➕ Algebra 1 · Systems of Equations

Solving Systems by Elimination practice questions

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Answer key with explanations

1. Solve by elimination: x + y = 17 and x − y = 5.

   Answer: (11, 6). The y-coefficients are opposites, so add: 2x = 22, so x = 11. Back-substitute: 11 + y = 17, so y = 6. Check: 11 + 6 = 17 and 11 − 6 = 5. The pair (6, 11) is the right numbers in the wrong order, (22, −5) forgets to divide by 2, and (12, 5) satisfies only the first equation.

2. Solve by elimination: 4x + 3y = 29 and 6x + 3y = 33.

   Answer: (2, 7). The y-coefficients are equal (both 3), so subtract the first equation from the second: (6x + 3y) − (4x + 3y) = 33 − 29, so 2x = 4 and x = 2. Back-substitute into the first equation: 4(2) + 3y = 29, so 3y = 21 and y = 7. Check the second equation: 6(2) + 3(7) = 33. The pair (2, 21) forgets to divide by 3 at the end, and (4, 13/3) forgets to divide by 2.

3. Solve by elimination: 2x + 5y = 4 and 2x − 3y = −12.

   Answer: (−3, 2). The x-coefficients are equal, so subtract the second equation from the first, and let the minus reach every term: 2x + 5y − 2x + 3y = 4 + 12. So 8y = 16 and y = 2. Back-substitute: 2x + 5(2) = 4, so 2x = −6 and x = −3. Check the second equation: 2(−3) − 3(2) = −12. The pair (−18, 8) comes from writing 5y − 3y = 2y (the minus did not flip the −3y), (12, −4) also writes 4 − 12 for the right side instead of 4 + 12, and (2, −3) has the numbers in the wrong order.

4. To solve 3x + 4y = −1 and 5x − 2y = 33 by elimination, you multiply the second equation by 2 so that the y-terms are opposites. What is the new second equation?

   Answer: 10x − 4y = 66. Multiply every term by 2, including the constant: 2(5x) = 10x, 2(−2y) = −4y and 2(33) = 66. Now the y-terms are +4y and −4y, so adding the equations gives 13x = 65, so x = 5, and then y = −4. The other choices leave the constant alone, leave the y-term alone, or add 2 instead of multiplying by 2.

5. Solve by elimination: x + y = 20 and 3x = y + 12.

   Answer: (8, 12). The second equation is not in standard form. Subtract y from both sides: 3x − y = 12. The y-coefficients are now 1 and −1, so add: 4x = 32 and x = 8. Back-substitute: 8 + y = 20, so y = 12. Check the second original equation: 3(8) = 24 and 12 + 12 = 24. The pair (−4, 24) comes from writing 3x + y = 12 (forgetting to change the sign when moving y), and (4, 16) and (12, 8) satisfy only the first equation.

6. How many solutions does the system 6x − 9y = 12 and 4x − 6y = 7 have?

   Answer: No solution. Multiply the first equation by 2 and the second by 3 (every term): 12x − 18y = 24 and 12x − 18y = 21. Subtract: 0 = 3, which is false. Both variables vanished and what is left is false, so there is no solution (the lines are parallel). The point (2, 0) makes only the first equation true, and (7/4, 0) makes only the second equation true.

7. Solve by elimination: 3x + 2y = −8 and 5x + 3y = −14.

   Answer: (−4, 2). Match the y-terms: multiply the first equation by 3 and the second by 2, every term each time. That gives 9x + 6y = −24 and 10x + 6y = −28. Subtract the first from the second: x = −4. Back-substitute into the first original equation: 3(−4) + 2y = −8, so 2y = 4 and y = 2. Check the second: 5(−4) + 3(2) = −14. The pair (−6, 5) comes from forgetting to multiply the constants, and (2, −4) has the numbers in the wrong order. The pair (4, −10) comes from subtracting the right sides in the opposite order (−24 − (−28) = 4), which gives x = 4 instead of −4. Then 3(4) + 2y = −8 gives y = −10.

8. For which value of k does the system x + 4y = 3 and 3x + 12y = k have infinitely many solutions?

   Answer: k = 9. Multiply the first equation by 3 (every term, including the 3 on the right): 3x + 12y = 9. Subtract it from the second equation: 0 = k − 9. If k = 9 the leftover 0 = 0 is true, so both equations describe the same line and there are infinitely many solutions. For any other k the leftover is false and there is no solution. The choice k = 3 forgets to multiply the constant.

9. Solve by elimination: 2x + 3y = 1 and 4x − 3y = 5.

   Answer: (1, −1/3). The y-coefficients are opposites, so add: 6x = 6 and x = 1. Back-substitute into the first equation: 2(1) + 3y = 1, so 3y = −1 and y = −1/3. Check the second equation: 4(1) − 3(−1/3) = 4 + 1 = 5. The pair (1, 1/3) has a sign error, (1, −3) multiplies by 3 instead of dividing by 3, and (6, −11/3) forgets to divide by 6 at the first step.