Algebra Quiz: Solving One Variable Linear Equations Inequalities

20 questions · exam conditions

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Solving One Variable Linear Equations InequalitiesQuestion 1 of 20

Solve for $x$ : $2(x - 3) + 5 = 15$

x = 11

x = 5

x = 8

x = 2

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Algebra Quiz: Solving One Variable Linear Equations Inequalities

Practice Solving One Variable Linear Equations Inequalities in Algebra 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 Solving One Variable Linear Equations Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

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 $x$ : $2(x - 3) + 5 = 15$

$x = 11$
$x = 5$
$x = 8$ (correct answer)
$x = 2$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. The properties of equality tell us that we can add, subtract, multiply, or divide the same number on both sides of an equation without changing the solution—think of it like keeping a balance scale level. In this equation, we start by distributing 2 across the parentheses: 2(x - 3) + 5 becomes 2x - 6 + 5. Then we combine like terms to get 2x - 1 = 15, and solving from there gives us 2x = 16, so x = 8. Choice C is correct because it follows all the steps properly: distributes correctly (2x - 6), combines like terms (-6 + 5 = -1), and isolates x accurately to get x = 8. Choice B makes a common mistake: it forgets to distribute 2 to both terms inside the parentheses, calculating 2x - 3 instead of 2x - 6. To check if your answer is right, plug it back into the original equation: substitute your x-value and see if both sides equal the same number. This is a great habit that catches mistakes!

Question 2

If $\frac{2p - 1}{p + 3} = \frac{3}{4}$ and $p \neq -3$ , what is the value of $p$ ?

$p = \frac{9}{5}$
$p = \frac{11}{5}$
$p = \frac{13}{5}$ (correct answer)
$p = \frac{7}{5}$

Explanation: When you encounter an equation with fractions set equal to each other, you're dealing with a rational equation. The most efficient approach is to cross-multiply, which eliminates the fractions and creates a linear equation you can solve directly. Starting with

\frac{2p - 1}{p + 3} = \frac{3}{4}

, cross-multiply by multiplying the numerator of each fraction by the denominator of the other:

4(2p - 1) = 3(p + 3)

. Expanding both sides gives you

8p - 4 = 3p + 9

. Now collect like terms: subtract

3p

from both sides to get

5p - 4 = 9

, then add

4

to both sides:

5p = 13

. Therefore,

p = \frac{13}{5}

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

p = \frac{9}{5}

) results from incorrectly distributing or making an arithmetic error when collecting terms. Choice B (

p = \frac{11}{5}

) typically comes from a sign error, such as writing

5p - 4 = 9

5p = 7

instead of

5p = 13

. Choice D (

p = \frac{7}{5}

) often results from incorrectly cross-multiplying or making multiple computational errors along the way. You can verify your answer by substituting

p = \frac{13}{5}

back into the original equation to confirm both sides equal

\frac{3}{4}

. Study tip: Always cross-multiply when solving rational equations with single fractions on each side—it's faster and less error-prone than finding common denominators. Double-check by substituting your answer back into the original equation.

Question 3

Which inequality represents all values of $x$ for which $2x - 3 \geq 4x + 1$ ?

$x \geq 2$
$x \geq -2$
$x \leq 2$
$x \leq -2$ (correct answer)

Explanation: When solving linear inequalities, you follow the same steps as solving equations, with one crucial exception: flipping the inequality sign when multiplying or dividing by a negative number. To solve

2x - 3 \geq 4x + 1

, start by collecting like terms. Subtract

2x

from both sides:

-3 \geq 2x + 1

. Then subtract 1 from both sides:

-4 \geq 2x

. Finally, divide both sides by 2:

-2 \geq x

, which is equivalent to

x \leq -2

. This confirms answer choice D is correct. Let's examine why the other options are wrong. Choice A (

x \geq 2

) represents the opposite inequality direction and uses the wrong boundary value - this would happen if you made sign errors throughout. Choice B (

x \geq -2

) has the correct boundary value but the wrong inequality direction - this occurs if you forget that

-2 \geq x

means

x \leq -2

. Choice C (

x \leq 2

) has the correct inequality direction but wrong boundary value - this mistake happens if you make arithmetic errors while solving but remember to handle the inequality correctly. Key strategy: When solving inequalities, double-check your work by testing a value from your solution set in the original inequality. For example, try

x = -3

2(-3) - 3 = -9

and

4(-3) + 1 = -11

. Since

-9 \geq -11

is true,

x = -3

works, confirming

x \leq -2

is correct.

