Solve for x, y
x=100
y=232
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2x-y=-32
Consider the first equation. Subtract y from both sides.
5x-36-2y=0
Consider the second equation. Subtract 2y from both sides.
5x-2y=36
Add 36 to both sides. Anything plus zero gives itself.
2x-y=-32,5x-2y=36
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.
2x-y=-32
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=y-32
Add y to both sides of the equation.
x=\frac{1}{2}\left(y-32\right)
Divide both sides by 2.
x=\frac{1}{2}y-16
Multiply \frac{1}{2} times y-32.
5\left(\frac{1}{2}y-16\right)-2y=36
Substitute \frac{y}{2}-16 for x in the other equation, 5x-2y=36.
\frac{5}{2}y-80-2y=36
Multiply 5 times \frac{y}{2}-16.
\frac{1}{2}y-80=36
Add \frac{5y}{2} to -2y.
\frac{1}{2}y=116
Add 80 to both sides of the equation.
y=232
Multiply both sides by 2.
x=\frac{1}{2}\times 232-16
Substitute 232 for y in x=\frac{1}{2}y-16. Because the resulting equation contains only one variable, you can solve for x directly.
x=116-16
Multiply \frac{1}{2} times 232.
x=100
Add -16 to 116.
x=100,y=232
The system is now solved.
2x-y=-32
Consider the first equation. Subtract y from both sides.
5x-36-2y=0
Consider the second equation. Subtract 2y from both sides.
5x-2y=36
Add 36 to both sides. Anything plus zero gives itself.
2x-y=-32,5x-2y=36
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-1\\5&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-32\\36\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-1\\5&-2\end{matrix}\right))\left(\begin{matrix}2&-1\\5&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\5&-2\end{matrix}\right))\left(\begin{matrix}-32\\36\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-1\\5&-2\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}2&-1\\5&-2\end{matrix}\right))\left(\begin{matrix}-32\\36\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}2&-1\\5&-2\end{matrix}\right))\left(\begin{matrix}-32\\36\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{2}{2\left(-2\right)-\left(-5\right)}&-\frac{-1}{2\left(-2\right)-\left(-5\right)}\\-\frac{5}{2\left(-2\right)-\left(-5\right)}&\frac{2}{2\left(-2\right)-\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}-32\\36\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}-2&1\\-5&2\end{matrix}\right)\left(\begin{matrix}-32\\36\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\left(-32\right)+36\\-5\left(-32\right)+2\times 36\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\232\end{matrix}\right)
Do the arithmetic.
x=100,y=232
Extract the matrix elements x and y.
2x-y=-32
Consider the first equation. Subtract y from both sides.
5x-36-2y=0
Consider the second equation. Subtract 2y from both sides.
5x-2y=36
Add 36 to both sides. Anything plus zero gives itself.
2x-y=-32,5x-2y=36
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 2x+5\left(-1\right)y=5\left(-32\right),2\times 5x+2\left(-2\right)y=2\times 36
To make 2x 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 2.
10x-5y=-160,10x-4y=72
Simplify.
10x-10x-5y+4y=-160-72
Subtract 10x-4y=72 from 10x-5y=-160 by subtracting like terms on each side of the equal sign.
-5y+4y=-160-72
Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.
-y=-160-72
Add -5y to 4y.
-y=-232
Add -160 to -72.
y=232
Divide both sides by -1.
5x-2\times 232=36
Substitute 232 for y in 5x-2y=36. Because the resulting equation contains only one variable, you can solve for x directly.
5x-464=36
Multiply -2 times 232.
5x=500
Add 464 to both sides of the equation.
x=100
Divide both sides by 5.
x=100,y=232
The system is now solved.
Examples
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{ 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}