Math 1 Quiz: Solving Systems By Substitution
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Solving Systems By SubstitutionQuestion 1 of 20

Consider the system 2x−3y=62x - 3y = 6 and y=x−4y = x - 4. When solved by substitution, what is the solution?

(12,8)(12, 8)
(8,4)(8, 4)
(9,5)(9, 5)
(6,2)(6, 2)
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Math 1 Quiz

Math 1 Quiz: Solving Systems By Substitution

Practice Solving Systems By Substitution in Math 1 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 Systems By Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

Consider the system 2x−3y=62x - 3y = 6 and y=x−4y = x - 4. When solved by substitution, what is the solution?

  1. (12,8)(12, 8)
  2. (8,4)(8, 4)
  3. (9,5)(9, 5)
  4. (6,2)(6, 2) (correct answer)
Explanation: When you encounter a system of linear equations like this, substitution is often the most efficient method when one equation is already solved for a variable. Here, the second equation gives you y=x−4y = x - 4, making it perfect for substitution. To solve by substitution, replace yy in the first equation with the expression from the second equation. Substituting y=x−4y = x - 4 into 2x−3y=62x - 3y = 6 gives you: 2x−3(x−4)=62x - 3(x - 4) = 6. Distribute the −3-3: 2x−3x+12=62x - 3x + 12 = 6. Combine like terms: −x+12=6-x + 12 = 6. Solving for xx: −x=−6-x = -6, so x=6x = 6. Now substitute back into y=x−4y = x - 4 to get y=6−4=2y = 6 - 4 = 2. The solution is (6,2)(6, 2). Let's check why the other answers are incorrect. Choice A) (12,8)(12, 8) would result from errors in distributing or combining like terms—perhaps forgetting the negative sign when distributing −3-3. Choice B) (8,4)(8, 4) might come from solving −x=−6-x = -6 incorrectly or making arithmetic mistakes in the substitution step. Choice C) (9,5)(9, 5) could result from sign errors or incorrectly setting up the substitution equation. Always verify your solution by substituting both values back into both original equations. For (6,2)(6, 2): 2(6)−3(2)=12−6=62(6) - 3(2) = 12 - 6 = 6 ✓ and 2=6−42 = 6 - 4 ✓. When solving systems by substitution, work methodically through each step and double-check your arithmetic—small errors compound quickly in multi-step problems.

Question 2

A system has equations y=2x−5y = 2x - 5 and 3x+4y=223x + 4y = 22. When the substitution method is used, what equation results after substitution but before simplification?

  1. y+4(2x−5)=22y + 4(2x - 5) = 22
  2. 3x+2x−5=223x + 2x - 5 = 22
  3. 3x+4(2x−5)=223x + 4(2x - 5) = 22 (correct answer)
  4. 3x+8x−20=223x + 8x - 20 = 22
Explanation: When you encounter a system of equations problem asking about substitution "before simplification," you need to identify the exact moment after substitution occurs but before any algebraic simplification takes place. The substitution method involves taking one equation solved for a variable and substituting that expression into the other equation. Here, the first equation y=2x−5y = 2x - 5 is already solved for yy. To use substitution, you take this expression for yy and replace every yy in the second equation 3x+4y=223x + 4y = 22. Substituting y=2x−5y = 2x - 5 into 3x+4y=223x + 4y = 22 gives you 3x+4(2x−5)=223x + 4(2x - 5) = 22. This is the equation immediately after substitution but before any simplification—you haven't distributed the 4 or combined like terms yet. Choice A, y+4(2x−5)=22y + 4(2x - 5) = 22, incorrectly keeps the yy term instead of replacing it with 3x3x. Choice B, 3x+2x−5=223x + 2x - 5 = 22, shows partial simplification where someone incorrectly multiplied 4×2x4 \times 2x but forgot to multiply 4×(−5)4 \times (-5). Choice D, 3x+8x−20=223x + 8x - 20 = 22, shows the equation after complete simplification (distributing the 4), which goes beyond what the question asks for. The correct answer is C because it shows the exact state after substitution but before any distributive property or combining of terms. Study tip: When a question asks for results "before simplification," stop immediately after the substitution step—don't distribute or combine terms.

