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3x-2y=0
Consider the first equation. Subtract 2y from both sides.
3x-2y=0,5x+4y=2200
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.
3x-2y=0
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=2y
Add 2y to both sides of the equation.
x=\frac{1}{3}\times 2y
Divide both sides by 3.
x=\frac{2}{3}y
Multiply \frac{1}{3} times 2y.
5\times \frac{2}{3}y+4y=2200
Substitute \frac{2y}{3} for x in the other equation, 5x+4y=2200.
\frac{10}{3}y+4y=2200
Multiply 5 times \frac{2y}{3}.
\frac{22}{3}y=2200
Add \frac{10y}{3} to 4y.
y=300
Divide both sides of the equation by \frac{22}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{2}{3}\times 300
Substitute 300 for y in x=\frac{2}{3}y. Because the resulting equation contains only one variable, you can solve for x directly.
x=200
Multiply \frac{2}{3} times 300.
x=200,y=300
The system is now solved.
3x-2y=0
Consider the first equation. Subtract 2y from both sides.
3x-2y=0,5x+4y=2200
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&-2\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\2200\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&-2\\5&4\end{matrix}\right))\left(\begin{matrix}3&-2\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-2\\5&4\end{matrix}\right))\left(\begin{matrix}0\\2200\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-2\\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}3&-2\\5&4\end{matrix}\right))\left(\begin{matrix}0\\2200\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}3&-2\\5&4\end{matrix}\right))\left(\begin{matrix}0\\2200\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}{3\times 4-\left(-2\times 5\right)}&-\frac{-2}{3\times 4-\left(-2\times 5\right)}\\-\frac{5}{3\times 4-\left(-2\times 5\right)}&\frac{3}{3\times 4-\left(-2\times 5\right)}\end{matrix}\right)\left(\begin{matrix}0\\2200\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}{11}&\frac{1}{11}\\-\frac{5}{22}&\frac{3}{22}\end{matrix}\right)\left(\begin{matrix}0\\2200\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{11}\times 2200\\\frac{3}{22}\times 2200\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}200\\300\end{matrix}\right)
Do the arithmetic.
x=200,y=300
Extract the matrix elements x and y.
3x-2y=0
Consider the first equation. Subtract 2y from both sides.
3x-2y=0,5x+4y=2200
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.
5\times 3x+5\left(-2\right)y=0,3\times 5x+3\times 4y=3\times 2200
To make 3x 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 3.
15x-10y=0,15x+12y=6600
Simplify.
15x-15x-10y-12y=-6600
Subtract 15x+12y=6600 from 15x-10y=0 by subtracting like terms on each side of the equal sign.
-10y-12y=-6600
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
-22y=-6600
Add -10y to -12y.
y=300
Divide both sides by -22.
5x+4\times 300=2200
Substitute 300 for y in 5x+4y=2200. Because the resulting equation contains only one variable, you can solve for x directly.
5x+1200=2200
Multiply 4 times 300.
5x=1000
Subtract 1200 from both sides of the equation.
x=200
Divide both sides by 5.
x=200,y=300
The system is now solved.