SSAT Upper Level Quantitative Quiz: One And Multi Step Equations
10 questions · exam conditions
0:00
One And Multi Step EquationsQuestion 1 of 10

If 8 is subtracted from 3 times a number, the result is 4 more than twice the number. Find it.

4
16
12
28
← Back to quizzes

SSAT Upper Level Quantitative Quiz

SSAT Upper Level Quantitative Quiz: One And Multi Step Equations

Practice One And Multi Step Equations in SSAT Upper Level Quantitative 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 And Multi Step Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level Quantitative.

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

If 8 is subtracted from 3 times a number, the result is 4 more than twice the number. Find it.

  1. 4
  2. 16
  3. 12 (correct answer)
  4. 28
Explanation: 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.

Question 2

If 5(x−2)+7=3(x+1)−85(x-2)+7=3(x+1)-8, what is xx?

  1. -1 (correct answer)
  2. 1
  3. -5
  4. -2
Explanation: 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.

Question 3

If 1.2x−3.6=0.4x+1.21.2x-3.6=0.4x+1.2, what is xx?

  1. -6
  2. 4
  3. 8
  4. 6 (correct answer)
Explanation: 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.

Question 4

If 2x+13=x−3\frac{2x+1}{3}=x-3, what is xx?

  1. 4
  2. -8
  3. -4
  4. 10 (correct answer)
Explanation: 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.

Question 5

If 4−3(2x−1)=134-3(2x-1)=13, what is xx?

  1. -2
  2. -1 (correct answer)
  3. 1
  4. 7
Explanation: 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.

Question 6

When the equation ax+1=3x−7ax + 1 = 3x - 7 has the solution x=4x = 4, the value of a2−3aa^2 - 3a is:

  1. −2-2 (correct answer)
  2. 00
  3. 22
  4. 1010
Explanation: Since x=4x = 4 is a solution to ax+1=3x−7ax + 1 = 3x - 7, we can substitute x=4x = 4 into the equation: a(4)+1=3(4)−7a(4) + 1 = 3(4) - 7. This gives 4a+1=12−7=54a + 1 = 12 - 7 = 5. Therefore, 4a=44a = 4, so a=1a = 1. Now we can find a2−3a=12−3(1)=1−3=−2a^2 - 3a = 1^2 - 3(1) = 1 - 3 = -2.

Question 7

If 4x−(x+3)=6+(x−3)4x - (x + 3) = 6 + (x - 3), then x2+xx^2 + x equals:

  1. 22
  2. 66
  3. 1212 (correct answer)
  4. 2020
Explanation: First, expand both sides of the equation 4x−(x+3)=6+(x−3)4x - (x + 3) = 6 + (x - 3). Left side: 4x−x−3=3x−34x - x - 3 = 3x - 3. Right side: 6+x−3=3+x6 + x - 3 = 3 + x. So the equation becomes 3x−3=3+x3x - 3 = 3 + x. Subtracting xx from both sides: 2x−3=32x - 3 = 3. Adding 3 to both sides: 2x=62x = 6, so x=3x = 3. Therefore, x2+x=32+3=9+3=12x^2 + x = 3^2 + 3 = 9 + 3 = 12.

Question 8

The equation ∣2x−6∣=4x−10|2x - 6| = 4x - 10 has how many solutions?

  1. 00
  2. 11 (correct answer)
  3. 22
  4. infinitely many
Explanation: To solve ∣2x−6∣=4x−10|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≥04x - 10 \geq 0, which means x≥2.5x \geq 2.5. Case 1: 2x−6≥02x - 6 \geq 0, i.e., x≥3x \geq 3. Then ∣2x−6∣=2x−6|2x - 6| = 2x - 6, so the equation becomes 2x−6=4x−102x - 6 = 4x - 10. Solving: −6+10=4x−2x-6 + 10 = 4x - 2x, so 4=2x4 = 2x and x=2x = 2. But we need x≥3x \geq 3 for this case, and 2<32 < 3, so x=2x = 2 is not valid for Case 1. Case 2: 2x−6<02x - 6 < 0, i.e., x<3x < 3. Then ∣2x−6∣=−(2x−6)=6−2x|2x - 6| = -(2x - 6) = 6 - 2x, so the equation becomes 6−2x=4x−106 - 2x = 4x - 10. Solving: 6+10=4x+2x6 + 10 = 4x + 2x, so 16=6x16 = 6x and x=83≈2.67x = \frac{8}{3} \approx 2.67. For this case, we need x<3x < 3, and indeed 83<3\frac{8}{3} < 3. Also, we need x≥2.5x \geq 2.5 for the right side to be non-negative: 83=83≈2.67>2.5\frac{8}{3} = \frac{8}{3} \approx 2.67 > 2.5, so this condition is satisfied. Let's verify: ∣2(83)−6∣=∣163−6∣=∣16−183∣=∣−23∣=23|2(\frac{8}{3}) - 6| = |\frac{16}{3} - 6| = |\frac{16 - 18}{3}| = |\frac{-2}{3}| = \frac{2}{3}. And 4(83)−10=323−10=32−303=234(\frac{8}{3}) - 10 = \frac{32}{3} - 10 = \frac{32 - 30}{3} = \frac{2}{3}. Since both sides equal 23\frac{2}{3}, x=83x = \frac{8}{3} is indeed a solution. Therefore, the equation has exactly one solution: x=83x = \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.

Question 9

If xx satisfies x4+x12=x6−4\frac{x}{4} + \frac{x}{12} = \frac{x}{6} - 4, then 2x3\frac{2x}{3} equals:

  1. −16-16 (correct answer)
  2. −8-8
  3. 88
  4. 1212
Explanation: To solve x4+x12=x6−4\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−483x + x = 2x - 48, which simplifies to 4x=2x−484x = 2x - 48. Subtracting 2x2x from both sides gives 2x=−482x = -48, so x=−24x = -24. Therefore, 2x3=2(−24)3=−483=−16\frac{2x}{3} = \frac{2(-24)}{3} = \frac{-48}{3} = -16.

Question 10

If xx and yy satisfy the system {2x+3y=12x−y=1\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}, then xyxy equals:

  1. 1515
  2. 99
  3. 1212
  4. 66 (correct answer)
Explanation: 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=1x - y = 1, you get x=y+1x = y + 1. Substituting this into the first equation: 2(y+1)+3y=122(y + 1) + 3y = 12, which simplifies to 2y+2+3y=122y + 2 + 3y = 12, then 5y=105y = 10, so y=2y = 2. Substituting back: x=2+1=3x = 2 + 1 = 3. Therefore, xy=3×2=6xy = 3 \times 2 = 6. Let's check why the wrong answers appear. Choice A (1515) likely comes from incorrectly adding the two equations and getting confused about what operation to perform. Choice B (99) might result from solving incorrectly and getting x=3,y=3x = 3, y = 3, then calculating 3×3=93 \times 3 = 9. Choice C (1212) 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=3x = 3 and y=2y = 2: the first equation gives 2(3)+3(2)=6+6=122(3) + 3(2) = 6 + 6 = 12 ✓, and the second gives 3−2=13 - 2 = 1 ✓. This confirmation step catches calculation errors and builds confidence in your final answer.