ISEE Upper Level Mathematics Achievement Quiz: Linear Equations And Inequalities

What this quiz covers

This quiz focuses on Linear Equations And Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper Level Mathematics Achievement.

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.

Question 1

1. $x\ge12$
2. $x\ge15$ (correct answer)
3. $x\le15$
4. $x=13$

Explanation: This question tests the ability to solve linear inequalities at the ISEE Upper Level, focusing on break-even analysis and profit calculations. Linear inequalities model real-world constraints where we need to find when a condition is met or exceeded. The shop has $120 in fixed costs and earns$8 profit per item sold, so the net profit is 8x - 120 where x is the number of items. For the shop to break even or make a profit, we need 8x - 120 ≥ 0. Solving: add 120 to both sides to get 8x ≥ 120, then divide by 8 to get x ≥ 15. A common error is confusing profit with revenue or misunderstanding what break-even means. To help students, teach them to identify fixed costs versus variable profits, understand that break-even means zero net profit, and verify their answer by substituting back into the original context.

Question 2

1. $k = 0$
2. $k = 2$ (correct answer)
3. $k = 4$
4. $k = 8$

Explanation: When you encounter a linear equation with a parameter that asks "for what value does the equation have no solution," you're dealing with a situation where the equation becomes inconsistent—meaning the variables cancel out but leave behind a false statement. Let's expand and simplify the equation $2(x + 3) = k(x - 1) + 14$. First, distribute on both sides: $2x + 6 = kx - k + 14$. Now collect all terms with $x$ on one side and constants on the other: $2x - kx = -k + 14 - 6$, which gives us $(2 - k)x = -k + 8$. For this equation to have no solution, the coefficient of $x$ must equal zero while the right side is non-zero. This means $2 - k = 0$ (so $k = 2$) and $-k + 8 \neq 0$. When $k = 2$, we get $0 \cdot x = -2 + 8 = 6$, or $0 = 6$, which is impossible. Looking at the wrong answers: Choice A ($k = 0$) gives us $2x = 8$, so $x = 4$—this has a solution. Choice C ($k = 4$) results in $-2x = 4$, so $x = -2$—also a solution. Choice D ($k = 8$) yields $-6x = 0$, so $x = 0$—again, a solution exists. Remember this pattern: a linear equation has no solution when the coefficient of the variable becomes zero but the constant term remains non-zero. Always check that both conditions are met when working with parametric equations.

Question 3

1. $-7$
2. $-5$
3. $-3$ (correct answer)
4. $-1$

Explanation: When you encounter a linear equation with variables on both sides, your goal is to isolate the variable by systematically eliminating terms through inverse operations. Starting with $3(x - 2) + 5 = 2(x + 1) - 4$, first distribute on both sides:

• Left side: $3x - 6 + 5 = 3x - 1$
• Right side: $2x + 2 - 4 = 2x - 2$

Question 4

1. $m = 3$
2. $m = 6$ (correct answer)
3. $m = 9$
4. $m = 12$

Explanation: When you encounter a system of linear equations asking about infinitely many solutions, you're dealing with the concept of dependent equations. A system has infinitely many solutions when the two equations are actually the same line written in different forms. For infinitely many solutions, one equation must be a scalar multiple of the other. Let's examine our system: $\begin{cases} 2x + 3y = 7 \\ 4x + my = 14 \end{cases}$ Notice that if you multiply the first equation by 2, you get: $2(2x + 3y) = 2(7)$, which simplifies to $4x + 6y = 14$. For the system to have infinitely many solutions, this transformed equation must be identical to the second equation $4x + my = 14$. Since the coefficients of $x$ and the constants already match (4 and 14), we need the coefficient of $y$ to match as well. Therefore, $m = 6$. Let's check why the other answers fail: Choice (A) $m = 3$ would give us $4x + 3y = 14$, which isn't a multiple of the first equation. Choice (C) $m = 9$ would create $4x + 9y = 14$, making the system inconsistent with no solution. Choice (D) $m = 12$ similarly creates an inconsistent system. Study tip: When checking for infinitely many solutions in linear systems, look for one equation being a scalar multiple of another. All coefficients and constants must scale by the same factor—if they don't, you'll either have no solution or exactly one solution.

