)$

$-3$
$0.3$
$3$
$-0.3$ (correct answer)

Explanation: When solving linear equations with decimals and parentheses, work systematically through distribution and combining like terms to isolate the variable. Start by distributing on both sides:

5(t - 0.6) + 1.8 = 3(2t - 0.3)

becomes

5t - 3.0 + 1.8 = 6t - 0.9

. Simplify the left side:

5t - 1.2 = 6t - 0.9

. Now collect all terms with

t

on one side and constants on the other. Subtract

5t

from both sides:

-1.2 = t - 0.9

. Add

0.9

to both sides:

-1.2 + 0.9 = t

, so

t = -0.3

. Let's examine why the other answers are incorrect. Choice A (

-3

) likely results from calculation errors when handling the decimals, perhaps multiplying incorrectly during distribution. Choice B (

0.3

) is the positive version of the correct answer—this happens when students make sign errors, especially when moving terms across the equals sign. Choice C (

3

) is the whole number version of choice B and represents similar sign confusion combined with decimal mistakes. The correct answer is D (

-0.3

). You can verify by substituting back into the original equation:

5(-0.3 - 0.6) + 1.8 = 5(-0.9) + 1.8 = -4.5 + 1.8 = -2.7

, and

3(2(-0.3) - 0.3) = 3(-0.6 - 0.3) = 3(-0.9) = -2.7

. Both sides equal

-2.7

. When solving equations with decimals, double-check your arithmetic at each step and always verify your answer by substitution—this catches sign errors and computational mistakes.

Question 4

Solve for $y$ : $\dfrac{5(2y - 3)}{4} = \dfrac{3(y + 1)}{2}$

$\dfrac{21}{8}$
$\dfrac{4}{21}$
$-\dfrac{21}{4}$
$\dfrac{21}{4}$ (correct answer)

Explanation: When you encounter an equation with fractions on both sides, your goal is to eliminate the fractions first, then solve systematically. This makes the algebra much cleaner and reduces calculation errors. To solve

\dfrac{5(2y - 3)}{4} = \dfrac{3(y + 1)}{2}

, start by finding the least common denominator of 4 and 2, which is 4. Multiply both sides by 4:

4 \cdot \dfrac{5(2y - 3)}{4} = 4 \cdot \dfrac{3(y + 1)}{2}

This simplifies to:

5(2y - 3) = 6(y + 1)

Now distribute on both sides:

10y - 15 = 6y + 6

Subtract

6y

from both sides:

4y - 15 = 6

Add 15 to both sides:

4y = 21

Therefore:

y = \dfrac{21}{4}

Looking at the wrong answers: Choice A gives

\dfrac{21}{8}

, which likely results from incorrectly multiplying by 2 instead of 4 when clearing fractions. Choice B gives

\dfrac{4}{21}

, which is the reciprocal of the correct answer—a common error when students flip the final division. Choice C gives

-\dfrac{21}{4}

, which occurs if you make a sign error during the distribution or combining like terms. The correct answer is D. Strategy tip: When solving fractional equations, always clear the fractions first by multiplying both sides by the LCD. This prevents messy fraction arithmetic and makes it easier to spot calculation errors. Double-check your work by substituting your answer back into the original equation.

Question 5

Solve for $x$ : $5x - 7 = 18$

$3$
$5$ (correct answer)
$25$
$-5$

Explanation: When you encounter a linear equation like this one, you're solving for the value of the variable that makes the equation true. The key is to isolate

x

by performing the same operations on both sides of the equation. Starting with

5x - 7 = 18

, you need to "undo" what's being done to

x

. First, eliminate the

-7

by adding 7 to both sides:

5x - 7 + 7 = 18 + 7

, which simplifies to

5x = 25

. Then divide both sides by 5 to isolate

x

x = \frac{25}{5} = 5

. You can verify this by substituting back:

5(5) - 7 = 25 - 7 = 18

✓ Looking at the wrong answers: Choice A (

3

) comes from a calculation error - if you substitute

x = 3

, you get

5(3) - 7 = 15 - 7 = 8

, not 18. Choice C (

25

) represents stopping too early in the solution process; this is the value of

5x

, but you must divide by 5 to find

x

. Choice D (

-5

) might result from sign errors during the solving process - perhaps incorrectly handling the subtraction when isolating the variable. Remember the golden rule for linear equations: whatever you do to one side, you must do to the other side. Always work systematically - first handle addition/subtraction, then multiplication/division. Double-check your answer by substituting it back into the original equation to ensure both sides are equal.

