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x+5y=775,4x+3y=890
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+5y=775
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-5y+775
Subtract 5y from both sides of the equation.
4\left(-5y+775\right)+3y=890
Substitute -5y+775 for x in the other equation, 4x+3y=890.
-20y+3100+3y=890
Multiply 4 times -5y+775.
-17y+3100=890
Add -20y to 3y.
-17y=-2210
Subtract 3100 from both sides of the equation.
y=130
Divide both sides by -17.
x=-5\times 130+775
Substitute 130 for y in x=-5y+775. Because the resulting equation contains only one variable, you can solve for x directly.
x=-650+775
Multiply -5 times 130.
x=125
Add 775 to -650.
x=125,y=130
The system is now solved.
x+5y=775,4x+3y=890
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&5\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}775\\890\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&5\\4&3\end{matrix}\right))\left(\begin{matrix}1&5\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&5\\4&3\end{matrix}\right))\left(\begin{matrix}775\\890\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&5\\4&3\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&5\\4&3\end{matrix}\right))\left(\begin{matrix}775\\890\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&5\\4&3\end{matrix}\right))\left(\begin{matrix}775\\890\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{3}{3-5\times 4}&-\frac{5}{3-5\times 4}\\-\frac{4}{3-5\times 4}&\frac{1}{3-5\times 4}\end{matrix}\right)\left(\begin{matrix}775\\890\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{3}{17}&\frac{5}{17}\\\frac{4}{17}&-\frac{1}{17}\end{matrix}\right)\left(\begin{matrix}775\\890\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{17}\times 775+\frac{5}{17}\times 890\\\frac{4}{17}\times 775-\frac{1}{17}\times 890\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}125\\130\end{matrix}\right)
Do the arithmetic.
x=125,y=130
Extract the matrix elements x and y.
x+5y=775,4x+3y=890
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\times 5y=4\times 775,4x+3y=890
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+20y=3100,4x+3y=890
Simplify.
4x-4x+20y-3y=3100-890
Subtract 4x+3y=890 from 4x+20y=3100 by subtracting like terms on each side of the equal sign.
20y-3y=3100-890
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
17y=3100-890
Add 20y to -3y.
17y=2210
Add 3100 to -890.
y=130
Divide both sides by 17.
4x+3\times 130=890
Substitute 130 for y in 4x+3y=890. Because the resulting equation contains only one variable, you can solve for x directly.
4x+390=890
Multiply 3 times 130.
4x=500
Subtract 390 from both sides of the equation.
x=125
Divide both sides by 4.
x=125,y=130
The system is now solved.