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x+y=111,5x+3y=27
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=111
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+111
Subtract y from both sides of the equation.
5\left(-y+111\right)+3y=27
Substitute -y+111 for x in the other equation, 5x+3y=27.
-5y+555+3y=27
Multiply 5 times -y+111.
-2y+555=27
Add -5y to 3y.
-2y=-528
Subtract 555 from both sides of the equation.
y=264
Divide both sides by -2.
x=-264+111
Substitute 264 for y in x=-y+111. Because the resulting equation contains only one variable, you can solve for x directly.
x=-153
Add 111 to -264.
x=-153,y=264
The system is now solved.
x+y=111,5x+3y=27
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}111\\27\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\5&3\end{matrix}\right))\left(\begin{matrix}1&1\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\5&3\end{matrix}\right))\left(\begin{matrix}111\\27\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\5&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&1\\5&3\end{matrix}\right))\left(\begin{matrix}111\\27\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\\5&3\end{matrix}\right))\left(\begin{matrix}111\\27\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}&-\frac{1}{3-5}\\-\frac{5}{3-5}&\frac{1}{3-5}\end{matrix}\right)\left(\begin{matrix}111\\27\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}{2}&\frac{1}{2}\\\frac{5}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}111\\27\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{2}\times 111+\frac{1}{2}\times 27\\\frac{5}{2}\times 111-\frac{1}{2}\times 27\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-153\\264\end{matrix}\right)
Do the arithmetic.
x=-153,y=264
Extract the matrix elements x and y.
x+y=111,5x+3y=27
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.
5x+5y=5\times 111,5x+3y=27
To make x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 1.
5x+5y=555,5x+3y=27
Simplify.
5x-5x+5y-3y=555-27
Subtract 5x+3y=27 from 5x+5y=555 by subtracting like terms on each side of the equal sign.
5y-3y=555-27
Add 5x to -5x. Terms 5x and -5x cancel out, leaving an equation with only one variable that can be solved.
2y=555-27
Add 5y to -3y.
2y=528
Add 555 to -27.
y=264
Divide both sides by 2.
5x+3\times 264=27
Substitute 264 for y in 5x+3y=27. Because the resulting equation contains only one variable, you can solve for x directly.
5x+792=27
Multiply 3 times 264.
5x=-765
Subtract 792 from both sides of the equation.
x=-153
Divide both sides by 5.
x=-153,y=264
The system is now solved.