Question 6

Solve for $p$ : $\dfrac{3}{2}(p + 4) = 6$

$0$ (correct answer)
$-4$
$1$
$4$

Explanation: This is a linear equation that requires you to isolate the variable

p

using inverse operations. When you see an equation with a fraction multiplied by an expression in parentheses, your goal is to systematically undo each operation. Start by eliminating the fraction coefficient

\frac{3}{2}

. Multiply both sides by the reciprocal

\frac{2}{3}

\frac{2}{3} \cdot \frac{3}{2}(p + 4) = 6 \cdot \frac{2}{3}

The left side simplifies to

(p + 4)

since

\frac{2}{3} \cdot \frac{3}{2} = 1

. The right side becomes

6 \cdot \frac{2}{3} = \frac{12}{3} = 4

(p + 4) = 4

Now subtract 4 from both sides to isolate

p

p = 4 - 4 = 0

Let's examine why the other answers are incorrect. Choice B (

-4

) likely comes from incorrectly subtracting 4 from the left side only, giving

p = -4

instead of properly isolating

p

. Choice C (

1

) might result from calculation errors when working with the fraction, perhaps incorrectly computing

6 \cdot \frac{2}{3}

as 1. Choice D (

4

) is what you get if you forget the final step of subtracting 4, leaving

p + 4 = 4

as your final answer instead of solving for

p

. Always verify your solution by substituting back into the original equation. With

p = 0

\frac{3}{2}(0 + 4) = \frac{3}{2} \cdot 4 = 6

✓ Remember: work systematically through inverse operations, and always check your answer by substitution.

Question 7

Solve for $x$ : $3(x - 2) = 2x + 5$

$11$ (correct answer)
$7$
$-7$
$-11$

Explanation: When you encounter a linear equation with parentheses and variables on both sides, your goal is to isolate the variable by systematically undoing operations using inverse operations. Start by distributing the 3 to everything inside the parentheses:

3(x - 2) = 3x - 6

. So the equation becomes

3x - 6 = 2x + 5

. Next, collect all terms with

x

on one side and all constants on the other. Subtract

2x

from both sides:

3x - 2x - 6 = 2x - 2x + 5

, which simplifies to

x - 6 = 5

. Finally, add 6 to both sides:

x - 6 + 6 = 5 + 6

, giving you

x = 11

. You can verify this by substituting back into the original equation:

3(11 - 2) = 3(9) = 27

and

2(11) + 5 = 22 + 5 = 27

✓ Choice A (

11

) is correct. Choice B (

7

) likely results from incorrectly adding 6 and 5 to get 11, then thinking that's your final answer without properly isolating

x

. Choice C (

-7

) probably comes from sign errors when moving terms across the equals sign. Choice D (

-11

) suggests you found the right absolute value but made a sign error, possibly when distributing the negative or moving terms. Always double-check linear equation solutions by substituting your answer back into the original equation. Both sides should yield the same value, confirming your solution is correct.

