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5x+6y=9500,3x+2y=4500
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.
5x+6y=9500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=-6y+9500
Subtract 6y from both sides of the equation.
x=\frac{1}{5}\left(-6y+9500\right)
Divide both sides by 5.
x=-\frac{6}{5}y+1900
Multiply \frac{1}{5} times -6y+9500.
3\left(-\frac{6}{5}y+1900\right)+2y=4500
Substitute -\frac{6y}{5}+1900 for x in the other equation, 3x+2y=4500.
-\frac{18}{5}y+5700+2y=4500
Multiply 3 times -\frac{6y}{5}+1900.
-\frac{8}{5}y+5700=4500
Add -\frac{18y}{5} to 2y.
-\frac{8}{5}y=-1200
Subtract 5700 from both sides of the equation.
y=750
Divide both sides of the equation by -\frac{8}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{6}{5}\times 750+1900
Substitute 750 for y in x=-\frac{6}{5}y+1900. Because the resulting equation contains only one variable, you can solve for x directly.
x=-900+1900
Multiply -\frac{6}{5} times 750.
x=1000
Add 1900 to -900.
x=1000,y=750
The system is now solved.
5x+6y=9500,3x+2y=4500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&6\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9500\\4500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&6\\3&2\end{matrix}\right))\left(\begin{matrix}5&6\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\3&2\end{matrix}\right))\left(\begin{matrix}9500\\4500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&6\\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}5&6\\3&2\end{matrix}\right))\left(\begin{matrix}9500\\4500\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}5&6\\3&2\end{matrix}\right))\left(\begin{matrix}9500\\4500\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}{5\times 2-6\times 3}&-\frac{6}{5\times 2-6\times 3}\\-\frac{3}{5\times 2-6\times 3}&\frac{5}{5\times 2-6\times 3}\end{matrix}\right)\left(\begin{matrix}9500\\4500\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}{4}&\frac{3}{4}\\\frac{3}{8}&-\frac{5}{8}\end{matrix}\right)\left(\begin{matrix}9500\\4500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{4}\times 9500+\frac{3}{4}\times 4500\\\frac{3}{8}\times 9500-\frac{5}{8}\times 4500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1000\\750\end{matrix}\right)
Do the arithmetic.
x=1000,y=750
Extract the matrix elements x and y.
5x+6y=9500,3x+2y=4500
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 5x+3\times 6y=3\times 9500,5\times 3x+5\times 2y=5\times 4500
To make 5x 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 5.
15x+18y=28500,15x+10y=22500
Simplify.
15x-15x+18y-10y=28500-22500
Subtract 15x+10y=22500 from 15x+18y=28500 by subtracting like terms on each side of the equal sign.
18y-10y=28500-22500
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
8y=28500-22500
Add 18y to -10y.
8y=6000
Add 28500 to -22500.
y=750
Divide both sides by 8.
3x+2\times 750=4500
Substitute 750 for y in 3x+2y=4500. Because the resulting equation contains only one variable, you can solve for x directly.
3x+1500=4500
Multiply 2 times 750.
3x=3000
Subtract 1500 from both sides of the equation.
x=1000
Divide both sides by 3.
x=1000,y=750
The system is now solved.