Skip to main content
Solve for x, y
Tick mark Image
Graph

Similar Problems from Web Search

Share

3x+2y=144,2x+5y=217
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=144
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-2y+144
Subtract 2y from both sides of the equation.
x=\frac{1}{3}\left(-2y+144\right)
Divide both sides by 3.
x=-\frac{2}{3}y+48
Multiply \frac{1}{3} times -2y+144.
2\left(-\frac{2}{3}y+48\right)+5y=217
Substitute -\frac{2y}{3}+48 for x in the other equation, 2x+5y=217.
-\frac{4}{3}y+96+5y=217
Multiply 2 times -\frac{2y}{3}+48.
\frac{11}{3}y+96=217
Add -\frac{4y}{3} to 5y.
\frac{11}{3}y=121
Subtract 96 from both sides of the equation.
y=33
Divide both sides of the equation by \frac{11}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{2}{3}\times 33+48
Substitute 33 for y in x=-\frac{2}{3}y+48. Because the resulting equation contains only one variable, you can solve for x directly.
x=-22+48
Multiply -\frac{2}{3} times 33.
x=26
Add 48 to -22.
x=26,y=33
The system is now solved.
3x+2y=144,2x+5y=217
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}144\\217\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\2&5\end{matrix}\right))\left(\begin{matrix}3&2\\2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\2&5\end{matrix}\right))\left(\begin{matrix}144\\217\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\2&5\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\\2&5\end{matrix}\right))\left(\begin{matrix}144\\217\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\\2&5\end{matrix}\right))\left(\begin{matrix}144\\217\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{5}{3\times 5-2\times 2}&-\frac{2}{3\times 5-2\times 2}\\-\frac{2}{3\times 5-2\times 2}&\frac{3}{3\times 5-2\times 2}\end{matrix}\right)\left(\begin{matrix}144\\217\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{5}{11}&-\frac{2}{11}\\-\frac{2}{11}&\frac{3}{11}\end{matrix}\right)\left(\begin{matrix}144\\217\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{11}\times 144-\frac{2}{11}\times 217\\-\frac{2}{11}\times 144+\frac{3}{11}\times 217\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}26\\33\end{matrix}\right)
Do the arithmetic.
x=26,y=33
Extract the matrix elements x and y.
3x+2y=144,2x+5y=217
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.
2\times 3x+2\times 2y=2\times 144,3\times 2x+3\times 5y=3\times 217
To make 3x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 3.
6x+4y=288,6x+15y=651
Simplify.
6x-6x+4y-15y=288-651
Subtract 6x+15y=651 from 6x+4y=288 by subtracting like terms on each side of the equal sign.
4y-15y=288-651
Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.
-11y=288-651
Add 4y to -15y.
-11y=-363
Add 288 to -651.
y=33
Divide both sides by -11.
2x+5\times 33=217
Substitute 33 for y in 2x+5y=217. Because the resulting equation contains only one variable, you can solve for x directly.
2x+165=217
Multiply 5 times 33.
2x=52
Subtract 165 from both sides of the equation.
x=26
Divide both sides by 2.
x=26,y=33
The system is now solved.