\left\{ \begin{array} { l } { x + y = 500 } \\ { 24 x + 23 y = 118 } \end{array} \right.
Solve for x, y
x=-11382
y=11882
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x+y=500,24x+23y=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=500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+500
Subtract y from both sides of the equation.
24\left(-y+500\right)+23y=118
Substitute -y+500 for x in the other equation, 24x+23y=118.
-24y+12000+23y=118
Multiply 24 times -y+500.
-y+12000=118
Add -24y to 23y.
-y=-11882
Subtract 12000 from both sides of the equation.
y=11882
Divide both sides by -1.
x=-11882+500
Substitute 11882 for y in x=-y+500. Because the resulting equation contains only one variable, you can solve for x directly.
x=-11382
Add 500 to -11882.
x=-11382,y=11882
The system is now solved.
x+y=500,24x+23y=118
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\24&23\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\118\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\24&23\end{matrix}\right))\left(\begin{matrix}1&1\\24&23\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\24&23\end{matrix}\right))\left(\begin{matrix}500\\118\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\24&23\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\\24&23\end{matrix}\right))\left(\begin{matrix}500\\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\\24&23\end{matrix}\right))\left(\begin{matrix}500\\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{23}{23-24}&-\frac{1}{23-24}\\-\frac{24}{23-24}&\frac{1}{23-24}\end{matrix}\right)\left(\begin{matrix}500\\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}-23&1\\24&-1\end{matrix}\right)\left(\begin{matrix}500\\118\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-23\times 500+118\\24\times 500-118\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-11382\\11882\end{matrix}\right)
Do the arithmetic.
x=-11382,y=11882
Extract the matrix elements x and y.
x+y=500,24x+23y=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.
24x+24y=24\times 500,24x+23y=118
To make x and 24x equal, multiply all terms on each side of the first equation by 24 and all terms on each side of the second by 1.
24x+24y=12000,24x+23y=118
Simplify.
24x-24x+24y-23y=12000-118
Subtract 24x+23y=118 from 24x+24y=12000 by subtracting like terms on each side of the equal sign.
24y-23y=12000-118
Add 24x to -24x. Terms 24x and -24x cancel out, leaving an equation with only one variable that can be solved.
y=12000-118
Add 24y to -23y.
y=11882
Add 12000 to -118.
24x+23\times 11882=118
Substitute 11882 for y in 24x+23y=118. Because the resulting equation contains only one variable, you can solve for x directly.
24x+273286=118
Multiply 23 times 11882.
24x=-273168
Subtract 273286 from both sides of the equation.
x=-11382
Divide both sides by 24.
x=-11382,y=11882
The system is now solved.
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