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x-7y=135
Consider the first equation. Subtract 7y from both sides.
y-4x=0
Consider the second equation. Subtract 4x from both sides.
x-7y=135,-4x+y=0
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.
x-7y=135
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=7y+135
Add 7y to both sides of the equation.
-4\left(7y+135\right)+y=0
Substitute 7y+135 for x in the other equation, -4x+y=0.
-28y-540+y=0
Multiply -4 times 7y+135.
-27y-540=0
Add -28y to y.
-27y=540
Add 540 to both sides of the equation.
y=-20
Divide both sides by -27.
x=7\left(-20\right)+135
Substitute -20 for y in x=7y+135. Because the resulting equation contains only one variable, you can solve for x directly.
x=-140+135
Multiply 7 times -20.
x=-5
Add 135 to -140.
x=-5,y=-20
The system is now solved.
x-7y=135
Consider the first equation. Subtract 7y from both sides.
y-4x=0
Consider the second equation. Subtract 4x from both sides.
x-7y=135,-4x+y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}135\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right))\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right))\left(\begin{matrix}135\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-7\\-4&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right))\left(\begin{matrix}135\\0\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-7\\-4&1\end{matrix}\right))\left(\begin{matrix}135\\0\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-7\left(-4\right)\right)}&-\frac{-7}{1-\left(-7\left(-4\right)\right)}\\-\frac{-4}{1-\left(-7\left(-4\right)\right)}&\frac{1}{1-\left(-7\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}135\\0\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{27}&-\frac{7}{27}\\-\frac{4}{27}&-\frac{1}{27}\end{matrix}\right)\left(\begin{matrix}135\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{27}\times 135\\-\frac{4}{27}\times 135\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\\-20\end{matrix}\right)
Do the arithmetic.
x=-5,y=-20
Extract the matrix elements x and y.
x-7y=135
Consider the first equation. Subtract 7y from both sides.
y-4x=0
Consider the second equation. Subtract 4x from both sides.
x-7y=135,-4x+y=0
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
-4x-4\left(-7\right)y=-4\times 135,-4x+y=0
To make x and -4x equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by 1.
-4x+28y=-540,-4x+y=0
Simplify.
-4x+4x+28y-y=-540
Subtract -4x+y=0 from -4x+28y=-540 by subtracting like terms on each side of the equal sign.
28y-y=-540
Add -4x to 4x. Terms -4x and 4x cancel out, leaving an equation with only one variable that can be solved.
27y=-540
Add 28y to -y.
y=-20
Divide both sides by 27.
-4x-20=0
Substitute -20 for y in -4x+y=0. Because the resulting equation contains only one variable, you can solve for x directly.
-4x=20
Add 20 to both sides of the equation.
x=-5
Divide both sides by -4.
x=-5,y=-20
The system is now solved.