Question 3

A student attempts to solve the system 4x+y=104x + y = 10 and x−2y=3x - 2y = 3 by substitution. From the second equation, they solve for xx and get x=3+2yx = 3 + 2y. When they substitute this into the first equation, what coefficient will yy have after simplification?

  1. 77
  2. 88
  3. 99 (correct answer)
  4. 1111
Explanation: Substituting x=3+2yx = 3 + 2y into 4x+y=104x + y = 10: 4(3+2y)+y=104(3 + 2y) + y = 10. Expanding: 12+8y+y=1012 + 8y + y = 10, which simplifies to 12+9y=1012 + 9y = 10. The coefficient of yy is 9. Choice A would result from 4(2y)−y4(2y) - y, choice B from forgetting to add the yy term, and choice D from incorrectly expanding 4(3+2y)4(3 + 2y) as 12+11y12 + 11y.

Question 4

When solving the system 5x−2y=45x - 2y = 4 and y=3x+1y = 3x + 1 by substitution, a student gets the equation 5x−2(3x+1)=45x - 2(3x + 1) = 4. After correctly expanding and simplifying, what equation should they solve next?

  1. −x−2=4-x - 2 = 4 (correct answer)
  2. −x+2=4-x + 2 = 4
  3. 11x−2=411x - 2 = 4
  4. x−2=4x - 2 = 4
Explanation: Expanding 5x−2(3x+1)=45x - 2(3x + 1) = 4: 5x−6x−2=45x - 6x - 2 = 4, which simplifies to −x−2=4-x - 2 = 4. Choice B has the wrong sign on the constant term, choice C incorrectly adds instead of subtracting the xx terms (5x+6x5x + 6x), and choice D forgets the negative sign when combining 5x−6x5x - 6x.

Question 5

A system of equations ax+by=cax + by = c and x=dy+ex = dy + e is solved by substitution. After substituting the second equation into the first and collecting like terms, the resulting equation has the form (ad+b)y+ae=c(ad + b)y + ae = c. For the system 6x+7y=506x + 7y = 50 and x=2y+3x = 2y + 3, what is the coefficient of yy in the resulting equation?

  1. 1717
  2. 1818
  3. 1919 (correct answer)
  4. 2020
Explanation: Using the formula (ad+b)y+ae=c(ad + b)y + ae = c with a=6a = 6, b=7b = 7, d=2d = 2, and e=3e = 3: the coefficient of yy is ad+b=6(2)+7=12+7=19ad + b = 6(2) + 7 = 12 + 7 = 19. Choice A represents ad+b−2ad + b - 2, choice B represents ad+b−1ad + b - 1, and choice D represents ad+b+1ad + b + 1. We can verify by direct substitution: 6(2y+3)+7y=506(2y + 3) + 7y = 50 becomes 12y+18+7y=5012y + 18 + 7y = 50, or 19y+18=5019y + 18 = 50.

Question 6

The system 3x+4y=263x + 4y = 26 and x−y=1x - y = 1 is to be solved by substitution. If you solve the second equation for xx first, what expression would you substitute into the first equation?

  1. x+1x + 1
  2. y−1y - 1
  3. 1−y1 - y
  4. y+1y + 1 (correct answer)
Explanation: When solving a system of equations by substitution, you need to isolate one variable in terms of the other, then substitute that expression into the remaining equation. Starting with the second equation x−y=1x - y = 1, you solve for xx by adding yy to both sides: x=y+1x = y + 1. This expression for xx is what you'll substitute into the first equation 3x+4y=263x + 4y = 26, replacing every xx with (y+1)(y + 1). The correct answer is D) y+1y + 1 because this is exactly the expression we derived when solving x−y=1x - y = 1 for xx. Let's see why the other choices are wrong. Choice A) x+1x + 1 doesn't make sense because you're trying to eliminate xx, not create an expression that still contains it. Choice B) y−1y - 1 represents a sign error - if you incorrectly subtracted yy from both sides instead of adding it, you'd get x=1−yx = 1 - y, but that's not this expression. Choice C) 1−y1 - y is what you'd get if you solved for xx incorrectly by moving the yy term to the right side without changing its sign. Remember that substitution requires you to express one variable completely in terms of the other. Always double-check your algebra when isolating variables, and make sure your final expression contains only the variable you're keeping, plus any constants.

