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Algebra Help: Solving One Variable Linear Equations Inequalities

Review real example questions for Solving One Variable Linear Equations Inequalities in Algebra.

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

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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 $5p - 4 = 9$ as $5p = 7$ 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)
$x = 3$

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 $\frac{x}{3} + 2 = 7$ , 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!