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solve this by substitution :
5x-y =5
5x-3y=15

Sagot :

Selfie22
5x-y=5 ......1 5x-3y=15......2 From eqn 1 y=5x-5....3 Substitute y in eqn 2 5x-3(5x-5)=15 5x-15x-15=15 -10x=0 x=0 Sub it in eqn 3 y=5(0)-5 y=-5
[tex]\left\{\begin{array}{ccc}5x-y=5&|subtract\ 5x\ from\ both\ sides\\5x-3y=15\end{array}\right\\\left\{\begin{array}{ccc}-y=-5x+5&|change\ the\ signs-multiply\ both\ sides\ by\ (-1)\\5x-3y=15\end{array}\right\\\left\{\begin{array}{ccc}y=5x-5&(1^o)\\5x-3y=15&(2^o)\end{array}\right\\\\subtitute\ (1^o)\ to\ (2^o)\\\\5x-3(5x-5)=15\\5x-3(5x)-3(-5)=15\\5x-15x+15=15\\-10x+15=15\ \ \ \ |subtract\ 15\ from\ both\ sides\\-10x=0\ \ \ \ \ |divide\ both\ sides\ by\ (-10)\\\boxed{x=0}[/tex]

[tex]subtitute\ value\ of\ x\ to\ (1^o):\\\\y=5(0)-5=0-5=\boxed{-5}\\\\Answer:\boxed{\boxed{\left\{\begin{array}{ccc}x=0\\y=-5\end{array}\right}}[/tex]
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