The given system
-1/2x+5y=-20−12x+5y=−20 and 3x-2y=13x−2y=1
Let us use the method of substitution
From the second equation 3x-2y=13x−2y=1
x=(1+2y)/3x=1+2y3
We use the second in the first equation
-1/2x+5y=-20−12x+5y=−20
-1/2((1+2y)/3)+5y=-20−12(1+2y3)+5y=−20
-((1+2y)/3)+10y=-40−(1+2y3)+10y=−40
((1+2y)/3)-10y=40(1+2y3)−10y=40
1+2y-30y=1201+2y−30y=120
-28y=119−28y=119
y=-119/28y=−11928
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Solve for xx
x=(1+2(-119/28))/3x=1+2(−11928)3
x=(28-238)/84x=28−23884
x=-210/84x=−21084
x=-105/42x=−10542
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Checking:at x=-105/42x=−10542 and y=-119/28y=−11928
-1/2x+5y=-20−12x+5y=−20
-1/2(-105/42)+5(-119/28)=-20−12(−10542)+5(−11928)=−20
105/84-595/28=-2010584−59528=−20
(105-1785)/84=-20105−178584=−20
(-1680)/84=-20−168084=−20
#-20=-20
Also
3x-2y=13x−2y=1
3(-105/42)-2(-119/28)=13(−10542)−2(−11928)=1
-105/14+119/14=1−10514+11914=1
14/14=11414=1
1=11=1
Correct ! at x=-105/42x=−10542 and y=-119/28y=−11928
God bless....I hope the explanation is useful.
