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2x-5y=10,4x+y=120
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.
2x-5y=10
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=5y+10
Add 5y to both sides of the equation.
x=\frac{1}{2}\left(5y+10\right)
Divide both sides by 2.
x=\frac{5}{2}y+5
Multiply \frac{1}{2} times 10+5y.
4\left(\frac{5}{2}y+5\right)+y=120
Substitute 5+\frac{5y}{2} for x in the other equation, 4x+y=120.
10y+20+y=120
Multiply 4 times 5+\frac{5y}{2}.
11y+20=120
Add 10y to y.
11y=100
Subtract 20 from both sides of the equation.
y=\frac{100}{11}
Divide both sides by 11.
x=\frac{5}{2}\times \frac{100}{11}+5
Substitute \frac{100}{11} for y in x=\frac{5}{2}y+5. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{250}{11}+5
Multiply \frac{5}{2} times \frac{100}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{305}{11}
Add 5 to \frac{250}{11}.
x=\frac{305}{11},y=\frac{100}{11}
The system is now solved.
2x-5y=10,4x+y=120
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-5\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\120\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}2&-5\\4&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}10\\120\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-5\\4&1\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}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}10\\120\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}2&-5\\4&1\end{matrix}\right))\left(\begin{matrix}10\\120\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{1}{2-\left(-5\times 4\right)}&-\frac{-5}{2-\left(-5\times 4\right)}\\-\frac{4}{2-\left(-5\times 4\right)}&\frac{2}{2-\left(-5\times 4\right)}\end{matrix}\right)\left(\begin{matrix}10\\120\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{1}{22}&\frac{5}{22}\\-\frac{2}{11}&\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}10\\120\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{22}\times 10+\frac{5}{22}\times 120\\-\frac{2}{11}\times 10+\frac{1}{11}\times 120\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{305}{11}\\\frac{100}{11}\end{matrix}\right)
Do the arithmetic.
x=\frac{305}{11},y=\frac{100}{11}
Extract the matrix elements x and y.
2x-5y=10,4x+y=120
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.
4\times 2x+4\left(-5\right)y=4\times 10,2\times 4x+2y=2\times 120
To make 2x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 2.
8x-20y=40,8x+2y=240
Simplify.
8x-8x-20y-2y=40-240
Subtract 8x+2y=240 from 8x-20y=40 by subtracting like terms on each side of the equal sign.
-20y-2y=40-240
Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.
-22y=40-240
Add -20y to -2y.
-22y=-200
Add 40 to -240.
y=\frac{100}{11}
Divide both sides by -22.
4x+\frac{100}{11}=120
Substitute \frac{100}{11} for y in 4x+y=120. Because the resulting equation contains only one variable, you can solve for x directly.
4x=\frac{1220}{11}
Subtract \frac{100}{11} from both sides of the equation.
x=\frac{305}{11}
Divide both sides by 4.
x=\frac{305}{11},y=\frac{100}{11}
The system is now solved.