Question 4

For what value of $m$ does the equation $5x + m = 2(x + 4) + 3x$ have infinitely many solutions?

$m = 8$ (correct answer)
$m = -8$
$m = 0$
$m = 4$

Explanation: First simplify the right side:

2(x + 4) + 3x = 2x + 8 + 3x = 5x + 8

. So the equation becomes

5x + m = 5x + 8

. For this equation to have infinitely many solutions, both sides must be identical, which happens when

m = 8

. When

m = 8

, we get

5x + 8 = 5x + 8

, which is always true. Choice B would give

5x - 8 = 5x + 8

, leading to

-8 = 8

(no solution). Choice C would give

5x = 5x + 8

, leading to

0 = 8

(no solution). Choice D would give

5x + 4 = 5x + 8

, leading to

4 = 8

(no solution).

Question 5

Solve for $x$ : $\frac{x}{3} + 2 = 7$

$x = \frac{5}{3}$
$x = 9$
$x = 15$ (correct answer)
$x = 3$

\frac{x}{3} + 2 = 7

, we first subtract 2 from both sides to get

\frac{x}{3} = 5

. Then we multiply both sides by 3 to isolate x:

x = 15

. Choice C is correct because it follows all the steps properly: uses inverse operations and isolates x accurately to get

x = 15

. Choice B is a close attempt, but it doesn't quite finish the job: it stops at

\frac{x}{3} = 5

without multiplying both sides by 3 to fully isolate x. Here's a reliable strategy for solving linear equations: (1) simplify each side (distribute and combine like terms), (2) get all x-terms on one side and all numbers on the other side, (3) combine the x-terms and combine the numbers, (4) divide by the coefficient to isolate x. If you follow these steps in order, you'll get the right answer every time!

Question 6

Solve for $x$ : $3x + 7 = 22$

$x = 3$
$x = 5$ (correct answer)
$x = 29$
$x = -5$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. To solve an equation means to find the value of the variable that makes the equation true, and we do this by using inverse operations to isolate the variable on one side while keeping the equation balanced. Starting with 3x + 7 = 22, we first subtract 7 from both sides to get 3x = 15. Then we divide both sides by 3 to isolate x: x = 15 ÷ 3 = 5. Choice B is correct because it follows all the steps properly: subtracting 7 from both sides and then dividing by 3 to get x = 5. Choice A makes an error when dividing: 15 ÷ 3 equals 5, not 3. Here's a reliable strategy for solving linear equations: (1) simplify each side (distribute and combine like terms), (2) get all x-terms on one side and all numbers on the other side, (3) combine the x-terms and combine the numbers, (4) divide by the coefficient to isolate x. If you follow these steps in order, you'll get the right answer every time!

Question 7

Which value of $x$ satisfies $\frac{x - 2}{5} = \frac{3}{2}$ ?

$x = \frac{19}{2}$ (correct answer)
$x = \frac{11}{2}$
$x = \frac{7}{2}$
$x = \frac{25}{2}$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. The properties of equality tell us that we can add, subtract, multiply, or divide the same number on both sides of an equation without changing the solution—think of it like keeping a balance scale level. Starting with (x - 2)/5 = 3/2, we first multiply both sides by 5 to clear the fraction on the left: x - 2 = 5 × (3/2) = 15/2. Then we add 2 to both sides to isolate x: x = 15/2 + 2 = 15/2 + 4/2 = 19/2. Choice B is correct because it follows all the steps properly: multiplying both sides by 5 and then adding 2 (converted to 4/2) to get x = 19/2. Choice A makes an error when adding 2 to 15/2: it seems to get 11/2 instead of 19/2, forgetting that 2 = 4/2 when adding fractions. To check if your answer is right, plug it back into the original equation: substitute your x-value and see if both sides equal the same number. This is a great habit that catches mistakes!

