Solve for x, y
x=2000
y=4000
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9x-4y=2000,7x-3y=2000
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.
9x-4y=2000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
9x=4y+2000
Add 4y to both sides of the equation.
x=\frac{1}{9}\left(4y+2000\right)
Divide both sides by 9.
x=\frac{4}{9}y+\frac{2000}{9}
Multiply \frac{1}{9} times 2000+4y.
7\left(\frac{4}{9}y+\frac{2000}{9}\right)-3y=2000
Substitute \frac{2000+4y}{9} for x in the other equation, 7x-3y=2000.
\frac{28}{9}y+\frac{14000}{9}-3y=2000
Multiply 7 times \frac{2000+4y}{9}.
\frac{1}{9}y+\frac{14000}{9}=2000
Add \frac{28y}{9} to -3y.
\frac{1}{9}y=\frac{4000}{9}
Subtract \frac{14000}{9} from both sides of the equation.
y=4000
Multiply both sides by 9.
x=\frac{4}{9}\times 4000+\frac{2000}{9}
Substitute 4000 for y in x=\frac{4}{9}y+\frac{2000}{9}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{16000+2000}{9}
Multiply \frac{4}{9} times 4000.
x=2000
Add \frac{2000}{9} to \frac{16000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=2000,y=4000
The system is now solved.
9x-4y=2000,7x-3y=2000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}9&-4\\7&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\2000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}9&-4\\7&-3\end{matrix}\right))\left(\begin{matrix}9&-4\\7&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&-4\\7&-3\end{matrix}\right))\left(\begin{matrix}2000\\2000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&-4\\7&-3\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}9&-4\\7&-3\end{matrix}\right))\left(\begin{matrix}2000\\2000\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}9&-4\\7&-3\end{matrix}\right))\left(\begin{matrix}2000\\2000\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{3}{9\left(-3\right)-\left(-4\times 7\right)}&-\frac{-4}{9\left(-3\right)-\left(-4\times 7\right)}\\-\frac{7}{9\left(-3\right)-\left(-4\times 7\right)}&\frac{9}{9\left(-3\right)-\left(-4\times 7\right)}\end{matrix}\right)\left(\begin{matrix}2000\\2000\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}-3&4\\-7&9\end{matrix}\right)\left(\begin{matrix}2000\\2000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 2000+4\times 2000\\-7\times 2000+9\times 2000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\4000\end{matrix}\right)
Do the arithmetic.
x=2000,y=4000
Extract the matrix elements x and y.
9x-4y=2000,7x-3y=2000
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.
7\times 9x+7\left(-4\right)y=7\times 2000,9\times 7x+9\left(-3\right)y=9\times 2000
To make 9x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 9.
63x-28y=14000,63x-27y=18000
Simplify.
63x-63x-28y+27y=14000-18000
Subtract 63x-27y=18000 from 63x-28y=14000 by subtracting like terms on each side of the equal sign.
-28y+27y=14000-18000
Add 63x to -63x. Terms 63x and -63x cancel out, leaving an equation with only one variable that can be solved.
-y=14000-18000
Add -28y to 27y.
-y=-4000
Add 14000 to -18000.
y=4000
Divide both sides by -1.
7x-3\times 4000=2000
Substitute 4000 for y in 7x-3y=2000. Because the resulting equation contains only one variable, you can solve for x directly.
7x-12000=2000
Multiply -3 times 4000.
7x=14000
Add 12000 to both sides of the equation.
x=2000
Divide both sides by 7.
x=2000,y=4000
The system is now solved.
Examples
Quadratic equation
{ 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}