\left\{ \begin{array} { l } { 3 x + 2 y = 25 } \\ { 9 x + 5 y = 64 } \end{array} \right.
Solve for x, y
x=1
y=11
Graph
Share
Copied to clipboard
3x+2y=25,9x+5y=64
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=25
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-2y+25
Subtract 2y from both sides of the equation.
x=\frac{1}{3}\left(-2y+25\right)
Divide both sides by 3.
x=-\frac{2}{3}y+\frac{25}{3}
Multiply \frac{1}{3} times -2y+25.
9\left(-\frac{2}{3}y+\frac{25}{3}\right)+5y=64
Substitute \frac{-2y+25}{3} for x in the other equation, 9x+5y=64.
-6y+75+5y=64
Multiply 9 times \frac{-2y+25}{3}.
-y+75=64
Add -6y to 5y.
-y=-11
Subtract 75 from both sides of the equation.
y=11
Divide both sides by -1.
x=-\frac{2}{3}\times 11+\frac{25}{3}
Substitute 11 for y in x=-\frac{2}{3}y+\frac{25}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-22+25}{3}
Multiply -\frac{2}{3} times 11.
x=1
Add \frac{25}{3} to -\frac{22}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=1,y=11
The system is now solved.
3x+2y=25,9x+5y=64
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\9&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\64\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\9&5\end{matrix}\right))\left(\begin{matrix}3&2\\9&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\9&5\end{matrix}\right))\left(\begin{matrix}25\\64\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\9&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\\9&5\end{matrix}\right))\left(\begin{matrix}25\\64\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\\9&5\end{matrix}\right))\left(\begin{matrix}25\\64\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 9}&-\frac{2}{3\times 5-2\times 9}\\-\frac{9}{3\times 5-2\times 9}&\frac{3}{3\times 5-2\times 9}\end{matrix}\right)\left(\begin{matrix}25\\64\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}{3}&\frac{2}{3}\\3&-1\end{matrix}\right)\left(\begin{matrix}25\\64\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{3}\times 25+\frac{2}{3}\times 64\\3\times 25-64\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\11\end{matrix}\right)
Do the arithmetic.
x=1,y=11
Extract the matrix elements x and y.
3x+2y=25,9x+5y=64
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.
9\times 3x+9\times 2y=9\times 25,3\times 9x+3\times 5y=3\times 64
To make 3x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 3.
27x+18y=225,27x+15y=192
Simplify.
27x-27x+18y-15y=225-192
Subtract 27x+15y=192 from 27x+18y=225 by subtracting like terms on each side of the equal sign.
18y-15y=225-192
Add 27x to -27x. Terms 27x and -27x cancel out, leaving an equation with only one variable that can be solved.
3y=225-192
Add 18y to -15y.
3y=33
Add 225 to -192.
y=11
Divide both sides by 3.
9x+5\times 11=64
Substitute 11 for y in 9x+5y=64. Because the resulting equation contains only one variable, you can solve for x directly.
9x+55=64
Multiply 5 times 11.
9x=9
Subtract 55 from both sides of the equation.
x=1
Divide both sides by 9.
x=1,y=11
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}