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4x+4y=280,4x+y=124
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.
4x+4y=280
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x=-4y+280
Subtract 4y from both sides of the equation.
x=\frac{1}{4}\left(-4y+280\right)
Divide both sides by 4.
x=-y+70
Multiply \frac{1}{4} times -4y+280.
4\left(-y+70\right)+y=124
Substitute -y+70 for x in the other equation, 4x+y=124.
-4y+280+y=124
Multiply 4 times -y+70.
-3y+280=124
Add -4y to y.
-3y=-156
Subtract 280 from both sides of the equation.
y=52
Divide both sides by -3.
x=-52+70
Substitute 52 for y in x=-y+70. Because the resulting equation contains only one variable, you can solve for x directly.
x=18
Add 70 to -52.
x=18,y=52
The system is now solved.
4x+4y=280,4x+y=124
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&4\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}280\\124\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&4\\4&1\end{matrix}\right))\left(\begin{matrix}4&4\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&4\\4&1\end{matrix}\right))\left(\begin{matrix}280\\124\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&4\\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}4&4\\4&1\end{matrix}\right))\left(\begin{matrix}280\\124\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}4&4\\4&1\end{matrix}\right))\left(\begin{matrix}280\\124\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}{4-4\times 4}&-\frac{4}{4-4\times 4}\\-\frac{4}{4-4\times 4}&\frac{4}{4-4\times 4}\end{matrix}\right)\left(\begin{matrix}280\\124\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}{12}&\frac{1}{3}\\\frac{1}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}280\\124\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{12}\times 280+\frac{1}{3}\times 124\\\frac{1}{3}\times 280-\frac{1}{3}\times 124\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18\\52\end{matrix}\right)
Do the arithmetic.
x=18,y=52
Extract the matrix elements x and y.
4x+4y=280,4x+y=124
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-4x+4y-y=280-124
Subtract 4x+y=124 from 4x+4y=280 by subtracting like terms on each side of the equal sign.
4y-y=280-124
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
3y=280-124
Add 4y to -y.
3y=156
Add 280 to -124.
y=52
Divide both sides by 3.
4x+52=124
Substitute 52 for y in 4x+y=124. Because the resulting equation contains only one variable, you can solve for x directly.
4x=72
Subtract 52 from both sides of the equation.
x=18
Divide both sides by 4.
x=18,y=52
The system is now solved.