Q. Solve the system of equations by elimination.3x+3y−2z=−14x−y+2z=203x+y−3z=−15
Combine Equations to Eliminate z: First, let's eliminate z from the first and third equations by adding them.3x+3y−2z=−143x+y−3z=−15Add the equations:(3x+3y−2z)+(3x+y−3z)=−14+(−15)6x+4y−5z=−29
Multiply and Add Equations to Eliminate z: Now, let's eliminate z from the second and third equations by multiplying the second equation by 3 and adding it to the third equation.3(x−y+2z)=3(20)3x−3y+6z=60Now add this to the third equation:(3x−3y+6z)+(3x+y−3z)=60+(−15)6x−2y+3z=45
Subtract Equations to Eliminate x: We now have two new equations without z:6x+4y−5z=−296x−2y+3z=45Let's eliminate x by subtracting the second equation from the first.(6x+4y)−(6x−2y)=−29−456x+4y−6x+2y=−746y=−74y=6−74y=−12.333
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