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Solving Systems of Equations by Substitution — Easy Worksheet 2
Grade 8 · Whoobee Name: ____________________ Date: __________
Solve each system by substitution. Write the solution as an ordered pair (x, y).
How to solve
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Example
Solve the system.
| x + 3y = 25 | |
| y = 2x - 1 |
Answer: (4, 7)
| 4x + 4y = 8 | |
| y = 2x - 1 |
Not quite. The answer is
💡 The equation y = 2x − 1 already says what y is. Substitute 2x − 1 for y in 4x + 4y = 8: 4x + 4(2x − 1) = 8. Distribute: 4x + 8x − 4 = 8. Combine like terms: 12x − 4 = 8. Add 4 to both sides: 12x = 8 + 4 = 12. Divide both sides by 12: x = 12 ÷ 12 = 1. Substitute back: put x = 1 into y = 2x − 1, so y = 2(1) − 1 = 2 − 1 = 1. Check in both equations: 2(1) − 1 = 2 − 1 = 1, which is y, and 4(1) + 4(1) = 4 + 4 = 8. Both are true, so the solution is (1, 1).
✓ Correct!
Not quite. The answer is
💡 The equation y = 2x − 1 already says what y is. Substitute 2x − 1 for y in 4x + 4y = 8: 4x + 4(2x − 1) = 8. Distribute: 4x + 8x − 4 = 8. Combine like terms: 12x − 4 = 8. Add 4 to both sides: 12x = 8 + 4 = 12. Divide both sides by 12: x = 12 ÷ 12 = 1. Substitute back: put x = 1 into y = 2x − 1, so y = 2(1) − 1 = 2 − 1 = 1. Check in both equations: 2(1) − 1 = 2 − 1 = 1, which is y, and 4(1) + 4(1) = 4 + 4 = 8. Both are true, so the solution is (1, 1).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| y = x - 7 | |
| 2x + 3y = -6 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| y = 2x + 11 | |
| x + 2y = -3 |
Not quite. The answer is
💡 The equation y = 2x + 11 already says what y is. Substitute 2x + 11 for y in x + 2y = −3: x + 2(2x + 11) = −3. Distribute: x + 4x + 22 = −3. Combine like terms: 5x + 22 = −3. Subtract 22 from both sides: 5x = −3 − 22 = −25. Divide both sides by 5: x = −25 ÷ 5 = −5. Substitute back: put x = −5 into y = 2x + 11, so y = 2(−5) + 11 = −10 + 11 = 1. Check in both equations: 2(−5) + 11 = −10 + 11 = 1, which is y, and −5 + 2(1) = −5 + 2 = −3. Both are true, so the solution is (−5, 1).
✓ Correct!
Not quite. The answer is
💡 The equation y = 2x + 11 already says what y is. Substitute 2x + 11 for y in x + 2y = −3: x + 2(2x + 11) = −3. Distribute: x + 4x + 22 = −3. Combine like terms: 5x + 22 = −3. Subtract 22 from both sides: 5x = −3 − 22 = −25. Divide both sides by 5: x = −25 ÷ 5 = −5. Substitute back: put x = −5 into y = 2x + 11, so y = 2(−5) + 11 = −10 + 11 = 1. Check in both equations: 2(−5) + 11 = −10 + 11 = 1, which is y, and −5 + 2(1) = −5 + 2 = −3. Both are true, so the solution is (−5, 1).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| 2x + y = 1 | |
| y = 2x - 7 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| x + y = 1 | |
| x = y + 1 |
Not quite. The answer is
💡 The equation x = y + 1 already says what x is. Substitute y + 1 for x in x + y = 1: (y + 1) + y = 1. Combine like terms: 2y + 1 = 1. Subtract 1 from both sides: 2y = 1 − 1 = 0. Divide both sides by 2: y = 0 ÷ 2 = 0. Substitute back: put y = 0 into x = y + 1, so x = 0 + 1 = 1. Check in both equations: 0 + 1 = 1, which is x, and 1 + 0 = 1. Both are true, so the solution is (1, 0).
✓ Correct!
