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Solving a system of equations using substitution

y=3x+3

y=2x+8


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x+3y=17

2x+3y=22

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1.
\(y=3x+3 \\ y=2x+8 \\ \\ \hbox{substitute 3x+3 for y in the 2nd equation:} \\ 3x+3=2x+8 \\ 3x-2x=8-3 \\ x=5 \\ \\ y=2x+8=2 \times 5+8=10+8=18 \\ \\ \boxed{(x,y)=(5,18)}\)

2.
\(x+3y=17 \\ 2x+3y=22 \\ \\ \hbox{solve the 1st equation for x:} \\ x+3y=17 \\ x=17-3y \\ \\ \hbox{substitute 17-3y for x in the 2nd equation:} \\ 2(17-3y)+3y=22 \\ 34-6y+3y=22 \\ -6y+3y=22-34 \\ -3y=-12 \\ y=\frac{-12}{-3} \\ y=4 \\ \\ x=17-3y=17-3 \times 4=17-12=5 \\ \\ \boxed{(x,y)=(5,4)}\)

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