Verified Answer :

mariamikayla

\(\left \{ {{x+5y=6} \ \ |2\atop {6x-10y=12}} \right. \\\\\\ \left \{ {{2x+10y=12} \atop {6x-10y=12}} \right. \\ ======== \\\\ 8x=24 \ \ \ |:8 \\\\ \boxed{x=3} \\\\ x+5y=6 \\\\ 5y=6-x \\\\ y=\frac{6-x}{5} \\\\ y=\frac{6-3}{5} \\\\ \boxed{y=\frac{3}{5}}\)

apologiabiology

so assuming x and y remain constant, we can make one of the place holders negative and add the equations together to cancel each other out so

we can cancel out the y terms since 5 is close to 10
(x+5y=6) times 2=2x+10y=12
add this to 6x-10y=12
2x+6x+10y-10y=12+12
8x=24
divide both sides by 8
x=3
subsitute into first equation
x+5y=6
3+5y=6
5y=3
y=3/5
x=3

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