\left\{ \begin{array} { l } { 5 a + 4 y = 1 } \\ { x + 3 y = 17 } \end{array} \right.
Solve for x, y
x=\frac{15a+65}{4}
y=\frac{1-5a}{4}
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4y+5a=1,3y+x=17
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.
4y+5a=1
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.
4y=1-5a
Subtract 5a from both sides of the equation.
y=\frac{1-5a}{4}
Divide both sides by 4.
3\times \frac{1-5a}{4}+x=17
Substitute \frac{1-5a}{4} for y in the other equation, 3y+x=17.
\frac{3-15a}{4}+x=17
Multiply 3 times \frac{1-5a}{4}.
x=\frac{15a+65}{4}
Subtract \frac{3-15a}{4} from both sides of the equation.
y=\frac{1-5a}{4},x=\frac{15a+65}{4}
The system is now solved.
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Limits
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