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2x+3y=1500,3x+2y=17000
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=1500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-3y+1500
Subtract 3y from both sides of the equation.
x=\frac{1}{2}\left(-3y+1500\right)
Divide both sides by 2.
x=-\frac{3}{2}y+750
Multiply \frac{1}{2} times -3y+1500.
3\left(-\frac{3}{2}y+750\right)+2y=17000
Substitute -\frac{3y}{2}+750 for x in the other equation, 3x+2y=17000.
-\frac{9}{2}y+2250+2y=17000
Multiply 3 times -\frac{3y}{2}+750.
-\frac{5}{2}y+2250=17000
Add -\frac{9y}{2} to 2y.
-\frac{5}{2}y=14750
Subtract 2250 from both sides of the equation.
y=-5900
Divide both sides of the equation by -\frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3}{2}\left(-5900\right)+750
Substitute -5900 for y in x=-\frac{3}{2}y+750. Because the resulting equation contains only one variable, you can solve for x directly.
x=8850+750
Multiply -\frac{3}{2} times -5900.
x=9600
Add 750 to 8850.
x=9600,y=-5900
The system is now solved.
2x+3y=1500,3x+2y=17000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&3\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1500\\17000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&3\\3&2\end{matrix}\right))\left(\begin{matrix}2&3\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\3&2\end{matrix}\right))\left(\begin{matrix}1500\\17000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\3&2\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\\3&2\end{matrix}\right))\left(\begin{matrix}1500\\17000\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\\3&2\end{matrix}\right))\left(\begin{matrix}1500\\17000\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{2}{2\times 2-3\times 3}&-\frac{3}{2\times 2-3\times 3}\\-\frac{3}{2\times 2-3\times 3}&\frac{2}{2\times 2-3\times 3}\end{matrix}\right)\left(\begin{matrix}1500\\17000\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{2}{5}&\frac{3}{5}\\\frac{3}{5}&-\frac{2}{5}\end{matrix}\right)\left(\begin{matrix}1500\\17000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5}\times 1500+\frac{3}{5}\times 17000\\\frac{3}{5}\times 1500-\frac{2}{5}\times 17000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9600\\-5900\end{matrix}\right)
Do the arithmetic.
x=9600,y=-5900
Extract the matrix elements x and y.
2x+3y=1500,3x+2y=17000
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.
3\times 2x+3\times 3y=3\times 1500,2\times 3x+2\times 2y=2\times 17000
To make 2x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 2.
6x+9y=4500,6x+4y=34000
Simplify.
6x-6x+9y-4y=4500-34000
Subtract 6x+4y=34000 from 6x+9y=4500 by subtracting like terms on each side of the equal sign.
9y-4y=4500-34000
Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.
5y=4500-34000
Add 9y to -4y.
5y=-29500
Add 4500 to -34000.
y=-5900
Divide both sides by 5.
3x+2\left(-5900\right)=17000
Substitute -5900 for y in 3x+2y=17000. Because the resulting equation contains only one variable, you can solve for x directly.
3x-11800=17000
Multiply 2 times -5900.
3x=28800
Add 11800 to both sides of the equation.
x=9600
Divide both sides by 3.
x=9600,y=-5900
The system is now solved.