\left\{ \begin{array} { l } { 3 = x 50 + y \cdot 20 } \\ { 9 = x + y } \end{array} \right.
Solve for x, y
x=\frac{x_{50}+177}{20}
y=\frac{3-x_{50}}{20}
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x_{50}+y\times 20=3
Consider the first equation. Swap sides so that all variable terms are on the left hand side.
y\times 20=3-x_{50}
Subtract x_{50} from both sides.
x+y=9
Consider the second equation. Swap sides so that all variable terms are on the left hand side.
20y=3-x_{50},y+x=9
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.
20y=3-x_{50}
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.
y=\frac{3-x_{50}}{20}
Divide both sides by 20.
\frac{3-x_{50}}{20}+x=9
Substitute \frac{3-x_{50}}{20} for y in the other equation, y+x=9.
x=\frac{x_{50}+177}{20}
Subtract \frac{3-x_{50}}{20} from both sides of the equation.
y=\frac{3-x_{50}}{20},x=\frac{x_{50}+177}{20}
The system is now solved.
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