Question 7

A system of equations is given by 2x+3y=112x + 3y = 11 and x=4y−5x = 4y - 5. When solving this system by substitution, Maria substituted the expression for xx into the first equation and obtained 2(4y−5)+3y=112(4y - 5) + 3y = 11. After simplifying, what equation did she get before solving for yy?

  1. 11y−10=1111y - 10 = 11 (correct answer)
  2. 8y−10+3y=118y - 10 + 3y = 11
  3. 11y+10=1111y + 10 = 11
  4. 5y−10=115y - 10 = 11
Explanation: After substituting x=4y−5x = 4y - 5 into 2x+3y=112x + 3y = 11, Maria gets 2(4y−5)+3y=112(4y - 5) + 3y = 11. Distributing: 8y−10+3y=118y - 10 + 3y = 11. Combining like terms: 11y−10=1111y - 10 = 11. Choice B shows the equation before combining like terms, choice C has the wrong sign on 10, and choice D incorrectly combines 8y+3y=5y8y + 3y = 5y.

Question 8

A system of equations is given by 3x−2y=73x - 2y = 7 and y=mx+4y = mx + 4. After substitution, the resulting equation in xx is 3x−2(mx+4)=73x - 2(mx + 4) = 7. If this simplifies to −5x−8=7-5x - 8 = 7, what is the value of mm?

  1. m=−4m = -4
  2. m=−3m = -3
  3. m=4m = 4 (correct answer)
  4. m=3m = 3
Explanation: Expanding 3x−2(mx+4)=73x - 2(mx + 4) = 7 gives 3x−2mx−8=73x - 2mx - 8 = 7, which rearranges to (3−2m)x−8=7(3 - 2m)x - 8 = 7. For this to equal −5x−8=7-5x - 8 = 7, we need 3−2m=−53 - 2m = -5. Solving: −2m=−8-2m = -8, so m=4m = 4. Choice A results from solving 3−2m=−53 - 2m = -5 incorrectly as m=−4m = -4. Choice B results from setting 3+2m=−53 + 2m = -5. Choice D results from solving 2m−3=52m - 3 = 5.

Question 9

The system {2x+y=84x−3y=6\begin{cases} 2x + y = 8 \\ 4x - 3y = 6 \end{cases} is being solved by substitution. From the first equation, y=8−2xy = 8 - 2x. After substituting this into the second equation and solving, which ordered pair represents the solution?

  1. (3,2)(3, 2) (correct answer)
  2. (2,4)(2, 4)
  3. (4,0)(4, 0)
  4. (1,6)(1, 6)
Explanation: Substituting y=8−2xy = 8 - 2x into 4x−3y=64x - 3y = 6: 4x−3(8−2x)=64x - 3(8 - 2x) = 6, which gives 4x−24+6x=64x - 24 + 6x = 6, so 10x=3010x = 30 and x=3x = 3. Then y=8−2(3)=2y = 8 - 2(3) = 2. The solution is (3,2)(3, 2). Choice B results from solving 10x=2010x = 20 instead of 10x=3010x = 30. Choice C results from setting y=0y = 0 and solving only the first equation. Choice D results from an arithmetic error giving x=1x = 1.

Question 10

The system {2x−y=56x−3y=12\begin{cases} 2x - y = 5 \\ 6x - 3y = 12 \end{cases} is being solved by substitution. From the first equation, y=2x−5y = 2x - 5. After substituting into the second equation, what conclusion can be drawn?

  1. The system has infinitely many solutions since both equations are equivalent
  2. The system has no solution since substitution leads to 0=−30 = -3 (correct answer)
  3. The system has one solution at (x,y)=(2.5,0)(x, y) = (2.5, 0)
  4. The system has no solution since the equations represent parallel lines
Explanation: Substituting y=2x−5y = 2x - 5 into 6x−3y=126x - 3y = 12: 6x−3(2x−5)=126x - 3(2x - 5) = 12, which gives 6x−6x+15=126x - 6x + 15 = 12, simplifying to 15=1215 = 12 or 0=−30 = -3. This contradiction means no solution exists. Choice A is wrong because the equations aren't equivalent. Choice C assumes a solution exists. Choice D correctly identifies no solution but gives an incomplete geometric reason.