Not quite. The answer is
💡 The equation x = y + 1 already says what x is. Substitute y + 1 for x in x + y = 1: (y + 1) + y = 1. Combine like terms: 2y + 1 = 1. Subtract 1 from both sides: 2y = 1 − 1 = 0. Divide both sides by 2: y = 0 ÷ 2 = 0. Substitute back: put y = 0 into x = y + 1, so x = 0 + 1 = 1. Check in both equations: 0 + 1 = 1, which is x, and 1 + 0 = 1. Both are true, so the solution is (1, 0).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| y = 2x - 2 | |
| 4x + 4y = 28 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| y = 3x - 13 | |
| 3x + 3y = 21 |
Not quite. The answer is
💡 The equation y = 3x − 13 already says what y is. Substitute 3x − 13 for y in 3x + 3y = 21: 3x + 3(3x − 13) = 21. Distribute: 3x + 9x − 39 = 21. Combine like terms: 12x − 39 = 21. Add 39 to both sides: 12x = 21 + 39 = 60. Divide both sides by 12: x = 60 ÷ 12 = 5. Substitute back: put x = 5 into y = 3x − 13, so y = 3(5) − 13 = 15 − 13 = 2. Check in both equations: 3(5) − 13 = 15 − 13 = 2, which is y, and 3(5) + 3(2) = 15 + 6 = 21. Both are true, so the solution is (5, 2).
✓ Correct!
Not quite. The answer is
💡 The equation y = 3x − 13 already says what y is. Substitute 3x − 13 for y in 3x + 3y = 21: 3x + 3(3x − 13) = 21. Distribute: 3x + 9x − 39 = 21. Combine like terms: 12x − 39 = 21. Add 39 to both sides: 12x = 21 + 39 = 60. Divide both sides by 12: x = 60 ÷ 12 = 5. Substitute back: put x = 5 into y = 3x − 13, so y = 3(5) − 13 = 15 − 13 = 2. Check in both equations: 3(5) − 13 = 15 − 13 = 2, which is y, and 3(5) + 3(2) = 15 + 6 = 21. Both are true, so the solution is (5, 2).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| 2x + 4y = -18 | |
| y = x |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| 4x + 3y = 13 | |
| y = 4x - 1 |
Not quite. The answer is
💡 The equation y = 4x − 1 already says what y is. Substitute 4x − 1 for y in 4x + 3y = 13: 4x + 3(4x − 1) = 13. Distribute: 4x + 12x − 3 = 13. Combine like terms: 16x − 3 = 13. Add 3 to both sides: 16x = 13 + 3 = 16. Divide both sides by 16: x = 16 ÷ 16 = 1. Substitute back: put x = 1 into y = 4x − 1, so y = 4(1) − 1 = 4 − 1 = 3. Check in both equations: 4(1) − 1 = 4 − 1 = 3, which is y, and 4(1) + 3(3) = 4 + 9 = 13. Both are true, so the solution is (1, 3).
✓ Correct!
Not quite. The answer is
💡 The equation y = 4x − 1 already says what y is. Substitute 4x − 1 for y in 4x + 3y = 13: 4x + 3(4x − 1) = 13. Distribute: 4x + 12x − 3 = 13. Combine like terms: 16x − 3 = 13. Add 3 to both sides: 16x = 13 + 3 = 16. Divide both sides by 16: x = 16 ÷ 16 = 1. Substitute back: put x = 1 into y = 4x − 1, so y = 4(1) − 1 = 4 − 1 = 3. Check in both equations: 4(1) − 1 = 4 − 1 = 3, which is y, and 4(1) + 3(3) = 4 + 9 = 13. Both are true, so the solution is (1, 3).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| y = 3x + 2 | |
| 2x + 3y = -16 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| y = 3x + 2 | |
| 3x + y = -4 |
Not quite. The answer is
💡 The equation y = 3x + 2 already says what y is. Substitute 3x + 2 for y in 3x + y = −4: 3x + (3x + 2) = −4. Combine like terms: 6x + 2 = −4. Subtract 2 from both sides: 6x = −4 − 2 = −6. Divide both sides by 6: x = −6 ÷ 6 = −1. Substitute back: put x = −1 into y = 3x + 2, so y = 3(−1) + 2 = −3 + 2 = −1. Check in both equations: 3(−1) + 2 = −3 + 2 = −1, which is y, and 3(−1) + (−1) = −3 − 1 = −4. Both are true, so the solution is (−1, −1).
✓ Correct!