Question 8

Solve for $x$ : $\frac{1}{2}(x + 6) = 5$

$x = 4$ (correct answer)
$x = 16$
$x = -4$
$x = 7$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. The properties of equality tell us that we can add, subtract, multiply, or divide the same number on both sides of an equation without changing the solution—think of it like keeping a balance scale level. Starting with (1/2)(x + 6) = 5, we can either distribute first or multiply both sides by 2; let's multiply both sides by 2 to clear the fraction: x + 6 = 10. Then we subtract 6 from both sides to isolate x: x = 10 - 6 = 4. Choice A is correct because it follows all the steps properly: multiplies both sides by 2 to get x + 6 = 10, then subtracts 6 to get x = 4. Choice B makes a common mistake: it might have incorrectly multiplied to get x + 6 = 22 instead of x + 6 = 10, leading to x = 16. To check if your answer is right, plug it back into the original equation: substitute your x-value and see if both sides equal the same number. This is a great habit that catches mistakes!

Question 9

Solve the inequality: $3x - 5 > 16$

$x > 7$ (correct answer)
$x < 7$
$x > \frac{11}{3}$
$x \ge 7$

Explanation: This question tests your ability to solve linear inequalities in one variable using inverse operations and properties of equality. To solve an inequality means to find all values of the variable that make the inequality true, and we do this by using inverse operations to isolate the variable on one side while keeping the inequality balanced. Starting with 3x - 5 > 16, we first add 5 to both sides to get 3x > 21. Then we divide both sides by 3 to isolate x: x > 21 ÷ 3 = 7. Choice A is correct because it follows all the steps properly: adds 5 to both sides and then divides by 3 to get x > 7. Choice C makes an arithmetic error: when we have 3x > 21, dividing both sides by 3 gives x > 7, not x > 11/3 (which would be approximately 3.67). For inequalities, there's just one extra rule to remember beyond regular equation solving: if you multiply or divide by a negative number (like going from -2x < 6 to x > -3), flip the inequality direction. Everything else works exactly the same as equations!

Question 10

What is the solution set for $3x - 5 > 16$ ?

$x > \frac{11}{3}$
$x < 7$
$x \ge 7$
$x > 7$ (correct answer)

Explanation: This question tests your ability to solve linear inequalities in one variable using inverse operations and properties of equality. To solve an inequality means to find all values of the variable that make the inequality true, and we do this by using inverse operations to isolate the variable on one side while keeping the inequality balanced. Starting with 3x - 5 > 16, we first add 5 to both sides to get 3x > 21. Then we divide both sides by 3 to isolate x: x > 21 ÷ 3 = 7. Choice A is correct because it follows all the steps properly: adding 5 to both sides and then dividing by 3 to get x > 7. Choice C makes an error when adding: 16 + 5 equals 21, not 11, so we get 3x > 21 and x > 7, not x > 11/3. The key to avoiding sign errors: when you move a term to the other side of the equation, its sign changes. If you have -5 on the left and want to move it right, it becomes +5 on the right. Think 'move and flip the sign' to keep track!

Question 11

What is $x$ in the equation $5x - 8 = 2x + 10$ ?

$x = 2$
$x = -6$
$x = \frac{2}{3}$
$x = 6$ (correct answer)

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. A linear equation in one variable has exactly one solution (unless it's a special case), and we find it by systematically undoing the operations around the variable using inverse operations in reverse order. When the variable appears on both sides like 5x - 8 = 2x + 10, a good strategy is to move all x-terms to one side (let's move them to the left by subtracting 2x from both sides): 3x - 8 = 10. Then move constants to the other side by adding 8: 3x = 18. This gives us x = 18 ÷ 3 = 6. Choice A is correct because it properly moves the x-terms to one side, the constants to the other, and divides to get x = 6. Choice B has a sign error—when we get 3x = 18 and divide by 3, we get positive 6, not negative 6. When solving equations with variables on both sides, you can move the variable to either side—choose whichever keeps your coefficient positive to avoid extra negative signs and reduce errors!

Question 12

If $ax + 5 = 3x - 7$ and $a \neq 3$ , solve for $x$ in terms of $a$ .

