\left\{ \begin{array} { l } { x + y = 15 } \\ { 250 x + 80 y = 2900 } \end{array} \right.
Solve for x, y
x=10
y=5
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x+y=15,250x+80y=2900
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=15
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+15
Subtract y from both sides of the equation.
250\left(-y+15\right)+80y=2900
Substitute -y+15 for x in the other equation, 250x+80y=2900.
-250y+3750+80y=2900
Multiply 250 times -y+15.
-170y+3750=2900
Add -250y to 80y.
-170y=-850
Subtract 3750 from both sides of the equation.
y=5
Divide both sides by -170.
x=-5+15
Substitute 5 for y in x=-y+15. Because the resulting equation contains only one variable, you can solve for x directly.
x=10
Add 15 to -5.
x=10,y=5
The system is now solved.
x+y=15,250x+80y=2900
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\250&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\2900\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\250&80\end{matrix}\right))\left(\begin{matrix}1&1\\250&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\250&80\end{matrix}\right))\left(\begin{matrix}15\\2900\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\250&80\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\\250&80\end{matrix}\right))\left(\begin{matrix}15\\2900\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\\250&80\end{matrix}\right))\left(\begin{matrix}15\\2900\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{80}{80-250}&-\frac{1}{80-250}\\-\frac{250}{80-250}&\frac{1}{80-250}\end{matrix}\right)\left(\begin{matrix}15\\2900\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{8}{17}&\frac{1}{170}\\\frac{25}{17}&-\frac{1}{170}\end{matrix}\right)\left(\begin{matrix}15\\2900\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{17}\times 15+\frac{1}{170}\times 2900\\\frac{25}{17}\times 15-\frac{1}{170}\times 2900\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\5\end{matrix}\right)
Do the arithmetic.
x=10,y=5
Extract the matrix elements x and y.
x+y=15,250x+80y=2900
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.
250x+250y=250\times 15,250x+80y=2900
To make x and 250x equal, multiply all terms on each side of the first equation by 250 and all terms on each side of the second by 1.
250x+250y=3750,250x+80y=2900
Simplify.
250x-250x+250y-80y=3750-2900
Subtract 250x+80y=2900 from 250x+250y=3750 by subtracting like terms on each side of the equal sign.
250y-80y=3750-2900
Add 250x to -250x. Terms 250x and -250x cancel out, leaving an equation with only one variable that can be solved.
170y=3750-2900
Add 250y to -80y.
170y=850
Add 3750 to -2900.
y=5
Divide both sides by 170.
250x+80\times 5=2900
Substitute 5 for y in 250x+80y=2900. Because the resulting equation contains only one variable, you can solve for x directly.
250x+400=2900
Multiply 80 times 5.
250x=2500
Subtract 400 from both sides of the equation.
x=10
Divide both sides by 250.
x=10,y=5
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
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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}