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  5y(2x−3)=y(x+2)+20 where x=3

Sagot :

here x=3
so, putting 3 instead of x
[tex]5y(2(3)-3)=y(3+2)+20 \\ 5y(3)=5y+20 \\ 15y-5y=20 \\ 10y=20 \\ y=20/10=2[/tex]
5y(2(3)-3)=y(3+2)+20
5y(6-3)=y(5)+20
5y(3)=5y+20
15y=5y+20
10y=20
y=2