➕ Algebra 1 · Systems of Equations

Solving Systems by Substitution practice questions

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Answer key with explanations

1. Solve the system y = 2x + 5 and 3x + y = 15.

   Answer: (2, 9). y is already alone, so substitute (2x + 5) for y in the other equation: 3x + (2x + 5) = 15. Combine: 5x + 5 = 15. Subtract 5: 5x = 10, so x = 2. Back-substitute: y = 2(2) + 5 = 9. Check: 3(2) + 9 = 15. The pair (9, 2) swaps the coordinates. The pair (3, 11) divides without first subtracting 5. The pair (4, 13) adds 5 instead of subtracting it. Both of those fail the second equation: 3(3) + 11 = 20 and 3(4) + 13 = 25.

2. Substituting y = 2x − 7 into 4x − 3y = 9 gives which equation?

   Answer: 4x − 3(2x − 7) = 9. Replace y with (2x − 7), keeping the parentheses so the −3 multiplies both terms: 4x − 3(2x − 7) = 9. Solving it gives 4x − 6x + 21 = 9, so −2x = −12, x = 6, and y = 2(6) − 7 = 5. Check: 4(6) − 3(5) = 9. Without parentheses only 2x gets multiplied by −3, and the −7 is left alone. In 4x − (2x − 7) = 9 the 3 was dropped. In 4x − 3(2x + 7) = 9 the sign of the 7 changed.

3. Solve the system x = 2y + 1 and 3x − 4y = −1.

   Answer: (−3, −2). Substitute (2y + 1) for x in the other equation: 3(2y + 1) − 4y = −1. Distribute: 6y + 3 − 4y = −1. Combine: 2y + 3 = −1. Subtract 3: 2y = −4, so y = −2. Back-substitute: x = 2(−2) + 1 = −3. Check: 3(−3) − 4(−2) = −9 + 8 = −1. The pair (−2, −3) swaps the coordinates. The pair (3, 1) comes from adding 3 instead of subtracting it. The pair (−1, −1) comes from forgetting the parentheses, which leaves 3 · 2y + 1 − 4y = −1.

4. Solve the system 5x − 2y = 7 and −3x + y = −5.

   Answer: (3, 4). No variable is alone, but y has coefficient 1 in the second equation. Add 3x to both sides: y = 3x − 5. Substitute, with parentheses, into the first equation: 5x − 2(3x − 5) = 7. Distribute: 5x − 6x + 10 = 7. Combine: −x + 10 = 7. Subtract 10: −x = −3, so x = 3. Back-substitute: y = 3(3) − 5 = 4. Check: 5(3) − 2(4) = 7 and −3(3) + 4 = −5. The pair (4, 3) swaps the coordinates. The pair (3, 14) uses +5 instead of −5 to find y. The pair (−12, −41) forgets the parentheses: 5x − 2 · 3x − 5 = 7 gives x = −12.

5. How many solutions does the system y = 5x − 2 and 10x − 2y = 9 have?

   Answer: No solution. Substitute (5x − 2) for y: 10x − 2(5x − 2) = 9. Distribute: 10x − 10x + 4 = 9. Combine: 4 = 9. The x-terms vanished and the leftover is false, so the system has no solution. (The slopes are both 5 and the y-intercepts differ, so the lines are parallel.) The pair (0, −2) is only the y-intercept of the first line: it gives 10(0) − 2(−2) = 4, not 9. The numbers 4 and 9 in 4 = 9 are not coordinates.

6. How many solutions does the system x = 4 − 3y and 2x + 6y = 8 have?

   Answer: Infinitely many solutions. Substitute (4 − 3y) for x: 2(4 − 3y) + 6y = 8. Distribute: 8 − 6y + 6y = 8. Combine: 8 = 8. The y-terms vanished and the leftover is true for every y, so every pair (4 − 3y, y) is a solution: infinitely many solutions. (Dividing 2x + 6y = 8 by 2 gives x + 3y = 4, the same line as x = 4 − 3y.) A true leftover does not give one pair like (8, 8), and a system of two lines cannot have exactly two solutions.

7. The sum of two numbers is 18. The larger number is 6 more than twice the smaller number. What is the larger number?

   Answer: 14. Let s be the smaller number and l the larger number. The clues are s + l = 18 and l = 2s + 6. Substitute (2s + 6) for l: s + (2s + 6) = 18. Combine: 3s + 6 = 18. Subtract 6: 3s = 12, so s = 4. Back-substitute: l = 2(4) + 6 = 14. Check: 4 + 14 = 18. The number 4 is the smaller number, and 12 is just 3s. The value 22 comes from adding 6 instead of subtracting it: it would make s = 8 and s + l = 30.

8. Solve the system x = (2/3)y − 1 and 6x − 5y = −12.

   Answer: (3, 6). Substitute, with parentheses: 6((2/3)y − 1) − 5y = −12. Multiply 6 by both terms (6 · (2/3) = 4 and 6 · 1 = 6): 4y − 6 − 5y = −12. Combine: −y − 6 = −12. Add 6: −y = −6, so y = 6. Back-substitute: x = (2/3)(6) − 1 = 4 − 1 = 3. Check: 3 = (2/3)(6) − 1 and 6(3) − 5(6) = −12. The pair (6, 3) swaps the coordinates. The pair (19/3, 11) forgets to multiply the −1 by 6, giving 4y − 1 − 5y = −12. The pair (5, 6) uses +1 instead of −1 to find x.