\left\{ \begin{array} { l } { x + y = 9 } \\ { 100 x + 50 y = 4500 } \end{array} \right.
Solve for x, y
x=81
y=-72
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x+y=9,100x+50y=4500
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=9
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+9
Subtract y from both sides of the equation.
100\left(-y+9\right)+50y=4500
Substitute -y+9 for x in the other equation, 100x+50y=4500.
-100y+900+50y=4500
Multiply 100 times -y+9.
-50y+900=4500
Add -100y to 50y.
-50y=3600
Subtract 900 from both sides of the equation.
y=-72
Divide both sides by -50.
x=-\left(-72\right)+9
Substitute -72 for y in x=-y+9. Because the resulting equation contains only one variable, you can solve for x directly.
x=72+9
Multiply -1 times -72.
x=81
Add 9 to 72.
x=81,y=-72
The system is now solved.
x+y=9,100x+50y=4500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\100&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\\4500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\100&50\end{matrix}\right))\left(\begin{matrix}1&1\\100&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\100&50\end{matrix}\right))\left(\begin{matrix}9\\4500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\100&50\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\\100&50\end{matrix}\right))\left(\begin{matrix}9\\4500\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\\100&50\end{matrix}\right))\left(\begin{matrix}9\\4500\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{50}{50-100}&-\frac{1}{50-100}\\-\frac{100}{50-100}&\frac{1}{50-100}\end{matrix}\right)\left(\begin{matrix}9\\4500\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}-1&\frac{1}{50}\\2&-\frac{1}{50}\end{matrix}\right)\left(\begin{matrix}9\\4500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-9+\frac{1}{50}\times 4500\\2\times 9-\frac{1}{50}\times 4500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}81\\-72\end{matrix}\right)
Do the arithmetic.
x=81,y=-72
Extract the matrix elements x and y.
x+y=9,100x+50y=4500
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.
100x+100y=100\times 9,100x+50y=4500
To make x and 100x equal, multiply all terms on each side of the first equation by 100 and all terms on each side of the second by 1.
100x+100y=900,100x+50y=4500
Simplify.
100x-100x+100y-50y=900-4500
Subtract 100x+50y=4500 from 100x+100y=900 by subtracting like terms on each side of the equal sign.
100y-50y=900-4500
Add 100x to -100x. Terms 100x and -100x cancel out, leaving an equation with only one variable that can be solved.
50y=900-4500
Add 100y to -50y.
50y=-3600
Add 900 to -4500.
y=-72
Divide both sides by 50.
100x+50\left(-72\right)=4500
Substitute -72 for y in 100x+50y=4500. Because the resulting equation contains only one variable, you can solve for x directly.
100x-3600=4500
Multiply 50 times -72.
100x=8100
Add 3600 to both sides of the equation.
x=81
Divide both sides by 100.
x=81,y=-72
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}