\left\{ \begin{array} { l } { x + y = 190 } \\ { 2 x 8 x = 22 y } \end{array} \right.
Solve for x, y
x=\frac{2090}{x_{8}+11}
y=\frac{190x_{8}}{x_{8}+11}
x_{8}\neq -11
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2x_{8}x-22y=0
Consider the second equation. Subtract 22y from both sides.
x+y=190,2x_{8}x-22y=0
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=190
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+190
Subtract y from both sides of the equation.
2x_{8}\left(-y+190\right)-22y=0
Substitute -y+190 for x in the other equation, 2x_{8}x-22y=0.
\left(-2x_{8}\right)y+380x_{8}-22y=0
Multiply 2x_{8} times -y+190.
\left(-2x_{8}-22\right)y+380x_{8}=0
Add -2x_{8}y to -22y.
\left(-2x_{8}-22\right)y=-380x_{8}
Subtract 380x_{8} from both sides of the equation.
y=\frac{190x_{8}}{x_{8}+11}
Divide both sides by -2x_{8}-22.
x=-\frac{190x_{8}}{x_{8}+11}+190
Substitute \frac{190x_{8}}{11+x_{8}} for y in x=-y+190. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{2090}{x_{8}+11}
Add 190 to -\frac{190x_{8}}{11+x_{8}}.
x=\frac{2090}{x_{8}+11},y=\frac{190x_{8}}{x_{8}+11}
The system is now solved.
2x_{8}x-22y=0
Consider the second equation. Subtract 22y from both sides.
x+y=190,2x_{8}x-22y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\2x_{8}&-22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}190\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\2x_{8}&-22\end{matrix}\right))\left(\begin{matrix}1&1\\2x_{8}&-22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2x_{8}&-22\end{matrix}\right))\left(\begin{matrix}190\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2x_{8}&-22\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\\2x_{8}&-22\end{matrix}\right))\left(\begin{matrix}190\\0\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\\2x_{8}&-22\end{matrix}\right))\left(\begin{matrix}190\\0\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{22}{-22-2x_{8}}&-\frac{1}{-22-2x_{8}}\\-\frac{2x_{8}}{-22-2x_{8}}&\frac{1}{-22-2x_{8}}\end{matrix}\right)\left(\begin{matrix}190\\0\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{11}{x_{8}+11}&\frac{1}{2\left(x_{8}+11\right)}\\\frac{x_{8}}{x_{8}+11}&-\frac{1}{2\left(x_{8}+11\right)}\end{matrix}\right)\left(\begin{matrix}190\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{x_{8}+11}\times 190\\\frac{x_{8}}{x_{8}+11}\times 190\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2090}{x_{8}+11}\\\frac{190x_{8}}{x_{8}+11}\end{matrix}\right)
Do the arithmetic.
x=\frac{2090}{x_{8}+11},y=\frac{190x_{8}}{x_{8}+11}
Extract the matrix elements x and y.
2x_{8}x-22y=0
Consider the second equation. Subtract 22y from both sides.
x+y=190,2x_{8}x-22y=0
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.
2x_{8}x+2x_{8}y=2x_{8}\times 190,2x_{8}x-22y=0
To make x and 2x_{8}x equal, multiply all terms on each side of the first equation by 2x_{8} and all terms on each side of the second by 1.
2x_{8}x+2x_{8}y=380x_{8},2x_{8}x-22y=0
Simplify.
2x_{8}x+\left(-2x_{8}\right)x+2x_{8}y+22y=380x_{8}
Subtract 2x_{8}x-22y=0 from 2x_{8}x+2x_{8}y=380x_{8} by subtracting like terms on each side of the equal sign.
2x_{8}y+22y=380x_{8}
Add 2x_{8}x to -2x_{8}x. Terms 2x_{8}x and -2x_{8}x cancel out, leaving an equation with only one variable that can be solved.
\left(2x_{8}+22\right)y=380x_{8}
Add 2x_{8}y to 22y.
y=\frac{190x_{8}}{x_{8}+11}
Divide both sides by 22+2x_{8}.
2x_{8}x-22\times \frac{190x_{8}}{x_{8}+11}=0
Substitute \frac{190x_{8}}{11+x_{8}} for y in 2x_{8}x-22y=0. Because the resulting equation contains only one variable, you can solve for x directly.
2x_{8}x-\frac{4180x_{8}}{x_{8}+11}=0
Multiply -22 times \frac{190x_{8}}{11+x_{8}}.
2x_{8}x=\frac{4180x_{8}}{x_{8}+11}
Add \frac{4180x_{8}}{11+x_{8}} to both sides of the equation.
x=\frac{2090}{x_{8}+11}
Divide both sides by 2x_{8}.
x=\frac{2090}{x_{8}+11},y=\frac{190x_{8}}{x_{8}+11}
The system is now solved.
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