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x+y=102,12x+10y=118
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=102
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+102
Subtract y from both sides of the equation.
12\left(-y+102\right)+10y=118
Substitute -y+102 for x in the other equation, 12x+10y=118.
-12y+1224+10y=118
Multiply 12 times -y+102.
-2y+1224=118
Add -12y to 10y.
-2y=-1106
Subtract 1224 from both sides of the equation.
y=553
Divide both sides by -2.
x=-553+102
Substitute 553 for y in x=-y+102. Because the resulting equation contains only one variable, you can solve for x directly.
x=-451
Add 102 to -553.
x=-451,y=553
The system is now solved.
x+y=102,12x+10y=118
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\12&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}102\\118\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\12&10\end{matrix}\right))\left(\begin{matrix}1&1\\12&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12&10\end{matrix}\right))\left(\begin{matrix}102\\118\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\12&10\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\\12&10\end{matrix}\right))\left(\begin{matrix}102\\118\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\\12&10\end{matrix}\right))\left(\begin{matrix}102\\118\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{10}{10-12}&-\frac{1}{10-12}\\-\frac{12}{10-12}&\frac{1}{10-12}\end{matrix}\right)\left(\begin{matrix}102\\118\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&\frac{1}{2}\\6&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}102\\118\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\times 102+\frac{1}{2}\times 118\\6\times 102-\frac{1}{2}\times 118\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-451\\553\end{matrix}\right)
Do the arithmetic.
x=-451,y=553
Extract the matrix elements x and y.
x+y=102,12x+10y=118
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.
12x+12y=12\times 102,12x+10y=118
To make x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 1.
12x+12y=1224,12x+10y=118
Simplify.
12x-12x+12y-10y=1224-118
Subtract 12x+10y=118 from 12x+12y=1224 by subtracting like terms on each side of the equal sign.
12y-10y=1224-118
Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.
2y=1224-118
Add 12y to -10y.
2y=1106
Add 1224 to -118.
y=553
Divide both sides by 2.
12x+10\times 553=118
Substitute 553 for y in 12x+10y=118. Because the resulting equation contains only one variable, you can solve for x directly.
12x+5530=118
Multiply 10 times 553.
12x=-5412
Subtract 5530 from both sides of the equation.
x=-451
Divide both sides by 12.
x=-451,y=553
The system is now solved.