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x=6y+30
x=3/2y
solve by substitution.

Sagot :

naǫ
[tex]x=6y+30 \\ x=\frac{3}{2}y \\ \\ \hbox{substitute } \frac{3}{2}y \hbox{ for x in the 1st equation:} \\ \frac{3}{2}y=6y+30 \\ \frac{3}{2}y-6y=30 \\ 3y-12y=60 \\ -9y=60 \\ y=-\frac{60}{9}=-\frac{20}{3}=-6\frac{2}{3} \\ \\ x=\frac{3}{2} \cdot (-\frac{20}{3})=-\frac{20}{2}=-10 \\ \\ \left \{ {{x=-10} \atop {y=-6\frac{2}{3}}} \right.[/tex]
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