Question 8

Solve for $q$ : $4(2q - 1) - 3(q + 5) = 0$

$\dfrac{19}{7}$
$\dfrac{5}{19}$
$-\dfrac{19}{5}$
$\dfrac{19}{5}$ (correct answer)

Explanation: When you see an equation with parentheses and multiple terms, your goal is to simplify systematically by distributing, combining like terms, and isolating the variable. Start by distributing each coefficient to the terms inside the parentheses:

4(2q - 1) - 3(q + 5) = 0

8q - 4 - 3q - 15 = 0

Next, combine like terms by grouping the

q

terms and the constants:

8q - 3q - 4 - 15 = 0

5q - 19 = 0

Finally, solve for

q

by adding 19 to both sides, then dividing by 5:

5q = 19

q = \frac{19}{5}

This confirms answer D is correct. Looking at the wrong answers: Choice A (

\frac{19}{7}

) likely comes from incorrectly combining the coefficients of

q

8 - 1 = 7

instead of

8 - 3 = 5

. Choice B (

\frac{5}{19}

) flips the correct fraction, which happens when students confuse which number goes in the numerator versus denominator. Choice C (

-\frac{19}{5}

) results from a sign error, possibly subtracting 19 from both sides instead of adding it, or making a mistake when distributing the negative sign. For multi-step linear equations, always work methodically: distribute first, then combine like terms, and finally isolate the variable. Double-check your work by substituting your answer back into the original equation to verify it makes the equation true.

Question 9

Solve for $x$ : $\dfrac{2}{3}(6 - x) = 4 - \dfrac{1}{2}x$

$-6$
$1$
$6$
$0$ (correct answer)

Explanation: This problem tests your ability to solve linear equations with fractions and parentheses. When you encounter equations like this, your goal is to isolate the variable by systematically eliminating fractions and simplifying both sides. Start by distributing the

\frac{2}{3}

on the left side:

\frac{2}{3} \cdot 6 - \frac{2}{3} \cdot x = 4 - \frac{1}{2}x

, which gives you

4 - \frac{2}{3}x = 4 - \frac{1}{2}x

. Now subtract 4 from both sides:

-\frac{2}{3}x = -\frac{1}{2}x

. Add

\frac{1}{2}x

to both sides:

-\frac{2}{3}x + \frac{1}{2}x = 0

. To combine these fractions, find a common denominator of 6:

-\frac{4}{6}x + \frac{3}{6}x = 0

, which simplifies to

-\frac{1}{6}x = 0

. Therefore,

x = 0

. Choice A (

x = -6

) likely comes from sign errors during distribution or when combining fractions. Choice B (

x = 1

) might result from incorrectly handling the fraction arithmetic or making computational mistakes. Choice C (

x = 6

) could stem from mishandling the distributive property or confusing the coefficients during simplification. The correct answer is D. Strategy tip: When solving equations with fractions, work carefully through each step and double-check your fraction arithmetic. You can also verify your answer by substituting it back into the original equation—both sides should equal the same value.

Question 10

Solve for $x$ : $0.3x - 2 = 0.5x + 4$

$30$
$-30$ (correct answer)
$-6$
$6$

Explanation: When you encounter a linear equation with the variable on both sides, your goal is to isolate the variable by collecting all

x

terms on one side and all constants on the other. Starting with

0.3x - 2 = 0.5x + 4

, first move all

x

terms to one side by subtracting

0.5x

from both sides:

0.3x - 0.5x - 2 = 4

. This simplifies to

-0.2x - 2 = 4

. Next, isolate the

x

term by adding

2

to both sides:

-0.2x = 6

. Finally, divide both sides by

-0.2

x = \frac{6}{-0.2} = -30

. You can verify this by substituting back:

0.3(-30) - 2 = -9 - 2 = -11

and

0.5(-30) + 4 = -15 + 4 = -11

✓ Choice A (

30

) represents the most common error: forgetting the negative sign when dividing by

-0.2

. Choice C (

-6

) occurs if you incorrectly think

-0.2x = 6

means

x = -6

, perhaps by confusing the coefficient with the variable. Choice D (

6

) happens when you make both errors—dropping the negative from the division and mishandling the coefficient. Remember: when solving equations with decimals, pay extra attention to signs during division. It's also helpful to check your answer by substituting it back into the original equation—both sides should give you the same result.

