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x+y=1258,4x+5y=5426
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=1258
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+1258
Subtract y from both sides of the equation.
4\left(-y+1258\right)+5y=5426
Substitute -y+1258 for x in the other equation, 4x+5y=5426.
-4y+5032+5y=5426
Multiply 4 times -y+1258.
y+5032=5426
Add -4y to 5y.
y=394
Subtract 5032 from both sides of the equation.
x=-394+1258
Substitute 394 for y in x=-y+1258. Because the resulting equation contains only one variable, you can solve for x directly.
x=864
Add 1258 to -394.
x=864,y=394
The system is now solved.
x+y=1258,4x+5y=5426
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\4&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1258\\5426\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\4&5\end{matrix}\right))\left(\begin{matrix}1&1\\4&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\4&5\end{matrix}\right))\left(\begin{matrix}1258\\5426\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\4&5\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\\4&5\end{matrix}\right))\left(\begin{matrix}1258\\5426\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\\4&5\end{matrix}\right))\left(\begin{matrix}1258\\5426\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{5}{5-4}&-\frac{1}{5-4}\\-\frac{4}{5-4}&\frac{1}{5-4}\end{matrix}\right)\left(\begin{matrix}1258\\5426\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}5&-1\\-4&1\end{matrix}\right)\left(\begin{matrix}1258\\5426\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\times 1258-5426\\-4\times 1258+5426\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}864\\394\end{matrix}\right)
Do the arithmetic.
x=864,y=394
Extract the matrix elements x and y.
x+y=1258,4x+5y=5426
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+4y=4\times 1258,4x+5y=5426
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+4y=5032,4x+5y=5426
Simplify.
4x-4x+4y-5y=5032-5426
Subtract 4x+5y=5426 from 4x+4y=5032 by subtracting like terms on each side of the equal sign.
4y-5y=5032-5426
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-y=5032-5426
Add 4y to -5y.
-y=-394
Add 5032 to -5426.
y=394
Divide both sides by -1.
4x+5\times 394=5426
Substitute 394 for y in 4x+5y=5426. Because the resulting equation contains only one variable, you can solve for x directly.
4x+1970=5426
Multiply 5 times 394.
4x=3456
Subtract 1970 from both sides of the equation.
x=864
Divide both sides by 4.
x=864,y=394
The system is now solved.