$x = \frac{12}{a - 3}$
$x = \frac{-12}{a - 3}$ (correct answer)
$x = \frac{12}{3 - a}$
$x = \frac{2}{a - 3}$

Explanation: Starting with

ax + 5 = 3x - 7

, subtract

3x

from both sides:

ax - 3x + 5 = -7

. Factor out

x

(a - 3)x + 5 = -7

. Subtract 5 from both sides:

(a - 3)x = -12

. Divide by

(a - 3)

x = \frac{-12}{a - 3}

. Choice A has the wrong sign in the numerator. Choice C has the wrong sign in the denominator (note that

\frac{-12}{a-3} = \frac{12}{3-a}

, but the question asks for the form with

(a-3)

). Choice D has the wrong numerator entirely.

Question 13

Solve for $x$ : $3x + 7 = 22$

$x = 29$
$x = 5$ (correct answer)
$x = \frac{15}{3}$
$x = -5$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. To solve an equation means to find the value of the variable that makes the equation true, and we do this by using inverse operations to isolate the variable on one side while keeping the equation balanced. Starting with 3x + 7 = 22, we first subtract 7 from both sides to get 3x = 15. Then we divide both sides by 3 to isolate x: x = 15 ÷ 3 = 5. Choice B is correct because it follows all the steps properly: subtracting 7 from both sides and then dividing by 3 to get x = 5. Choice C makes a common mistake: it stops at x = 15/3 without simplifying the fraction, which equals 5. Here's a reliable strategy for solving linear equations: (1) simplify each side (distribute and combine like terms), (2) get all x-terms on one side and all numbers on the other side, (3) combine the x-terms and combine the numbers, (4) divide by the coefficient to isolate x. If you follow these steps in order, you'll get the right answer every time!

Question 14

What is $x$ in the equation $5x - 8 = 2x + 10$ ?

$x = 6$ (correct answer)
$x = \frac{2}{3}$
$x = \frac{18}{3}$
$x = -6$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. A linear equation in one variable has exactly one solution (unless it's a special case), and we find it by systematically undoing the operations around the variable using inverse operations in reverse order. When the variable appears on both sides like 5x - 8 = 2x + 10, a good strategy is to move all x-terms to one side (let's move them to the left by subtracting 2x from both sides): 3x - 8 = 10. Then move constants to the other side by adding 8: 3x = 18. This gives us 3x = 18, so x = 6. Choice A is correct because it follows all the steps properly: subtracting 2x from both sides, adding 8 to both sides, and dividing by 3 to get x = 6. Choice B has a sign error—a tricky spot! It somehow gets x = -6, but when we check: 5(-6) - 8 = -38 and 2(-6) + 10 = -2, which aren't equal. When solving equations with variables on both sides, you can move the variable to either side—choose whichever keeps your coefficient positive to avoid extra negative signs and reduce errors!

Question 15

What value of $x$ makes the equation true? $\frac{x}{3} + 2 = 7$

$x = 15$ (correct answer)
$x = \frac{5}{3}$
$x = 9$
$x = 3$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. To solve an equation means to find the value of the variable that makes the equation true, and we do this by using inverse operations to isolate the variable on one side while keeping the equation balanced. Starting with x/3 + 2 = 7, we first subtract 2 from both sides to get x/3 = 5. Then we multiply both sides by 3 to isolate x: x = 5 × 3 = 15. Choice A is correct because it follows all the steps properly: subtracting 2 from both sides and then multiplying by 3 to get x = 15. Choice B makes a common mistake: it forgets the final step and stops at x/3 = 5, giving x = 3 instead of multiplying both sides by 3. To check if your answer is right, plug it back into the original equation: substitute your x-value and see if both sides equal the same number. This is a great habit that catches mistakes!

Question 16

Solve for $x$ : $6x+4=2(3x+2)$

No solution
$x=0$
$x=1$
Infinitely many solutions (correct answer)

6x+4=2(3x+2)

, we distribute to get

6x+4=6x+4

, subtract 6x from both sides:

4=4

, which is always true, so there are infinitely many solutions. Choice D is correct because it recognizes this as an identity after simplifying, where both sides are equal for all x. Choice A makes a common mistake: it might assume no solution if overlooking that

4=4

is true, but actually it's true for every x. Here's a reliable strategy for solving linear equations: (1) simplify each side (distribute and combine like terms), (2) get all x-terms on one side and all numbers on the other side, (3) combine the x-terms and combine the numbers, (4) divide by the coefficient to isolate x. If you follow these steps in order, you'll get the right answer every time!

