\left\{ \begin{array} { l } { 9 m - 5 y = 39 } \\ { 15 x - 3 y = 81 } \end{array} \right.
Solve for x, y
x=\frac{9m+96}{25}
y=\frac{9m-39}{5}
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-5y+9m=39,-3y+15x=81
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
-5y+9m=39
Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.
-5y=39-9m
Subtract 9m from both sides of the equation.
y=\frac{9m-39}{5}
Divide both sides by -5.
-3\times \frac{9m-39}{5}+15x=81
Substitute \frac{-39+9m}{5} for y in the other equation, -3y+15x=81.
\frac{117-27m}{5}+15x=81
Multiply -3 times \frac{-39+9m}{5}.
15x=\frac{27m+288}{5}
Subtract \frac{117-27m}{5} from both sides of the equation.
x=\frac{9m+96}{25}
Divide both sides by 15.
y=\frac{9m-39}{5},x=\frac{9m+96}{25}
The system is now solved.
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