Question 11

The system {4x+y=152x−3y=−7\begin{cases} 4x + y = 15 \\ 2x - 3y = -7 \end{cases} has solution (x,y)=(a,b)(x, y) = (a, b). If instead we solve the equivalent system {y=15−4x2x−3y=−7\begin{cases} y = 15 - 4x \\ 2x - 3y = -7 \end{cases} by substitution, what equation in xx results?

  1. 14x−45=−714x - 45 = -7
  2. 4x+(15−4x)=−74x + (15 - 4x) = -7
  3. 2x−45+12x=−72x - 45 + 12x = -7
  4. 2x−3(15−4x)=−72x - 3(15 - 4x) = -7 (correct answer)
Explanation: When solving systems of equations by substitution, you're replacing one variable with an expression from another equation. This method requires careful attention to which equation you substitute into. The key insight here is understanding what "substitution" means. You start with the equivalent system where the first equation is already solved for yy: y=15−4xy = 15 - 4x. To use substitution, you take this expression for yy and substitute it into the second equation wherever you see yy. The second equation is 2x−3y=−72x - 3y = -7. When you substitute y=15−4xy = 15 - 4x into this equation, you replace the yy with (15−4x)(15 - 4x), giving you: 2x−3(15−4x)=−72x - 3(15 - 4x) = -7. This matches answer choice D. Let's see why the other options are wrong. Choice A (14x−45=−714x - 45 = -7) shows what you'd get after expanding and simplifying the substitution, but the question asks for the equation that "results" from substitution, meaning the immediate result before simplification. Choice B (4x+(15−4x)=−74x + (15 - 4x) = -7) incorrectly substitutes into the first equation instead of the second. Choice C (2x−45+12x=−72x - 45 + 12x = -7) appears to be a partially expanded version with sign errors. Study tip: In substitution problems, always substitute the solved expression into the equation that still contains that variable. Double-check that you're using the correct equation and maintaining proper parentheses around the substituted expression.

Question 12

Two equations are given: x−2y=8x - 2y = 8 and 3x+ky=63x + ky = 6, where kk is unknown. When solved by substitution using x=2y+8x = 2y + 8, the second equation becomes 3(2y+8)+ky=63(2y + 8) + ky = 6. If this simplifies to (6+k)y=−18(6 + k)y = -18, and the system has the solution y=−2y = -2, what is the value of kk?

  1. k=−3k = -3
  2. k=6k = 6
  3. k=9k = 9
  4. k=3k = 3 (correct answer)
Explanation: This problem tests your understanding of systems of linear equations and how substitution works to find unknown parameters. When you substitute x=2y+8x = 2y + 8 into the second equation 3x+ky=63x + ky = 6, you get 3(2y+8)+ky=63(2y + 8) + ky = 6. Expanding this: 6y+24+ky=66y + 24 + ky = 6, which simplifies to (6+k)y=6−24=−18(6 + k)y = 6 - 24 = -18. So the simplified form is (6+k)y=−18(6 + k)y = -18. Since the system has the solution y=−2y = -2, you can substitute this value: (6+k)(−2)=−18(6 + k)(-2) = -18. Solving for kk: −2(6+k)=−18-2(6 + k) = -18, so −12−2k=−18-12 - 2k = -18. Adding 12 to both sides: −2k=−6-2k = -6, therefore k=3k = 3. Let's examine why the other answers are incorrect. Choice A (k=−3k = -3) would give you (6+(−3))y=3y=−18(6 + (-3))y = 3y = -18, making y=−6y = -6, not y=−2y = -2. Choice B (k=6k = 6) would result in (6+6)y=12y=−18(6 + 6)y = 12y = -18, so y=−32y = -\frac{3}{2}, which doesn't match. Choice C (k=9k = 9) gives (6+9)y=15y=−18(6 + 9)y = 15y = -18, making y=−65y = -\frac{6}{5}, also incorrect. When working with systems involving unknown parameters, always use the given solution values to create equations you can solve for those parameters. The key is recognizing that if you know what yy equals, you can work backwards to find any unknowns in your simplified equation.