Question 5

1. $x \leq 1$
2. $x \geq 1$
3. $x \leq \frac{7}{11}$ (correct answer)
4. $x \geq \frac{7}{11}$

Explanation: When you encounter an inequality with variables on both sides, your goal is to isolate the variable while carefully tracking the direction of the inequality sign. Start by distributing the -2 on the left side: $-2(3x - 4) = -6x + 8$. So the inequality becomes $-6x + 8 \geq 5x + 1$. Next, collect all terms with $x$ on one side and constants on the other. Subtract $5x$ from both sides: $-6x - 5x + 8 \geq 1$, which simplifies to $-11x + 8 \geq 1$. Then subtract 8 from both sides: $-11x \geq -7$. Here's the crucial step: when you divide both sides by a negative number, you must flip the inequality sign. Dividing by -11 gives you $x \leq \frac{7}{11}$, which matches answer choice C. Answer A ($x \leq 1$) represents an error in arithmetic—likely miscalculating $\frac{7}{11}$ as 1. Answer B ($x \geq 1$) combines both the arithmetic error and forgetting to flip the inequality sign. Answer D ($x \geq \frac{7}{11}$) gets the fraction right but fails to reverse the inequality when dividing by the negative coefficient. The key strategy to remember: whenever you multiply or divide an inequality by a negative number, always flip the inequality sign. This is one of the most common mistakes on standardized tests, so double-check your work whenever you see negative coefficients in inequality problems.

Question 6

1. $x = -2$ only
2. $x = 3$ only
3. $x = -3$ and $x = 3$ (correct answer)
4. $x = -3, x = -2,$ and $x = 3$

Explanation: When you encounter a rational expression (a fraction with variables), it's undefined whenever the denominator equals zero, since division by zero is impossible in mathematics. To find where $\frac{x + 2}{x^2 - 9}$ is undefined, you need to determine when the denominator $x^2 - 9 = 0$. Notice that $x^2 - 9$ is a difference of squares, which factors as $(x - 3)(x + 3) = 0$. Using the zero product property, either $x - 3 = 0$ or $x + 3 = 0$, giving you $x = 3$ or $x = -3$. At both these values, the denominator becomes zero, making the expression undefined. The numerator $x + 2$ doesn't affect where the expression is undefined—only the denominator matters for this question. Looking at the answer choices: Choice A suggests only $x = -2$ makes the expression undefined, but substituting $x = -2$ gives us $\frac{0}{(-2)^2 - 9} = \frac{0}{-5} = 0$, which is perfectly defined. Choice B identifies only $x = 3$, missing the other critical value. Choice D incorrectly includes $x = -2$, which we just showed doesn't make the denominator zero. Choice C correctly identifies both $x = -3$ and $x = 3$ as the values that make the denominator zero. Remember: for rational expressions, always factor the denominator completely to find all values that make it zero. Don't be distracted by values that make the numerator zero—those typically just give you zeros of the function, not points where it's undefined.

Question 7

1. $4$
2. $6$
3. $8$ (correct answer)
4. $12$

Explanation: When you encounter a square root equation, your goal is to isolate the variable by eliminating the radical. The key strategy is to square both sides of the equation, which will remove the square root and give you a linear equation to solve. Starting with $\sqrt{3x + 1} = 5$, square both sides to get $(\sqrt{3x + 1})^2 = 5^2$. This simplifies to $3x + 1 = 25$. Now solve for $x$ by subtracting 1 from both sides: $3x = 24$. Dividing by 3 gives $x = 8$. You can verify this by substituting back: $\sqrt{3(8) + 1} = \sqrt{25} = 5$ ✓ Let's examine why the other answers don't work. Choice A) $4$ would give $\sqrt{3(4) + 1} = \sqrt{13} \approx 3.6$, not 5. Choice B) $6$ would give $\sqrt{3(6) + 1} = \sqrt{19} \approx 4.4$, still not 5. Choice D) $12$ would give $\sqrt{3(12) + 1} = \sqrt{37} \approx 6.1$, which overshoots the target. These incorrect answers likely represent common algebraic mistakes: perhaps forgetting to square the right side (leading to smaller values like A or B) or making arithmetic errors during the solving process (potentially leading to D). Remember: when solving radical equations, always square both sides completely and check your answer by substituting back into the original equation. This verification step catches any extraneous solutions that can arise from the squaring process.

