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3x-4y=0
Consider the first equation. Subtract 4y from both sides.
3x-4y=0,5x+4y=16000
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-4y=0
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=4y
Add 4y to both sides of the equation.
x=\frac{1}{3}\times 4y
Divide both sides by 3.
x=\frac{4}{3}y
Multiply \frac{1}{3} times 4y.
5\times \frac{4}{3}y+4y=16000
Substitute \frac{4y}{3} for x in the other equation, 5x+4y=16000.
\frac{20}{3}y+4y=16000
Multiply 5 times \frac{4y}{3}.
\frac{32}{3}y=16000
Add \frac{20y}{3} to 4y.
y=1500
Divide both sides of the equation by \frac{32}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{4}{3}\times 1500
Substitute 1500 for y in x=\frac{4}{3}y. Because the resulting equation contains only one variable, you can solve for x directly.
x=2000
Multiply \frac{4}{3} times 1500.
x=2000,y=1500
The system is now solved.
3x-4y=0
Consider the first equation. Subtract 4y from both sides.
3x-4y=0,5x+4y=16000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&-4\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\16000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&-4\\5&4\end{matrix}\right))\left(\begin{matrix}3&-4\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\5&4\end{matrix}\right))\left(\begin{matrix}0\\16000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-4\\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&-4\\5&4\end{matrix}\right))\left(\begin{matrix}0\\16000\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&-4\\5&4\end{matrix}\right))\left(\begin{matrix}0\\16000\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(-4\times 5\right)}&-\frac{-4}{3\times 4-\left(-4\times 5\right)}\\-\frac{5}{3\times 4-\left(-4\times 5\right)}&\frac{3}{3\times 4-\left(-4\times 5\right)}\end{matrix}\right)\left(\begin{matrix}0\\16000\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}{8}&\frac{1}{8}\\-\frac{5}{32}&\frac{3}{32}\end{matrix}\right)\left(\begin{matrix}0\\16000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}\times 16000\\\frac{3}{32}\times 16000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\1500\end{matrix}\right)
Do the arithmetic.
x=2000,y=1500
Extract the matrix elements x and y.
3x-4y=0
Consider the first equation. Subtract 4y from both sides.
3x-4y=0,5x+4y=16000
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(-4\right)y=0,3\times 5x+3\times 4y=3\times 16000
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-20y=0,15x+12y=48000
Simplify.
15x-15x-20y-12y=-48000
Subtract 15x+12y=48000 from 15x-20y=0 by subtracting like terms on each side of the equal sign.
-20y-12y=-48000
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
-32y=-48000
Add -20y to -12y.
y=1500
Divide both sides by -32.
5x+4\times 1500=16000
Substitute 1500 for y in 5x+4y=16000. Because the resulting equation contains only one variable, you can solve for x directly.
5x+6000=16000
Multiply 4 times 1500.
5x=10000
Subtract 6000 from both sides of the equation.
x=2000
Divide both sides by 5.
x=2000,y=1500
The system is now solved.