Question 13

Marcus is solving the system {x+2y=103x+y=11\begin{cases} x + 2y = 10 \\ 3x + y = 11 \end{cases} by substitution. He isolates xx from the first equation and substitutes into the second equation, obtaining 3(10−2y)+y=113(10 - 2y) + y = 11. What is his next step to find yy?

  1. Simplify to get 30−6y+y=1130 - 6y + y = 11, then solve −5y=−19-5y = -19 (correct answer)
  2. Simplify to get 30−5y=1130 - 5y = 11, then solve −5y=−19-5y = -19
  3. Simplify to get 13−6y+y=1113 - 6y + y = 11, then solve −5y=−2-5y = -2
  4. Simplify to get 30−6y+y=1130 - 6y + y = 11, then solve 5y=195y = 19
Explanation: From x+2y=10x + 2y = 10, Marcus gets x=10−2yx = 10 - 2y. Substituting: 3(10−2y)+y=113(10 - 2y) + y = 11 becomes 30−6y+y=1130 - 6y + y = 11, which simplifies to 30−5y=1130 - 5y = 11, giving −5y=−19-5y = -19. Choice B incorrectly combines −6y+y-6y + y as −5y-5y in the expansion step. Choice C uses 13 instead of 30 when distributing 3. Choice D has the wrong sign when moving terms.

Question 14

The system x+3y=8x + 3y = 8 and 2x−y=22x - y = 2 is solved by substitution. What is the solution?

  1. (2,2)(2, 2) (correct answer)
  2. (1,3)(1, 3)
  3. (3,1)(3, 1)
  4. (4,0)(4, 0)
Explanation: From x+3y=8x + 3y = 8, we get x=8−3yx = 8 - 3y. Substituting into 2x−y=22x - y = 2: 2(8−3y)−y=22(8 - 3y) - y = 2. Expanding: 16−6y−y=216 - 6y - y = 2, so 16−7y=216 - 7y = 2, giving 7y=147y = 14, thus y=2y = 2. Then x=8−3(2)=2x = 8 - 3(2) = 2. Choice B gives x+3y=10x + 3y = 10, choice C gives 2x−y=52x - y = 5, and choice D gives x+3y=4x + 3y = 4.

Question 15

The system 3x−2y=03x - 2y = 0 and y=2x−1y = 2x - 1 is solved by substitution. What is the value of x+yx + y?

  1. 44
  2. 55 (correct answer)
  3. 66
  4. 77
Explanation: Substituting y=2x−1y = 2x - 1 into 3x−2y=03x - 2y = 0: 3x−2(2x−1)=03x - 2(2x - 1) = 0. Expanding: 3x−4x+2=03x - 4x + 2 = 0, so −x+2=0-x + 2 = 0, giving x=2x = 2. Then y=2(2)−1=3y = 2(2) - 1 = 3. Therefore x+y=2+3=5x + y = 2 + 3 = 5. Choice A represents xyxy, choice C represents 3x3x, and choice D represents 2x+12x + 1.

Question 16

A linear system has the property that when solved by substitution, one of the variables is eliminated completely, resulting in the equation 0=00 = 0. This means the system has:

  1. exactly one solution that can be expressed as an ordered pair
  2. no solution because the equations contradict each other
  3. infinitely many solutions because the equations represent the same line (correct answer)
  4. exactly two solutions because each equation represents a different line
Explanation: When substitution leads to 0=00 = 0, this is always true, indicating that the original equations are equivalent and represent the same line. Therefore, every point on that line is a solution, giving infinitely many solutions. Choice A describes a typical system with one solution, choice B describes an inconsistent system (which would give 0=k0 = k where k≠0k ≠ 0), and choice D is impossible since two different lines can intersect at most once.

Question 17

Consider the system 2x+5y=252x + 5y = 25 and x=y+2x = y + 2. When this system is solved by substitution, what is the value of yy?

