QUESTION IMAGE
Question
solving systems using elimination
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period
date
solve each system of equations using the elimination method. check the solution.
- $x - 2y = 10$
$- x + 3y = 5$
- $x + y = -4$
$x - y = 2$
- $3x + 4y = 23$
$3x - 2y = 11$
- $10x - 6y = 54$
$-4x + 3y = -24$
- $4x + 2y = 6$
$4x - 3y = -19$
- $4x - y = -2$
$2x + y = 2$
- $3x + 8y = 2$
$4x - 2y = -10$
- $2x - 4y = 10$
$-4x + 3y = -20$
Step1: Solve system 2
We have the system:
Add the two equations to eliminate \(y\):
\((x + y)+(x - y)=-4 + 2\)
\(2x=-2\)
Step2: Find \(x\)
Divide both sides by 2:
\(x=\frac{-2}{2}=-1\)
Step3: Find \(y\)
Substitute \(x = - 1\) into \(x + y=-4\):
\(-1+y=-4\)
Add 1 to both sides:
\(y=-4 + 1=-3\)
Step4: Check the solution
Substitute \(x=-1,y = - 3\) into the first equation: \(x + y=-1+(-3)=-4\) (matches).
Substitute into the second equation: \(x - y=-1-(-3)=-1 + 3 = 2\) (matches).
Step1: Solve system 3
We have the system:
Subtract the second equation from the first to eliminate \(x\):
\((3x + 4y)-(3x - 2y)=23-11\)
\(3x+4y - 3x + 2y=12\)
\(6y=12\)
Step2: Find \(y\)
Divide both sides by 6:
\(y = \frac{12}{6}=2\)
Step3: Find \(x\)
Substitute \(y = 2\) into \(3x-2y=11\):
\(3x-2\times2=11\)
\(3x-4 = 11\)
Add 4 to both sides: \(3x=11 + 4=15\)
Divide by 3: \(x=\frac{15}{3}=5\)
Step4: Check the solution
Substitute \(x = 5,y=2\) into \(3x + 4y\): \(3\times5+4\times2=15 + 8=23\) (matches).
Substitute into \(3x-2y\): \(3\times5-2\times2=15 - 4 = 11\) (matches).
Step1: Solve system 4
We have the system:
Multiply the second equation by 2: \(-8x + 6y=-48\)
Now add this new equation to the first equation:
\((10x-6y)+(-8x + 6y)=54+(-48)\)
\(10x-6y-8x + 6y=6\)
\(2x=6\)
Step2: Find \(x\)
Divide by 2: \(x = 3\)
Step3: Find \(y\)
Substitute \(x = 3\) into \(10x-6y=54\):
\(10\times3-6y=54\)
\(30-6y=54\)
Subtract 30 from both sides: \(-6y=54 - 30=24\)
Divide by - 6: \(y=\frac{24}{-6}=-4\)
Step4: Check the solution
Substitute \(x = 3,y=-4\) into \(10x-6y\): \(10\times3-6\times(-4)=30 + 24=54\) (matches).
Substitute into \(-4x + 3y\): \(-4\times3+3\times(-4)=-12-12=-24\) (matches).
Step1: Solve system 6
We have the system:
Add the two equations to eliminate \(y\):
\((4x-y)+(2x + y)=-2 + 2\)
\(4x-y+2x + y=0\)
\(6x=0\)
Step2: Find \(x\)
Divide by 6: \(x = 0\)
Step3: Find \(y\)
Substitute \(x = 0\) into \(2x + y=2\):
\(0+y=2\)
\(y = 2\)
Step4: Check the solution
Substitute \(x = 0,y = 2\) into \(4x-y\): \(4\times0-2=-2\) (matches).
Substitute into \(2x + y\): \(2\times0+2=2\) (matches).
Step1: Solve system 7
We have the system:
Multiply the second equation by 4: \(16x-8y=-40\)
Add this new equation to the first equation:
\((3x + 8y)+(16x-8y)=2+(-40)\)
\(3x + 8y+16x-8y=-38\)
\(19x=-38\)
Step2: Find \(x\)
Divide by 19: \(x=\frac{-38}{19}=-2\)
Step3: Find \(y\)
Substitute \(x=-2\) into \(3x + 8y=2\):
\(3\times(-2)+8y=2\)
\(-6 + 8y=2\)
Add 6 to both sides: \(8y=2 + 6=8\)
Divide by 8: \(y = 1\)
Step4: Check the solution
Substitute \(x=-2,y = 1\) into \(3x + 8y\): \(3\times(-2)+8\times1=-6 + 8=2\) (matches).
Substitute into \(4x-2y\): \(4\times(-2)-2\times1=-8-2=-10\) (matches).
Step1: Solve system 8
We have the system:
Multiply the first equation by 2: \(4x-8y=20\)
Add this new equation to the second equation:
\((4x-8y)+(-4x + 3y)=20+(-20)\)
\(4x-8y-4x + 3y=0\)
\(-5y=0\)
Step2: Find \(y\)
Divide by - 5: \(y = 0\)
Step3: Find \(x\)
Substitute \(y = 0\) into \(2x-4y=10\):
\(2x-4\times0=10\)
\(2x=10\)
Divide by 2: \(x = 5\)
Step4: Check the solution
Substitute \(x = 5,y = 0\) into \(2x-4y\): \(2\times5-4\times0=10\) (matches).
Substitute into \(-4x + 3y\): \(-4\times5…
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