\left\{ \begin{array} { l } { x + y = 50 } \\ { 10 x + 20 y = 500 } \end{array} \right.
Solve for x, y
x=50
y=0
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x+y=50,10x+20y=500
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=50
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+50
Subtract y from both sides of the equation.
10\left(-y+50\right)+20y=500
Substitute -y+50 for x in the other equation, 10x+20y=500.
-10y+500+20y=500
Multiply 10 times -y+50.
10y+500=500
Add -10y to 20y.
10y=0
Subtract 500 from both sides of the equation.
y=0
Divide both sides by 10.
x=50
Substitute 0 for y in x=-y+50. Because the resulting equation contains only one variable, you can solve for x directly.
x=50,y=0
The system is now solved.
x+y=50,10x+20y=500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\10&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}1&1\\10&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\10&20\end{matrix}\right))\left(\begin{matrix}50\\500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\10&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\\10&20\end{matrix}\right))\left(\begin{matrix}50\\500\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\\10&20\end{matrix}\right))\left(\begin{matrix}50\\500\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}{20-10}&-\frac{1}{20-10}\\-\frac{10}{20-10}&\frac{1}{20-10}\end{matrix}\right)\left(\begin{matrix}50\\500\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&-\frac{1}{10}\\-1&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}50\\500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 50-\frac{1}{10}\times 500\\-50+\frac{1}{10}\times 500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\0\end{matrix}\right)
Do the arithmetic.
x=50,y=0
Extract the matrix elements x and y.
x+y=50,10x+20y=500
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.
10x+10y=10\times 50,10x+20y=500
To make x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 1.
10x+10y=500,10x+20y=500
Simplify.
10x-10x+10y-20y=500-500
Subtract 10x+20y=500 from 10x+10y=500 by subtracting like terms on each side of the equal sign.
10y-20y=500-500
Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.
-10y=500-500
Add 10y to -20y.
-10y=0
Add 500 to -500.
y=0
Divide both sides by -10.
10x=500
Substitute 0 for y in 10x+20y=500. Because the resulting equation contains only one variable, you can solve for x directly.
x=50
Divide both sides by 10.
x=50,y=0
The system is now solved.
Examples
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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
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Limits
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