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3x+2y=205,5x+3y=330
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=205
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-2y+205
Subtract 2y from both sides of the equation.
x=\frac{1}{3}\left(-2y+205\right)
Divide both sides by 3.
x=-\frac{2}{3}y+\frac{205}{3}
Multiply \frac{1}{3} times -2y+205.
5\left(-\frac{2}{3}y+\frac{205}{3}\right)+3y=330
Substitute \frac{-2y+205}{3} for x in the other equation, 5x+3y=330.
-\frac{10}{3}y+\frac{1025}{3}+3y=330
Multiply 5 times \frac{-2y+205}{3}.
-\frac{1}{3}y+\frac{1025}{3}=330
Add -\frac{10y}{3} to 3y.
-\frac{1}{3}y=-\frac{35}{3}
Subtract \frac{1025}{3} from both sides of the equation.
y=35
Multiply both sides by -3.
x=-\frac{2}{3}\times 35+\frac{205}{3}
Substitute 35 for y in x=-\frac{2}{3}y+\frac{205}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-70+205}{3}
Multiply -\frac{2}{3} times 35.
x=45
Add \frac{205}{3} to -\frac{70}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=45,y=35
The system is now solved.
3x+2y=205,5x+3y=330
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}205\\330\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}3&2\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}205\\330\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\5&3\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&3\end{matrix}\right))\left(\begin{matrix}205\\330\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&3\end{matrix}\right))\left(\begin{matrix}205\\330\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{3}{3\times 3-2\times 5}&-\frac{2}{3\times 3-2\times 5}\\-\frac{5}{3\times 3-2\times 5}&\frac{3}{3\times 3-2\times 5}\end{matrix}\right)\left(\begin{matrix}205\\330\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}-3&2\\5&-3\end{matrix}\right)\left(\begin{matrix}205\\330\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 205+2\times 330\\5\times 205-3\times 330\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}45\\35\end{matrix}\right)
Do the arithmetic.
x=45,y=35
Extract the matrix elements x and y.
3x+2y=205,5x+3y=330
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\times 2y=5\times 205,3\times 5x+3\times 3y=3\times 330
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=1025,15x+9y=990
Simplify.
15x-15x+10y-9y=1025-990
Subtract 15x+9y=990 from 15x+10y=1025 by subtracting like terms on each side of the equal sign.
10y-9y=1025-990
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
y=1025-990
Add 10y to -9y.
y=35
Add 1025 to -990.
5x+3\times 35=330
Substitute 35 for y in 5x+3y=330. Because the resulting equation contains only one variable, you can solve for x directly.
5x+105=330
Multiply 3 times 35.
5x=225
Subtract 105 from both sides of the equation.
x=45
Divide both sides by 5.
x=45,y=35
The system is now solved.