Solve for x, y
x=4387
y=-3287
Graph
Share
Copied to clipboard
x+y=1100,15x+20y=65
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.
x+y=1100
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+1100
Subtract y from both sides of the equation.
15\left(-y+1100\right)+20y=65
Substitute -y+1100 for x in the other equation, 15x+20y=65.
-15y+16500+20y=65
Multiply 15 times -y+1100.
5y+16500=65
Add -15y to 20y.
5y=-16435
Subtract 16500 from both sides of the equation.
y=-3287
Divide both sides by 5.
x=-\left(-3287\right)+1100
Substitute -3287 for y in x=-y+1100. Because the resulting equation contains only one variable, you can solve for x directly.
x=3287+1100
Multiply -1 times -3287.
x=4387
Add 1100 to 3287.
x=4387,y=-3287
The system is now solved.
x+y=1100,15x+20y=65
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\15&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1100\\65\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\15&20\end{matrix}\right))\left(\begin{matrix}1&1\\15&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&20\end{matrix}\right))\left(\begin{matrix}1100\\65\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\15&20\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}1&1\\15&20\end{matrix}\right))\left(\begin{matrix}1100\\65\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}1&1\\15&20\end{matrix}\right))\left(\begin{matrix}1100\\65\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{20}{20-15}&-\frac{1}{20-15}\\-\frac{15}{20-15}&\frac{1}{20-15}\end{matrix}\right)\left(\begin{matrix}1100\\65\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}4&-\frac{1}{5}\\-3&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}1100\\65\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\times 1100-\frac{1}{5}\times 65\\-3\times 1100+\frac{1}{5}\times 65\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4387\\-3287\end{matrix}\right)
Do the arithmetic.
x=4387,y=-3287
Extract the matrix elements x and y.
x+y=1100,15x+20y=65
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.
15x+15y=15\times 1100,15x+20y=65
To make x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 1.
15x+15y=16500,15x+20y=65
Simplify.
15x-15x+15y-20y=16500-65
Subtract 15x+20y=65 from 15x+15y=16500 by subtracting like terms on each side of the equal sign.
15y-20y=16500-65
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
-5y=16500-65
Add 15y to -20y.
-5y=16435
Add 16500 to -65.
y=-3287
Divide both sides by -5.
15x+20\left(-3287\right)=65
Substitute -3287 for y in 15x+20y=65. Because the resulting equation contains only one variable, you can solve for x directly.
15x-65740=65
Multiply 20 times -3287.
15x=65805
Add 65740 to both sides of the equation.
x=4387
Divide both sides by 15.
x=4387,y=-3287
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}