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x+y=50,6x+5y=264
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=50
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+50
Subtract y from both sides of the equation.
6\left(-y+50\right)+5y=264
Substitute -y+50 for x in the other equation, 6x+5y=264.
-6y+300+5y=264
Multiply 6 times -y+50.
-y+300=264
Add -6y to 5y.
-y=-36
Subtract 300 from both sides of the equation.
y=36
Divide both sides by -1.
x=-36+50
Substitute 36 for y in x=-y+50. Because the resulting equation contains only one variable, you can solve for x directly.
x=14
Add 50 to -36.
x=14,y=36
The system is now solved.
x+y=50,6x+5y=264
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\6&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\264\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\6&5\end{matrix}\right))\left(\begin{matrix}1&1\\6&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\6&5\end{matrix}\right))\left(\begin{matrix}50\\264\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\6&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\\6&5\end{matrix}\right))\left(\begin{matrix}50\\264\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\\6&5\end{matrix}\right))\left(\begin{matrix}50\\264\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}{5-6}&-\frac{1}{5-6}\\-\frac{6}{5-6}&\frac{1}{5-6}\end{matrix}\right)\left(\begin{matrix}50\\264\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\\6&-1\end{matrix}\right)\left(\begin{matrix}50\\264\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\times 50+264\\6\times 50-264\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\36\end{matrix}\right)
Do the arithmetic.
x=14,y=36
Extract the matrix elements x and y.
x+y=50,6x+5y=264
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.
6x+6y=6\times 50,6x+5y=264
To make x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 1.
6x+6y=300,6x+5y=264
Simplify.
6x-6x+6y-5y=300-264
Subtract 6x+5y=264 from 6x+6y=300 by subtracting like terms on each side of the equal sign.
6y-5y=300-264
Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.
y=300-264
Add 6y to -5y.
y=36
Add 300 to -264.
6x+5\times 36=264
Substitute 36 for y in 6x+5y=264. Because the resulting equation contains only one variable, you can solve for x directly.
6x+180=264
Multiply 5 times 36.
6x=84
Subtract 180 from both sides of the equation.
x=14
Divide both sides by 6.
x=14,y=36
The system is now solved.