Question 8

1. $x = 1$ or $x = 7$
2. $x = -1$ or $x = 7$ (correct answer)
3. $x = 1$ or $x = -7$
4. $x = -1$ or $x = -7$

Explanation: When you encounter an absolute value equation, remember that the expression inside the absolute value bars can equal either the positive or negative version of the number on the other side. This is because absolute value measures distance from zero, so both positive and negative numbers at the same distance give the same absolute value. To solve $|3x - 9| = 12$, you need to consider two cases: Case 1: $3x - 9 = 12$ Adding 9 to both sides: $3x = 21$ Dividing by 3: $x = 7$ Case 2: $3x - 9 = -12$ Adding 9 to both sides: $3x = -3$ Dividing by 3: $x = -1$ You can verify: when $x = 7$, we get $|3(7) - 9| = |21 - 9| = |12| = 12$ ✓ When $x = -1$, we get $|3(-1) - 9| = |-3 - 9| = |-12| = 12$ ✓ Looking at the wrong answers: Choice A gives $x = 1$, but substituting yields $|3(1) - 9| = |-6| = 6 \neq 12$. Choice C incorrectly uses $x = -7$, which gives $|3(-7) - 9| = |-30| = 30 \neq 12$. Choice D has both wrong values from the other incorrect options. The correct answer is B: $x = -1$ or $x = 7$. Study tip: For absolute value equations $|expression| = positive\ number$, always set up two equations: one where the expression equals the positive number, and another where it equals the negative number. Then solve both and check your answers.

Question 9

1. $-\frac{7}{3} \leq x \leq 7$ (correct answer)
2. $-\frac{5}{3} \leq x \leq \frac{21}{3}$
3. $-\frac{7}{3} \leq x \leq \frac{21}{3}$
4. $-\frac{5}{3} \leq x \leq 7$

Explanation: When you encounter compound inequalities like this one, you're solving for the range of values that satisfy both conditions simultaneously. The key is to treat this as two separate inequalities connected by "and" - both must be true at the same time. To solve $-2 \leq \frac{3x - 1}{4} \leq 5$, you need to isolate $x$ by performing the same operations on all three parts of the inequality. First, multiply everything by 4: $-8 \leq 3x - 1 \leq 20$. Then add 1 to all parts: $-7 \leq 3x \leq 21$. Finally, divide everything by 3: $-\frac{7}{3} \leq x \leq 7$. This matches answer choice A. Let's examine why the other options are incorrect. Choice B gives $-\frac{5}{3} \leq x \leq \frac{21}{3}$, which suggests errors in solving the left inequality - you'd get $-\frac{5}{3}$ if you forgot to add 1 after multiplying by 4. Choice C shows $-\frac{7}{3} \leq x \leq \frac{21}{3}$, where the left side is correct but the right side shows $\frac{21}{3}$ instead of simplifying to 7. Choice D combines both types of errors: the incorrect left boundary from choice B and the correct right boundary. Remember that with compound inequalities, you must perform identical operations on all three parts simultaneously. Always double-check your arithmetic at each step, and don't forget to simplify fractions when possible - $\frac{21}{3} = 7$ in this case.

Question 10

1. $x \geq -17$
2. $x \leq -17$ (correct answer)
3. $x \geq 17$
4. $x \leq 17$

Explanation: When you encounter an inequality with parentheses and variables on both sides, your approach should mirror solving regular equations, but with one crucial difference: flipping the inequality sign when multiplying or dividing by a negative number. Let's solve $2x - 3(x + 4) \geq 5$ step by step. First, distribute the -3: $2x - 3x - 12 \geq 5$. Combine like terms: $-x - 12 \geq 5$. Add 12 to both sides: $-x \geq 17$. Now comes the critical step: multiply both sides by -1 to solve for $x$. Since you're multiplying by a negative number, you must flip the inequality sign: $x \leq -17$. Choice A ($x \geq -17$) represents the most common error—forgetting to flip the inequality sign when multiplying by -1. Students often solve correctly up to $-x \geq 17$ but then write $x \geq -17$, missing this crucial rule. Choice C ($x \geq 17$) likely comes from sign errors during the distribution step, possibly treating $-3(x + 4)$ as $-3x + 12$ instead of $-3x - 12$. Choice D ($x \leq 17$) shows another sign error—correctly flipping the inequality but arriving at positive 17 instead of negative 17, again suggesting mistakes in the algebraic manipulation. Study tip: Whenever you multiply or divide an inequality by a negative number, always flip the inequality sign. Double-check your work by testing a value from your solution set in the original inequality to verify it makes the statement true.

