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5x+3y=450,3x+4y=413
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+3y=450
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=-3y+450
Subtract 3y from both sides of the equation.
x=\frac{1}{5}\left(-3y+450\right)
Divide both sides by 5.
x=-\frac{3}{5}y+90
Multiply \frac{1}{5} times -3y+450.
3\left(-\frac{3}{5}y+90\right)+4y=413
Substitute -\frac{3y}{5}+90 for x in the other equation, 3x+4y=413.
-\frac{9}{5}y+270+4y=413
Multiply 3 times -\frac{3y}{5}+90.
\frac{11}{5}y+270=413
Add -\frac{9y}{5} to 4y.
\frac{11}{5}y=143
Subtract 270 from both sides of the equation.
y=65
Divide both sides of the equation by \frac{11}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3}{5}\times 65+90
Substitute 65 for y in x=-\frac{3}{5}y+90. Because the resulting equation contains only one variable, you can solve for x directly.
x=-39+90
Multiply -\frac{3}{5} times 65.
x=51
Add 90 to -39.
x=51,y=65
The system is now solved.
5x+3y=450,3x+4y=413
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&3\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}450\\413\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&3\\3&4\end{matrix}\right))\left(\begin{matrix}5&3\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&3\\3&4\end{matrix}\right))\left(\begin{matrix}450\\413\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&3\\3&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}5&3\\3&4\end{matrix}\right))\left(\begin{matrix}450\\413\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&3\\3&4\end{matrix}\right))\left(\begin{matrix}450\\413\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}{5\times 4-3\times 3}&-\frac{3}{5\times 4-3\times 3}\\-\frac{3}{5\times 4-3\times 3}&\frac{5}{5\times 4-3\times 3}\end{matrix}\right)\left(\begin{matrix}450\\413\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{4}{11}&-\frac{3}{11}\\-\frac{3}{11}&\frac{5}{11}\end{matrix}\right)\left(\begin{matrix}450\\413\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{11}\times 450-\frac{3}{11}\times 413\\-\frac{3}{11}\times 450+\frac{5}{11}\times 413\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}51\\65\end{matrix}\right)
Do the arithmetic.
x=51,y=65
Extract the matrix elements x and y.
5x+3y=450,3x+4y=413
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 3y=3\times 450,5\times 3x+5\times 4y=5\times 413
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+9y=1350,15x+20y=2065
Simplify.
15x-15x+9y-20y=1350-2065
Subtract 15x+20y=2065 from 15x+9y=1350 by subtracting like terms on each side of the equal sign.
9y-20y=1350-2065
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
-11y=1350-2065
Add 9y to -20y.
-11y=-715
Add 1350 to -2065.
y=65
Divide both sides by -11.
3x+4\times 65=413
Substitute 65 for y in 3x+4y=413. Because the resulting equation contains only one variable, you can solve for x directly.
3x+260=413
Multiply 4 times 65.
3x=153
Subtract 260 from both sides of the equation.
x=51
Divide both sides by 3.
x=51,y=65
The system is now solved.