Question 11

Solve for $m$ : $9 = 2(5 - m) + m$

$1$ (correct answer)
$-1$
$3$
$-3$

Explanation: When you encounter a linear equation with parentheses and variables on both sides, your goal is to isolate the variable through systematic algebraic steps. Start by distributing the 2 to both terms inside the parentheses:

9 = 2(5) - 2(m) + m

, which simplifies to

9 = 10 - 2m + m

. Next, combine like terms on the right side:

-2m + m = -m

, giving you

9 = 10 - m

. To isolate

m

, subtract 10 from both sides:

9 - 10 = -m

, so

-1 = -m

. Finally, multiply both sides by -1 to get

m = 1

. You can verify this by substituting back:

9 = 2(5 - 1) + 1 = 2(4) + 1 = 8 + 1 = 9

✓ Looking at the wrong answers: Choice B (

-1

) is what you get if you forget the final step of multiplying by -1 to eliminate the negative sign in front of

m

. Choice C (

3

) results from incorrectly combining like terms, perhaps calculating

-2m + m = -3m

instead of

-m

. Choice D (

-3

) compounds multiple errors, likely from both incorrect distribution and sign mistakes. When solving equations with parentheses, always work methodically: distribute first, combine like terms, then isolate the variable. Double-check your work by substituting your answer back into the original equation—this catches most algebraic errors and builds confidence in your solution.

Question 12

Four times a number decreased by 7 equals the number increased by 5. What is the number?

$12$
$-4$
$4$ (correct answer)
$-12$

Explanation: When you encounter a word problem asking you to find an unknown number, your goal is to translate the English into a mathematical equation, then solve for the variable. Let's call the unknown number

x

. Breaking down the sentence: "Four times a number decreased by 7" translates to

4x - 7

. "The number increased by 5" translates to

x + 5

. Since these expressions are equal, we write:

4x - 7 = x + 5

. To solve, we'll collect like terms. Subtract

x

from both sides:

3x - 7 = 5

. Add 7 to both sides:

3x = 12

. Divide by 3:

x = 4

. Let's verify: four times 4 decreased by 7 gives us

16 - 7 = 9

, and 4 increased by 5 also gives us

4 + 5 = 9

. ✓ Looking at the wrong answers: Choice A (

12

) is what you get if you forget to divide by 3 in the final step—you'd stop at

3x = 12

and mistakenly think

x = 12

. Choice B (

-4

) occurs if you make a sign error when moving terms across the equals sign. Choice D (

-12

) combines both mistakes—getting the wrong sign and forgetting the final division. For word problems like this, always set up your equation carefully by identifying what each phrase means mathematically, solve step by step, and then check your answer by substituting it back into the original problem statement. This verification step catches most algebra errors.

Question 13

Solve for $y$ : $\dfrac{4}{5}y + \dfrac{3}{10} = \dfrac{1}{2}$

$0.25$ (correct answer)
$0.40$
$1.25$
$2$

Explanation: When you encounter an equation with fractions, your goal is to isolate the variable by systematically eliminating the fractions and constants. Start by subtracting

\frac{3}{10}

from both sides to isolate the term with

y

\frac{4}{5}y = \frac{1}{2} - \frac{3}{10}

To subtract these fractions, find a common denominator. The LCD of 2 and 10 is 10:

\frac{1}{2} = \frac{5}{10}

So:

\frac{4}{5}y = \frac{5}{10} - \frac{3}{10} = \frac{2}{10} = \frac{1}{5}

Now solve for

y

by dividing both sides by

\frac{4}{5}

, which is the same as multiplying by

\frac{5}{4}

y = \frac{1}{5} \times \frac{5}{4} = \frac{5}{20} = \frac{1}{4} = 0.25

The correct answer is A)

0.25

. Looking at the wrong answers: B)

0.40

likely comes from incorrectly handling the fraction arithmetic or confusing

\frac{2}{5}

with the final answer. C)

1.25

results from multiplying instead of dividing by

\frac{4}{5}

, or making sign errors. D)

2

suggests major computational mistakes, possibly treating fractions as whole numbers. Study tip: When solving fraction equations, always convert to a common denominator before adding or subtracting, and remember that dividing by a fraction means multiplying by its reciprocal. Double-check by substituting your answer back into the original equation.