Question 11

1. $x\ge15$ (correct answer)
2. $x\le15$
3. $x\ge10$
4. $x=10$

Explanation: This question tests the ability to solve linear equations and inequalities at the ISEE Upper Level, focusing on algebraic manipulation and logical reasoning. Linear equations and inequalities involve finding the value of a variable that satisfies a given condition. The foundational concept is applying inverse operations to isolate the variable. In this context, the problem describes a real-world situation where the inequality models covering rent with per-item earnings for a business. The correct answer is obtained by adding 135 to both sides, then dividing by 9, yielding x ≥ 15, indicating the minimum items needed. A common misconception is sign errors, where students might incorrectly apply the operations, leading to incorrect solutions. To help students, teach the systematic approach to solving inequalities—identify the variable, isolate using inverse operations, and check the solution in context. Encourage consistent practice with real-world problems to reinforce concepts and avoid pitfalls like sign errors or operation mistakes.

Question 12

1. $10$
2. $15$ (correct answer)
3. $20$
4. $25$

Explanation: When you encounter an equation with a square root, your goal is to isolate the radical term first, then eliminate it by squaring both sides. Starting with $\sqrt{2x - 5} + 3 = 8$, subtract 3 from both sides to isolate the square root: $\sqrt{2x - 5} = 5$. Now square both sides to eliminate the radical: $(\sqrt{2x - 5})^2 = 5^2$, which gives you $2x - 5 = 25$. Add 5 to both sides: $2x = 30$, then divide by 2: $x = 15$. You can verify this by substituting back: $\sqrt{2(15) - 5} + 3 = \sqrt{25} + 3 = 5 + 3 = 8$ ✓ Looking at the wrong answers: Choice (A) $x = 10$ would give $\sqrt{15} + 3 \approx 6.9$, not 8. This might result from incorrectly solving $2x - 5 = 15$ instead of $2x - 5 = 25$. Choice (C) $x = 20$ would give $\sqrt{35} + 3 \approx 8.9$, possibly from solving $2x = 35$ and forgetting the $-5$. Choice (D) $x = 25$ would give $\sqrt{45} + 3 \approx 9.7$, which might come from setting $x = 25$ directly without proper algebraic steps. Remember to always check your answer in radical equations, as squaring can sometimes introduce extraneous solutions. The key pattern is: isolate the radical, square both sides, solve the resulting linear equation, then verify.

Question 13

1. No solutions
2. Exactly one solution (correct answer)
3. Exactly two solutions
4. Infinitely many solutions

Explanation: When you encounter rational equations like this one, the key insight is recognizing when the denominators are identical on both sides. This creates a special situation that simplifies the solving process significantly. Since both sides have the same denominator $(x - 3)$, you can multiply both sides by $(x - 3)$ to eliminate the fractions, provided that $x \neq 3$ (since division by zero is undefined). This gives you: $2x + 1 = x + 5$ Solving this linear equation: $2x - x = 5 - 1$, so $x = 4$. Now you must check whether this solution is valid by substituting back into the original equation. Since $x = 4$ doesn't make the denominator zero ($4 - 3 = 1 \neq 0$), this solution is valid. Looking at the wrong answers: (A) No solutions would occur if our algebraic manipulation led to a contradiction like $0 = 5$, or if our solution made a denominator zero. (C) Exactly two solutions might tempt you if you mistakenly think rational equations always yield quadratic equations, but this one reduces to linear. (D) Infinitely many solutions would only happen if both the numerators and denominators were proportional, creating an identity that's true for all valid values of $x$. Remember this pattern: when rational equations have identical denominators, focus on whether the resulting equation after clearing denominators has solutions, and always verify that your solutions don't create zero denominators in the original equation.