  1. 11
  2. 22
  3. 33 (correct answer)
  4. 44
Explanation: Substituting x=y+2x = y + 2 into 2x+5y=252x + 5y = 25: 2(y+2)+5y=252(y + 2) + 5y = 25. Expanding: 2y+4+5y=252y + 4 + 5y = 25, so 7y+4=257y + 4 = 25, giving 7y=217y = 21, thus y=3y = 3. Choice A would result from solving 7y=77y = 7, choice B from 7y=147y = 14, and choice D from 7y=287y = 28.

Question 18

The system 10x+3y=2910x + 3y = 29 and x+y=7x + y = 7 is solved by substitution. If you solve for yy in the second equation first, what simplified equation do you get after substitution?

  1. x+21=29x + 21 = 29
  2. 7x+21=297x + 21 = 29 (correct answer)
  3. 13x+21=2913x + 21 = 29
  4. 10x+21=2910x + 21 = 29
Explanation: From x+y=7x + y = 7, we get y=7−xy = 7 - x. Substituting into 10x+3y=2910x + 3y = 29: 10x+3(7−x)=2910x + 3(7 - x) = 29. Expanding: 10x+21−3x=2910x + 21 - 3x = 29, which simplifies to 7x+21=297x + 21 = 29. Choice A would come from 4x+3y=294x + 3y = 29, choice C would come from 16x+3y=2916x + 3y = 29, and choice D shows the equation before combining like terms.

Question 19

A student solving {y=x+32x−y=−1\begin{cases} y = x + 3 \\ 2x - y = -1 \end{cases} by substitution writes: "2x−(x+3)=−12x - (x + 3) = -1, so 2x−x+3=−12x - x + 3 = -1, giving x=−4x = -4." What error did the student make?

  1. Failed to distribute the negative sign when substituting y=x+3y = x + 3 (correct answer)
  2. Incorrectly isolated yy from the first equation as y=x+3y = x + 3
  3. Made an arithmetic error when solving x+3=−1x + 3 = -1 for xx
  4. Substituted into the wrong equation in the system of equations
Explanation: The student wrote 2x−(x+3)=−12x - (x + 3) = -1 but then incorrectly simplified it as 2x−x+3=−12x - x + 3 = -1. The correct simplification should be 2x−x−3=−12x - x - 3 = -1, giving x−3=−1x - 3 = -1 and x=2x = 2. The student failed to distribute the negative sign. Choice B is wrong because the isolation was correct. Choice C is wrong because that's not the arithmetic error made. Choice D is wrong because the substitution was into the correct equation.

Question 20

Consider the system where the first equation is x=3y+1x = 3y + 1 and the second equation becomes 2(3y+1)+5y=232(3y + 1) + 5y = 23 after substitution. What was the original second equation before substitution?

  1. 6y+2+5y=236y + 2 + 5y = 23
  2. 2x+5y=232x + 5y = 23 (correct answer)
  3. 2x+5x=232x + 5x = 23
  4. x+5y=23x + 5y = 23
Explanation: When you encounter a substitution problem, you're working backwards from the result to find the original equation. The key is understanding that substitution means replacing one variable with an equivalent expression from another equation. You're given that x=3y+1x = 3y + 1 and that after substitution, the second equation becomes 2(3y+1)+5y=232(3y + 1) + 5y = 23. To find the original equation, you need to reverse the substitution process. Since x=3y+1x = 3y + 1, wherever you see (3y+1)(3y + 1) in the substituted equation, that's where xx originally appeared. Looking at 2(3y+1)+5y=232(3y + 1) + 5y = 23, you can see that (3y+1)(3y + 1) was substituted for xx. Replacing it back gives you 2x+5y=232x + 5y = 23, which is answer B. Answer A (6y+2+5y=236y + 2 + 5y = 23) shows what you'd get if you distributed the 2 in the substituted equation, but this isn't the original form before substitution occurred. Answer C (2x+5x=232x + 5x = 23) incorrectly suggests that both terms contained xx, but the 5y5y term remained unchanged during substitution. Answer D (x+5y=23x + 5y = 23) is missing the coefficient 2 that appears in front of the substituted expression. Remember: when reversing substitution, look for the complete expression that was replaced (including any coefficients in front of it) and substitute the original variable back in its place.