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2x+y=20
6x-5y=12
substitution


Sagot :

[tex]2x+y=20 \\ 6x-5y=12 \\ \\ \hbox{solve the first equation for y:} \\ 2x+y=20 \ \ \ |-2x \\ y=20-2x \\ \\ \hbox{substitute 20-2x for y in the second equation:} \\ 6x-5(20-2x)=12 \\ 6x-100+10x=12 \\ 16x-100=12 \ \ \ |+100 \\ 16x=112 \ \ \ |\div 16 \\ x=7 \\ \\ y=20-2x=20-2 \times 7=20-14=6 \\ \\ \boxed{(x,y)=(7,6)}[/tex]
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