Question 14

1. $x=6$
2. $x=9$ (correct answer)
3. $x=3$
4. $x=12$

Explanation: This question tests the ability to solve linear equations at the ISEE Upper Level, focusing on mixture problems with different concentrations. Linear equations model conservation of solute in mixture problems where solutions of different strengths are combined. The chemist needs 12L of 25% solution by mixing x liters of 20% solution with (12-x) liters of 40% solution. The equation 0.20x + 0.40(12-x) = 0.25(12) represents conservation of pure substance. Expanding: 0.20x + 4.8 - 0.40x = 3, which simplifies to -0.20x + 4.8 = 3, then -0.20x = -1.8, so x = 9. A common mistake is setting up the wrong equation or making arithmetic errors with decimals. To help students, emphasize that the amount of pure substance before mixing equals the amount after, practice converting percentages to decimals, and verify by checking that 9L of 20% plus 3L of 40% gives the correct final concentration.

Question 15

1. $x\ge7$ (correct answer)
2. $x\le7$
3. $x\ge12$
4. $x=12$

Explanation: This question tests the ability to solve linear equations and inequalities at the ISEE Upper Level, focusing on algebraic manipulation and logical reasoning. Linear equations and inequalities involve finding the value of a variable that satisfies a given condition. The foundational concept is applying inverse operations to isolate the variable. In this context, the problem describes a real-world situation where the inequality models the break-even point for a business with per-item charges and fixed costs. The correct answer is obtained by adding 98 to both sides, then dividing by 14, yielding x ≥ 7, indicating the minimum items for non-negative result. A common misconception is sign errors, where students might incorrectly apply the operations, leading to incorrect solutions. To help students, teach the systematic approach to solving inequalities—identify the variable, isolate using inverse operations, and check the solution in context. Encourage consistent practice with real-world problems to reinforce concepts and avoid pitfalls like sign errors or operation mistakes.

Question 16

1. $\frac{7}{2}$ (correct answer)
2. $\frac{5}{2}$
3. $\frac{3}{2}$
4. $\frac{1}{2}$

Explanation: When you encounter a rational equation like this, your goal is to eliminate the fraction by cross-multiplying, then solve the resulting linear equation. Starting with $\frac{3x + 2}{x - 1} = 5$, multiply both sides by $(x - 1)$ to clear the denominator: $3x + 2 = 5(x - 1)$ Distribute the 5 on the right side: $3x + 2 = 5x - 5$ Now solve for $x$ by collecting like terms. Subtract $3x$ from both sides: $2 = 2x - 5$ Add 5 to both sides: $7 = 2x$ Therefore: $x = \frac{7}{2}$ You can verify this by substituting back: $\frac{3(\frac{7}{2}) + 2}{\frac{7}{2} - 1} = \frac{\frac{21}{2} + 2}{\frac{5}{2}} = \frac{\frac{25}{2}}{\frac{5}{2}} = 5$ ✓ Looking at the wrong answers: Choice B ($\frac{5}{2}$) might result from incorrectly distributing or sign errors. Choice C ($\frac{3}{2}$) could come from mixing up coefficients during algebraic manipulation. Choice D ($\frac{1}{2}$) might arise from calculation mistakes when combining like terms or solving the final linear equation. Study tip: Always verify your solution in rational equations by substituting back into the original equation. Also, remember to check that your answer doesn't make any denominator zero—though that's already given here with the condition $x \neq 1$.

Question 17

1. $-1 \leq x \leq 7$ (correct answer)
2. $-2 \leq x \leq 8$
3. $x \leq -1$ or $x \geq 7$
4. $x \leq -2$ or $x \geq 8$

Explanation: When you encounter an absolute value inequality like $|2x - 6| \leq 8$, you're dealing with a compound inequality that describes all values within a certain distance from a central point. The absolute value inequality $|A| \leq B$ is equivalent to $-B \leq A \leq B$. This means the expression inside the absolute value bars must be between the negative and positive values of the number on the right side. Applying this rule: $|2x - 6| \leq 8$ becomes $-8 \leq 2x - 6 \leq 8$. Now solve both parts of this compound inequality by adding 6 to all three parts: $-8 + 6 \leq 2x \leq 8 + 6$, which gives us $-2 \leq 2x \leq 14$. Finally, divide everything by 2: $-1 \leq x \leq 7$. Choice A gives us exactly this result: $-1 \leq x \leq 7$. Choice B ($-2 \leq x \leq 8$) results from forgetting to divide by 2 in the final step. Choice C ($x \leq -1$ or $x \geq 7$) represents the solution to $|2x - 6| \geq 8$ instead of $|2x - 6| \leq 8$ - this is the common error of confusing "less than or equal to" with "greater than or equal to" in absolute value inequalities. Choice D makes both mistakes: using the wrong inequality direction and forgetting to divide by 2. Remember: $|A| \leq B$ means "between" (compound inequality with "and"), while $|A| \geq B$ means "outside" (compound inequality with "or").

