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x-y=0
Consider the first equation. Subtract y from both sides.
x-y=0,68x+74y=426
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-y=0
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=y
Add y to both sides of the equation.
68y+74y=426
Substitute y for x in the other equation, 68x+74y=426.
142y=426
Add 68y to 74y.
y=3
Divide both sides by 142.
x=3
Substitute 3 for y in x=y. Because the resulting equation contains only one variable, you can solve for x directly.
x=3,y=3
The system is now solved.
x-y=0
Consider the first equation. Subtract y from both sides.
x-y=0,68x+74y=426
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-1\\68&74\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\426\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-1\\68&74\end{matrix}\right))\left(\begin{matrix}1&-1\\68&74\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\68&74\end{matrix}\right))\left(\begin{matrix}0\\426\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\68&74\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&-1\\68&74\end{matrix}\right))\left(\begin{matrix}0\\426\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&-1\\68&74\end{matrix}\right))\left(\begin{matrix}0\\426\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{74}{74-\left(-68\right)}&-\frac{-1}{74-\left(-68\right)}\\-\frac{68}{74-\left(-68\right)}&\frac{1}{74-\left(-68\right)}\end{matrix}\right)\left(\begin{matrix}0\\426\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{37}{71}&\frac{1}{142}\\-\frac{34}{71}&\frac{1}{142}\end{matrix}\right)\left(\begin{matrix}0\\426\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{142}\times 426\\\frac{1}{142}\times 426\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\3\end{matrix}\right)
Do the arithmetic.
x=3,y=3
Extract the matrix elements x and y.
x-y=0
Consider the first equation. Subtract y from both sides.
x-y=0,68x+74y=426
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.
68x+68\left(-1\right)y=0,68x+74y=426
To make x and 68x equal, multiply all terms on each side of the first equation by 68 and all terms on each side of the second by 1.
68x-68y=0,68x+74y=426
Simplify.
68x-68x-68y-74y=-426
Subtract 68x+74y=426 from 68x-68y=0 by subtracting like terms on each side of the equal sign.
-68y-74y=-426
Add 68x to -68x. Terms 68x and -68x cancel out, leaving an equation with only one variable that can be solved.
-142y=-426
Add -68y to -74y.
y=3
Divide both sides by -142.
68x+74\times 3=426
Substitute 3 for y in 68x+74y=426. Because the resulting equation contains only one variable, you can solve for x directly.
68x+222=426
Multiply 74 times 3.
68x=204
Subtract 222 from both sides of the equation.
x=3
Divide both sides by 68.
x=3,y=3
The system is now solved.