\left\{ \begin{array} { l } { x + y = 25 } \\ { 80 x + 90 y = 2150 } \end{array} \right.
Solve for x, y
x=10
y=15
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x+y=25,80x+90y=2150
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=25
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+25
Subtract y from both sides of the equation.
80\left(-y+25\right)+90y=2150
Substitute -y+25 for x in the other equation, 80x+90y=2150.
-80y+2000+90y=2150
Multiply 80 times -y+25.
10y+2000=2150
Add -80y to 90y.
10y=150
Subtract 2000 from both sides of the equation.
y=15
Divide both sides by 10.
x=-15+25
Substitute 15 for y in x=-y+25. Because the resulting equation contains only one variable, you can solve for x directly.
x=10
Add 25 to -15.
x=10,y=15
The system is now solved.
x+y=25,80x+90y=2150
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\80&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\2150\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\80&90\end{matrix}\right))\left(\begin{matrix}1&1\\80&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\80&90\end{matrix}\right))\left(\begin{matrix}25\\2150\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\80&90\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\\80&90\end{matrix}\right))\left(\begin{matrix}25\\2150\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\\80&90\end{matrix}\right))\left(\begin{matrix}25\\2150\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{90}{90-80}&-\frac{1}{90-80}\\-\frac{80}{90-80}&\frac{1}{90-80}\end{matrix}\right)\left(\begin{matrix}25\\2150\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}9&-\frac{1}{10}\\-8&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}25\\2150\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\times 25-\frac{1}{10}\times 2150\\-8\times 25+\frac{1}{10}\times 2150\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\15\end{matrix}\right)
Do the arithmetic.
x=10,y=15
Extract the matrix elements x and y.
x+y=25,80x+90y=2150
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.
80x+80y=80\times 25,80x+90y=2150
To make x and 80x equal, multiply all terms on each side of the first equation by 80 and all terms on each side of the second by 1.
80x+80y=2000,80x+90y=2150
Simplify.
80x-80x+80y-90y=2000-2150
Subtract 80x+90y=2150 from 80x+80y=2000 by subtracting like terms on each side of the equal sign.
80y-90y=2000-2150
Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.
-10y=2000-2150
Add 80y to -90y.
-10y=-150
Add 2000 to -2150.
y=15
Divide both sides by -10.
80x+90\times 15=2150
Substitute 15 for y in 80x+90y=2150. Because the resulting equation contains only one variable, you can solve for x directly.
80x+1350=2150
Multiply 90 times 15.
80x=800
Subtract 1350 from both sides of the equation.
x=10
Divide both sides by 80.
x=10,y=15
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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Linear equation
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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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