Question 18

1. 87 (correct answer)
2. 90
3. 93
4. 96

Explanation: This problem tests your ability to solve a linear equation and then use that solution to evaluate a different algebraic expression. When you see this type of question, recognize that you need to find the value of the variable first, then substitute it into the target expression. To solve $3x - 7 = 2x + 11$, start by collecting like terms. Subtract $2x$ from both sides: $3x - 2x - 7 = 11$, which simplifies to $x - 7 = 11$. Then add 7 to both sides: $x = 18$. Now substitute $x = 18$ into the expression $5x - 3$: $5(18) - 3 = 90 - 3 = 87$. Looking at the wrong answers: Choice B (90) is what you'd get if you forgot to subtract the 3 after calculating $5x = 90$. Choice C (93) might result from an arithmetic error, perhaps adding 3 instead of subtracting it. Choice D (96) could come from solving the original equation incorrectly—if you mistakenly found $x = 19.8$ and rounded to 20, then $5(20) - 4 = 96$ (though this involves multiple errors). The correct answer is A) 87. Study tip: On multi-step algebra problems like this, always double-check your solution by substituting your $x$-value back into the original equation. Here, $3(18) - 7 = 54 - 7 = 47$ and $2(18) + 11 = 36 + 11 = 47$ ✓. This verification step catches calculation errors before you move to the final expression.

Question 19

1. $x=2$
2. $x=3$ (correct answer)
3. $x=4$
4. $x=-3$

Explanation: This question tests the ability to solve linear equations and inequalities at the ISEE Upper Level, focusing on algebraic manipulation and logical reasoning. Linear equations and inequalities involve finding the value of a variable that satisfies a given condition. The foundational concept is applying inverse operations to isolate the variable. In this context, the problem describes a real-world situation where the equation models mixing solutions to achieve 45% concentration in 6 liters. The correct answer is obtained by distributing, combining like terms, and solving -0.3x = -0.9 to get x = 3, representing liters of 30% solution. A common misconception is sign errors, where students might incorrectly apply the operations, leading to incorrect solutions. To help students, teach the systematic approach to solving equations—identify the variable, isolate using inverse operations, and check the solution in context. Encourage consistent practice with real-world problems to reinforce concepts and avoid pitfalls like sign errors or operation mistakes.

Question 20

1. Any value except $a = 1$ (correct answer)
2. Any value except $a = 0$
3. Only when $a = 1$
4. Only when $a = 0$

Explanation: When you encounter an equation with a parameter like $a$, you need to determine what values of that parameter give the equation exactly one solution, no solutions, or infinitely many solutions. Let's solve $x + 2 = a(x - 1) + 3$ by expanding and collecting like terms. First, distribute the $a$: $x + 2 = ax - a + 3$. Then rearrange to get all $x$ terms on one side: $x - ax = -a + 3 - 2$, which simplifies to $x(1 - a) = 1 - a$. Here's the key insight: when you have an equation in the form $x \cdot (\text{coefficient}) = \text{constant}$, the number of solutions depends on that coefficient. If the coefficient of $x$ is non-zero (meaning $1 - a \neq 0$, or $a \neq 1$), you can divide both sides to get exactly one solution: $x = \frac{1-a}{1-a} = 1$. But if $a = 1$, the coefficient becomes zero, giving you $0 \cdot x = 0$, which is true for all values of $x$ — infinitely many solutions. Choice A is correct because the equation has exactly one solution for any value of $a$ except $a = 1$. Choice B incorrectly identifies $a = 0$ as problematic, but when $a = 0$, we get $x = 1$, which is perfectly valid. Choice C suggests only $a = 1$ works, but that's when we get infinitely many solutions. Choice D incorrectly focuses on $a = 0$. Remember: in linear equations with parameters, look for when the coefficient of the variable becomes zero — that's usually where the special case occurs.