Let’s solve each system of equations step by step using substitution. We’ll go one problem at a time.
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Problem 1:
x = -y
3x + 4y = -8
Since x is already solved for (x = -y), we can plug that into the second equation:
3(-y) + 4y = -8
→ -3y + 4y = -8
→ y = -8
Now plug y = -8 back into x = -y:
x = -(-8) = 8
✔ Solution: x = 8, y = -8
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Problem 2:
y = -3x + 4
y = -5x - 17
Both equations are solved for y, so set them equal to each other:
-3x + 4 = -5x - 17
Add 5x to both sides:
2x + 4 = -17
Subtract 4 from both sides:
2x = -21
Divide by 2:
x = -10.5 or -21/2
Now plug into either equation to find y. Use first one:
y = -3(-21/2) + 4 = 63/2 + 8/2 = 71/2 = 35.5
✔ Solution: x = -21/2, y = 71/2 (or x = -10.5, y = 35.5)
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Problem 3:
y = -x - 6
y = x - 4
Set them equal:
-x - 6 = x - 4
Add x to both sides:
-6 = 2x - 4
Add 4 to both sides:
-2 = 2x → x = -1
Plug into y = x - 4:
y = -1 - 4 = -5
✔ Solution: x = -1, y = -5
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Problem 4:
2x - 3y = 1
-6x + 9y = -3
Notice: If you multiply the first equation by -3, you get the second equation:
-3*(2x - 3y) = -3*1 → -6x + 9y = -3 → same as second equation!
That means these two equations are actually the same line — infinitely many solutions.
But let’s try solving anyway. Solve first equation for x:
2x = 3y + 1 → x = (3y + 1)/2
Plug into second equation:
-6*((3y+1)/2) + 9y = -3
→ -3*(3y+1) + 9y = -3
→ -9y -3 + 9y = -3
→ -3 = -3 → always true!
So yes — infinite solutions. Any point on the line works.
✔ Solution: Infinitely many solutions (dependent system)
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Problem 5:
7x + 2y = -19
-x + 2y = 21
Solve second equation for x:
-x = 21 - 2y → x = 2y - 21
Plug into first equation:
7(2y - 21) + 2y = -19
→ 14y - 147 + 2y = -19
→ 16y = 128
→ y = 8
Now x = 2(8) - 21 = 16 - 21 = -5
✔ Solution: x = -5, y = 8
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Problem 6:
x = y - 4
-2x + 3y = 6
Plug x = y - 4 into second equation:
-2(y - 4) + 3y = 6
→ -2y + 8 + 3y = 6
→ y + 8 = 6
→ y = -2
Then x = (-2) - 4 = -6
✔ Solution: x = -6, y = -2
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Problem 7:
2y = 2x + 12
y = -2 - 3 → wait, this says y = -5? Let me check.
Actually, looking again: “y = -2 - 3” → that’s just y = -5. Probably a typo in the worksheet? But we’ll take it as written.
So y = -5
Plug into first equation:
2*(-5) = 2x + 12
→ -10 = 2x + 12
→ 2x = -22
→ x = -11
✔ Solution: x = -11, y = -5
*(Note: The second equation seems oddly written — maybe meant to be y = -2x - 3? But as written, it’s y = -5.)*
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Problem 8:
x - 3y = 0 → x = 3y
4x + 8y = 5
Plug x = 3y into second equation:
4*(3y) + 8y = 5
→ 12y + 8y = 5
→ 20y = 5
→ y = 5/20 = 1/4
Then x = 3*(1/4) = 3/4
✔ Solution: x = 3/4, y = 1/4
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Final Answer:
1. x = 8, y = -8
2. x = -21/2, y = 71/2
3. x = -1, y = -5
4. Infinitely many solutions
5. x = -5, y = 8
6. x = -6, y = -2
7. x = -11, y = -5
8. x = 3/4, y = 1/4
Parent Tip: Review the logic above to help your child master the concept of solving systems of linear equations by substitution worksheet.