Question 17

Solve the inequality: $3x - 5 > 16$

$x > 7$ (correct answer)
$x \ge 7$
$x > \frac{11}{3}$
$x < 7$

Explanation: This question tests your ability to solve linear inequalities in one variable using inverse operations and properties of equality. When solving inequalities, we follow the same steps as equations with one critical rule: if we multiply or divide both sides by a negative number, we must reverse the direction of the inequality symbol (< becomes >, ≤ becomes ≥). Starting with 3x - 5 > 16, we first add 5 to both sides to get 3x > 21. Then we divide both sides by 3 (which is positive, so no flip): x > 7. Choice A is correct because it follows all the steps properly: adds the constant and divides accurately to get x > 7. Choice D forgets to isolate correctly—a tricky spot! If you subtract instead of add the 5, you might get x < 7. For inequalities, there's just one extra rule to remember beyond regular equation solving: if you multiply or divide by a negative number (like going from -2x < 6 to x > -3), flip the inequality direction. Everything else works exactly the same as equations!

Question 18

Solve the inequality: $-2x+8 \le 4$

$x \le 2$
$x \ge 2$ (correct answer)
$x<2$
$x>-2$

-2x+8 \le 4

, we first subtract 8 from both sides to get

-2x \le -4

, then divide both sides by -2 and remember the key rule to flip the inequality direction:

x \ge 2

. Choice B is correct because it follows all the steps properly: subtracts correctly, divides by the negative coefficient, and flips the inequality accurately to get

x \ge 2

. Choice A forgets the critical inequality rule: when dividing both sides by a negative number, we must flip the inequality direction, so

x \le 2

should be

x \ge 2

. For inequalities, there's just one extra rule to remember beyond regular equation solving: if you multiply or divide by a negative number (like going from

-2x \le -4

x \ge 2

), flip the inequality direction. Everything else works exactly the same as equations!

Question 19

Solve the inequality: $-2x + 8 \le 4$

$x \le 2$
$x \ge -2$
$x \ge 2$ (correct answer)
$x \le -2$

Explanation: This question tests your ability to solve linear inequalities in one variable using inverse operations and properties of equality. When solving inequalities, we follow the same steps as equations with one critical rule: if we multiply or divide both sides by a negative number, we must reverse the direction of the inequality symbol (< becomes >, ≤ becomes ≥). Solving -2x + 8 ≤ 4, we first subtract 8 from both sides to get -2x ≤ -4. When we divide both sides by -2, remember the key rule: we must flip the inequality direction! So -2x ≤ -4 becomes x ≥ 2. Choice C is correct because it properly isolates x and remembers to flip the inequality sign when dividing by -2, giving us x ≥ 2. Choice A forgets the critical inequality rule: when dividing both sides by a negative number, we must flip the inequality direction, so x ≤ 2 should be x ≥ 2. For inequalities, there's just one extra rule to remember beyond regular equation solving: if you multiply or divide by a negative number (like going from -2x < 6 to x > -3), flip the inequality direction. Everything else works exactly the same as equations!

Question 20

What is $x$ in the equation $5x-8=2x+10$ ?

$x=6$ (correct answer)
$x=\frac{3}{2}$
$x=-6$
$x=\frac{2}{3}$

Explanation: This question tests your ability to solve linear equations in one variable using inverse operations and properties of equality. The properties of equality tell us that we can add, subtract, multiply, or divide the same number on both sides of an equation without changing the solution—think of it like keeping a balance scale level. When the variable appears on both sides like 5x-8=2x+10, a good strategy is to move all x-terms to one side (let's move them to the left by subtracting 2x): 3x-8=10. Then move constants to the other side: add 8 to get 3x=18. This gives us x=6. Choice A is correct because it follows all the steps properly to get x=6. Choice B has a sign error—a tricky spot! When moving the constants, perhaps they subtracted instead of adding, leading to a negative value like x=-6. The key to avoiding sign errors: when you move a term to the other side of the equation, its sign changes. If you have -3x on the left and want to move it right, it becomes +3x on the right. Think 'move and flip the sign' to keep track!