Not quite. The answer is
💡 The equation y = 3x + 2 already says what y is. Substitute 3x + 2 for y in 3x + y = −4: 3x + (3x + 2) = −4. Combine like terms: 6x + 2 = −4. Subtract 2 from both sides: 6x = −4 − 2 = −6. Divide both sides by 6: x = −6 ÷ 6 = −1. Substitute back: put x = −1 into y = 3x + 2, so y = 3(−1) + 2 = −3 + 2 = −1. Check in both equations: 3(−1) + 2 = −3 + 2 = −1, which is y, and 3(−1) + (−1) = −3 − 1 = −4. Both are true, so the solution is (−1, −1).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| 4x + 2y = 4 | |
| y = 4x - 4 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| 3x + 2y = 2 | |
| x = y - 1 |
Not quite. The answer is
💡 The equation x = y − 1 already says what x is. Substitute y − 1 for x in 3x + 2y = 2: 3(y − 1) + 2y = 2. Distribute: 3y + 2y − 3 = 2. Combine like terms: 5y − 3 = 2. Add 3 to both sides: 5y = 2 + 3 = 5. Divide both sides by 5: y = 5 ÷ 5 = 1. Substitute back: put y = 1 into x = y − 1, so x = 1 − 1 = 0. Check in both equations: 1 − 1 = 0, which is x, and 3(0) + 2(1) = 0 + 2 = 2. Both are true, so the solution is (0, 1).
✓ Correct!
Not quite. The answer is
💡 The equation x = y − 1 already says what x is. Substitute y − 1 for x in 3x + 2y = 2: 3(y − 1) + 2y = 2. Distribute: 3y + 2y − 3 = 2. Combine like terms: 5y − 3 = 2. Add 3 to both sides: 5y = 2 + 3 = 5. Divide both sides by 5: y = 5 ÷ 5 = 1. Substitute back: put y = 1 into x = y − 1, so x = 1 − 1 = 0. Check in both equations: 1 − 1 = 0, which is x, and 3(0) + 2(1) = 0 + 2 = 2. Both are true, so the solution is (0, 1).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| 2x + 2y = -12 | |
| y = x |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| y = 2x + 15 | |
| x + 3y = 10 |
Not quite. The answer is
💡 The equation y = 2x + 15 already says what y is. Substitute 2x + 15 for y in x + 3y = 10: x + 3(2x + 15) = 10. Distribute: x + 6x + 45 = 10. Combine like terms: 7x + 45 = 10. Subtract 45 from both sides: 7x = 10 − 45 = −35. Divide both sides by 7: x = −35 ÷ 7 = −5. Substitute back: put x = −5 into y = 2x + 15, so y = 2(−5) + 15 = −10 + 15 = 5. Check in both equations: 2(−5) + 15 = −10 + 15 = 5, which is y, and −5 + 3(5) = −5 + 15 = 10. Both are true, so the solution is (−5, 5).
✓ Correct!
Not quite. The answer is
💡 The equation y = 2x + 15 already says what y is. Substitute 2x + 15 for y in x + 3y = 10: x + 3(2x + 15) = 10. Distribute: x + 6x + 45 = 10. Combine like terms: 7x + 45 = 10. Subtract 45 from both sides: 7x = 10 − 45 = −35. Divide both sides by 7: x = −35 ÷ 7 = −5. Substitute back: put x = −5 into y = 2x + 15, so y = 2(−5) + 15 = −10 + 15 = 5. Check in both equations: 2(−5) + 15 = −10 + 15 = 5, which is y, and −5 + 3(5) = −5 + 15 = 10. Both are true, so the solution is (−5, 5).
💡 How to solve: Solving Systems of Equations by Substitution
Find an equation that says what one variable equals, such as y = 2x + 1. Substitute that expression for the variable in the other equation, so only one variable is left: 3x + (2x + 1) = 11, so 5x = 10 and x = 2. Put that value back into y = 2x + 1 to find the other variable: y = 2(2) + 1 = 5. The solution is (2, 5); check it in both equations. If no variable is alone, first solve one equation for a variable whose coefficient is 1 or -1.
Try a similar one
Solve the system.
| y = 3x - 9 | |
| 4x + 3y = 12 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Solving Systems of Equations by Substitution — Easy Worksheet 2
Solve each system by substitution. Write the solution as an ordered pair (x, y).