\left\{ \begin{array} { l } { x + y = 50 } \\ { 2100 x + 2500 y = 90000 } \end{array} \right.
Solve for x, y
x = \frac{175}{2} = 87\frac{1}{2} = 87.5
y = -\frac{75}{2} = -37\frac{1}{2} = -37.5
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x+y=50,2100x+2500y=90000
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.
2100\left(-y+50\right)+2500y=90000
Substitute -y+50 for x in the other equation, 2100x+2500y=90000.
-2100y+105000+2500y=90000
Multiply 2100 times -y+50.
400y+105000=90000
Add -2100y to 2500y.
400y=-15000
Subtract 105000 from both sides of the equation.
y=-\frac{75}{2}
Divide both sides by 400.
x=-\left(-\frac{75}{2}\right)+50
Substitute -\frac{75}{2} for y in x=-y+50. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{75}{2}+50
Multiply -1 times -\frac{75}{2}.
x=\frac{175}{2}
Add 50 to \frac{75}{2}.
x=\frac{175}{2},y=-\frac{75}{2}
The system is now solved.
x+y=50,2100x+2500y=90000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\2100&2500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\90000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\2100&2500\end{matrix}\right))\left(\begin{matrix}1&1\\2100&2500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2100&2500\end{matrix}\right))\left(\begin{matrix}50\\90000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2100&2500\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\\2100&2500\end{matrix}\right))\left(\begin{matrix}50\\90000\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\\2100&2500\end{matrix}\right))\left(\begin{matrix}50\\90000\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{2500}{2500-2100}&-\frac{1}{2500-2100}\\-\frac{2100}{2500-2100}&\frac{1}{2500-2100}\end{matrix}\right)\left(\begin{matrix}50\\90000\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{25}{4}&-\frac{1}{400}\\-\frac{21}{4}&\frac{1}{400}\end{matrix}\right)\left(\begin{matrix}50\\90000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{4}\times 50-\frac{1}{400}\times 90000\\-\frac{21}{4}\times 50+\frac{1}{400}\times 90000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{175}{2}\\-\frac{75}{2}\end{matrix}\right)
Do the arithmetic.
x=\frac{175}{2},y=-\frac{75}{2}
Extract the matrix elements x and y.
x+y=50,2100x+2500y=90000
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.
2100x+2100y=2100\times 50,2100x+2500y=90000
To make x and 2100x equal, multiply all terms on each side of the first equation by 2100 and all terms on each side of the second by 1.
2100x+2100y=105000,2100x+2500y=90000
Simplify.
2100x-2100x+2100y-2500y=105000-90000
Subtract 2100x+2500y=90000 from 2100x+2100y=105000 by subtracting like terms on each side of the equal sign.
2100y-2500y=105000-90000
Add 2100x to -2100x. Terms 2100x and -2100x cancel out, leaving an equation with only one variable that can be solved.
-400y=105000-90000
Add 2100y to -2500y.
-400y=15000
Add 105000 to -90000.
y=-\frac{75}{2}
Divide both sides by -400.
2100x+2500\left(-\frac{75}{2}\right)=90000
Substitute -\frac{75}{2} for y in 2100x+2500y=90000. Because the resulting equation contains only one variable, you can solve for x directly.
2100x-93750=90000
Multiply 2500 times -\frac{75}{2}.
2100x=183750
Add 93750 to both sides of the equation.
x=\frac{175}{2}
Divide both sides by 2100.
x=\frac{175}{2},y=-\frac{75}{2}
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
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Linear equation
y = 3x + 4
Arithmetic
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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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