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2x-3y=6,4x-5y=96
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-3y=6
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=3y+6
Add 3y to both sides of the equation.
x=\frac{1}{2}\left(3y+6\right)
Divide both sides by 2.
x=\frac{3}{2}y+3
Multiply \frac{1}{2} times 6+3y.
4\left(\frac{3}{2}y+3\right)-5y=96
Substitute \frac{3y}{2}+3 for x in the other equation, 4x-5y=96.
6y+12-5y=96
Multiply 4 times \frac{3y}{2}+3.
y+12=96
Add 6y to -5y.
y=84
Subtract 12 from both sides of the equation.
x=\frac{3}{2}\times 84+3
Substitute 84 for y in x=\frac{3}{2}y+3. Because the resulting equation contains only one variable, you can solve for x directly.
x=126+3
Multiply \frac{3}{2} times 84.
x=129
Add 3 to 126.
x=129,y=84
The system is now solved.
2x-3y=6,4x-5y=96
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-3\\4&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\96\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-3\\4&-5\end{matrix}\right))\left(\begin{matrix}2&-3\\4&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\4&-5\end{matrix}\right))\left(\begin{matrix}6\\96\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\4&-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}2&-3\\4&-5\end{matrix}\right))\left(\begin{matrix}6\\96\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&-3\\4&-5\end{matrix}\right))\left(\begin{matrix}6\\96\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}{2\left(-5\right)-\left(-3\times 4\right)}&-\frac{-3}{2\left(-5\right)-\left(-3\times 4\right)}\\-\frac{4}{2\left(-5\right)-\left(-3\times 4\right)}&\frac{2}{2\left(-5\right)-\left(-3\times 4\right)}\end{matrix}\right)\left(\begin{matrix}6\\96\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{5}{2}&\frac{3}{2}\\-2&1\end{matrix}\right)\left(\begin{matrix}6\\96\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{2}\times 6+\frac{3}{2}\times 96\\-2\times 6+96\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}129\\84\end{matrix}\right)
Do the arithmetic.
x=129,y=84
Extract the matrix elements x and y.
2x-3y=6,4x-5y=96
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(-3\right)y=4\times 6,2\times 4x+2\left(-5\right)y=2\times 96
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-12y=24,8x-10y=192
Simplify.
8x-8x-12y+10y=24-192
Subtract 8x-10y=192 from 8x-12y=24 by subtracting like terms on each side of the equal sign.
-12y+10y=24-192
Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.
-2y=24-192
Add -12y to 10y.
-2y=-168
Add 24 to -192.
y=84
Divide both sides by -2.
4x-5\times 84=96
Substitute 84 for y in 4x-5y=96. Because the resulting equation contains only one variable, you can solve for x directly.
4x-420=96
Multiply -5 times 84.
4x=516
Add 420 to both sides of the equation.
x=129
Divide both sides by 4.
x=129,y=84
The system is now solved.