One- and Multi-Step Equations - SSAT Upper Level Quantitative

If $x$ and $y$ satisfy the system $\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}$ , then $xy$ equals:
Answer: $6$ When you encounter a system of linear equations and need to find the product of the variables, you'll need to solve for each variable first, then multiply them together. To solve this system, substitution works well since the second equation can easily be rearranged. From $x - y = 1$ , you get $x = y + 1$ . Substituting this into the first equation: $2(y + 1) + 3y = 12$ , which simplifies to $2y + 2 + 3y = 12$ , then $5y = 10$ , so $y = 2$ . Substituting back: $x = 2 + 1 = 3$ . Therefore, $xy = 3 \times 2 = 6$ . Let's check why the wrong answers appear. Choice A ( $15$ ) likely comes from incorrectly adding the two equations and getting confused about what operation to perform. Choice B ( $9$ ) might result from solving incorrectly and getting $x = 3, y = 3$ , then calculating $3 \times 3 = 9$ . Choice C ( $12$ ) is tempting because it's the constant from the first equation—students sometimes grab familiar numbers without completing the calculation. The key strategy here is to always verify your solution by substituting both values back into both original equations. With $x = 3$ and $y = 2$ : the first equation gives $2(3) + 3(2) = 6 + 6 = 12$ ✓, and the second gives $3 - 2 = 1$ ✓. This confirmation step catches calculation errors and builds confidence in your final answer.
If $5(x-2)+7=3(x+1)-8$ , what is $x$ ?
Answer: -1 Simplify both sides: 5x - 10 + 7 = 3x + 3 - 8, so 5x - 3 = 3x - 5. Subtract 3x from both sides to get 2x - 3 = -5, then add 3 to get 2x = -2, so x = -1. A common wrong answer is 1, which comes from moving terms with the wrong sign; keeping signs correct gives -1.
If $4-3(2x-1)=13$ , what is $x$ ?
Answer: -1 Subtract 4 from both sides: -3(2x-1)=9. Divide by -3 to get 2x-1=-3, so add 1 and divide by 2: x=-1. A common wrong answer is 7, from combining 4-3 first to make 1(2x-1)=13; but you must multiply by -3 before subtracting.
When the equation $ax + 1 = 3x - 7$ has the solution $x = 4$ , the value of $a^2 - 3a$ is:
Answer: $-2$ Since $x = 4$ is a solution to $ax + 1 = 3x - 7$ , we can substitute $x = 4$ into the equation: $a(4) + 1 = 3(4) - 7$ . This gives $4a + 1 = 12 - 7 = 5$ . Therefore, $4a = 4$ , so $a = 1$ . Now we can find $a^2 - 3a = 1^2 - 3(1) = 1 - 3 = -2$ .
If $x$ satisfies $\frac{x}{4} + \frac{x}{12} = \frac{x}{6} - 4$ , then $\frac{2x}{3}$ equals:
Answer: $-16$ To solve $\frac{x}{4} + \frac{x}{12} = \frac{x}{6} - 4$ , first find a common denominator. The LCM of 4, 12, and 6 is 12. Multiply the entire equation by 12: $3x + x = 2x - 48$ , which simplifies to $4x = 2x - 48$ . Subtracting $2x$ from both sides gives $2x = -48$ , so $x = -24$ . Therefore, $\frac{2x}{3} = \frac{2(-24)}{3} = \frac{-48}{3} = -16$ .
If $4x - (x + 3) = 6 + (x - 3)$ , then $x^2 + x$ equals:
Answer: $12$ First, expand both sides of the equation $4x - (x + 3) = 6 + (x - 3)$ . Left side: $4x - x - 3 = 3x - 3$ . Right side: $6 + x - 3 = 3 + x$ . So the equation becomes $3x - 3 = 3 + x$ . Subtracting $x$ from both sides: $2x - 3 = 3$ . Adding 3 to both sides: $2x = 6$ , so $x = 3$ . Therefore, $x^2 + x = 3^2 + 3 = 9 + 3 = 12$ .
If 8 is subtracted from 3 times a number, the result is 4 more than twice the number. Find it.
Answer: 12 Let the number be x. The equation is 3x - 8 = 2x + 4. Subtract 2x from both sides to get x - 8 = 4, then add 8 to get x = 12. A common trap is reading "4 more than twice the number" as 2(x + 4), which would give 16, but it means 2x + 4.
If $\frac{2x+1}{3}=x-3$ , what is $x$ ?
Answer: 10 Multiply both sides by 3: 2x + 1 = 3x - 9. Then subtract 2x and add 9, giving x = 10. The trap is choosing 4 by writing the right side as 3x - 3 instead of 3x - 9.
If $1.2x-3.6=0.4x+1.2$ , what is $x$ ?
Answer: 6 Subtract 0.4x from both sides to get 0.8x - 3.6 = 1.2. Then add 3.6 to both sides, giving 0.8x = 4.8. Divide by 0.8, so x = 6. A tempting wrong answer is -6, which comes from reversing signs when moving terms across the equals sign.
The equation $|2x - 6| = 4x - 10$ has how many solutions?
Answer: $1$ To solve $|2x - 6| = 4x - 10$ , we need to consider two cases based on the definition of absolute value. First, note that for the equation to have real solutions, we need $4x - 10 \geq 0$ , which means $x \geq 2.5$ . Case 1: $2x - 6 \geq 0$ , i.e., $x \geq 3$ . Then $|2x - 6| = 2x - 6$ , so the equation becomes $2x - 6 = 4x - 10$ . Solving: $-6 + 10 = 4x - 2x$ , so $4 = 2x$ and $x = 2$ . But we need $x \geq 3$ for this case, and $2 < 3$ , so $x = 2$ is not valid for Case 1. Case 2: $2x - 6 < 0$ , i.e., $x < 3$ . Then $|2x - 6| = -(2x - 6) = 6 - 2x$ , so the equation becomes $6 - 2x = 4x - 10$ . Solving: $6 + 10 = 4x + 2x$ , so $16 = 6x$ and $x = \frac{8}{3} \approx 2.67$ . For this case, we need $x < 3$ , and indeed $\frac{8}{3} < 3$ . Also, we need $x \geq 2.5$ for the right side to be non-negative: $\frac{8}{3} = \frac{8}{3} \approx 2.67 > 2.5$ , so this condition is satisfied. Let's verify: $|2(\frac{8}{3}) - 6| = |\frac{16}{3} - 6| = |\frac{16 - 18}{3}| = |\frac{-2}{3}| = \frac{2}{3}$ . And $4(\frac{8}{3}) - 10 = \frac{32}{3} - 10 = \frac{32 - 30}{3} = \frac{2}{3}$ . Since both sides equal $\frac{2}{3}$ , $x = \frac{8}{3}$ is indeed a solution. Therefore, the equation has exactly one solution: $x = \frac{8}{3}$ . Choice A is wrong because we found a solution. Choice C is wrong because we only found one solution, not two. Choice D is wrong because the solution is unique, not infinitely many.

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