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x+y=25,5x+4y=115
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=25
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+25
Subtract y from both sides of the equation.
5\left(-y+25\right)+4y=115
Substitute -y+25 for x in the other equation, 5x+4y=115.
-5y+125+4y=115
Multiply 5 times -y+25.
-y+125=115
Add -5y to 4y.
-y=-10
Subtract 125 from both sides of the equation.
y=10
Divide both sides by -1.
x=-10+25
Substitute 10 for y in x=-y+25. Because the resulting equation contains only one variable, you can solve for x directly.
x=15
Add 25 to -10.
x=15,y=10
The system is now solved.
x+y=25,5x+4y=115
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\115\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\5&4\end{matrix}\right))\left(\begin{matrix}1&1\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\5&4\end{matrix}\right))\left(\begin{matrix}25\\115\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\5&4\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&4\end{matrix}\right))\left(\begin{matrix}25\\115\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&4\end{matrix}\right))\left(\begin{matrix}25\\115\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{4}{4-5}&-\frac{1}{4-5}\\-\frac{5}{4-5}&\frac{1}{4-5}\end{matrix}\right)\left(\begin{matrix}25\\115\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}-4&1\\5&-1\end{matrix}\right)\left(\begin{matrix}25\\115\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\times 25+115\\5\times 25-115\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\10\end{matrix}\right)
Do the arithmetic.
x=15,y=10
Extract the matrix elements x and y.
x+y=25,5x+4y=115
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 25,5x+4y=115
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=125,5x+4y=115
Simplify.
5x-5x+5y-4y=125-115
Subtract 5x+4y=115 from 5x+5y=125 by subtracting like terms on each side of the equal sign.
5y-4y=125-115
Add 5x to -5x. Terms 5x and -5x cancel out, leaving an equation with only one variable that can be solved.
y=125-115
Add 5y to -4y.
y=10
Add 125 to -115.
5x+4\times 10=115
Substitute 10 for y in 5x+4y=115. Because the resulting equation contains only one variable, you can solve for x directly.
5x+40=115
Multiply 4 times 10.
5x=75
Subtract 40 from both sides of the equation.
x=15
Divide both sides by 5.
x=15,y=10
The system is now solved.