Solve for x, y
x = \frac{800}{3} = 266\frac{2}{3} \approx 266.666666667
y = -\frac{100}{3} = -33\frac{1}{3} \approx -33.333333333
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2x+y=500,x-y=300
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.
2x+y=500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-y+500
Subtract y from both sides of the equation.
x=\frac{1}{2}\left(-y+500\right)
Divide both sides by 2.
x=-\frac{1}{2}y+250
Multiply \frac{1}{2} times -y+500.
-\frac{1}{2}y+250-y=300
Substitute -\frac{y}{2}+250 for x in the other equation, x-y=300.
-\frac{3}{2}y+250=300
Add -\frac{y}{2} to -y.
-\frac{3}{2}y=50
Subtract 250 from both sides of the equation.
y=-\frac{100}{3}
Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{2}\left(-\frac{100}{3}\right)+250
Substitute -\frac{100}{3} for y in x=-\frac{1}{2}y+250. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{50}{3}+250
Multiply -\frac{1}{2} times -\frac{100}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{800}{3}
Add 250 to \frac{50}{3}.
x=\frac{800}{3},y=-\frac{100}{3}
The system is now solved.
2x+y=500,x-y=300
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\300\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&1\\1&-1\end{matrix}\right))\left(\begin{matrix}2&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\1&-1\end{matrix}\right))\left(\begin{matrix}500\\300\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&1\\1&-1\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}2&1\\1&-1\end{matrix}\right))\left(\begin{matrix}500\\300\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}2&1\\1&-1\end{matrix}\right))\left(\begin{matrix}500\\300\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{1}{2\left(-1\right)-1}&-\frac{1}{2\left(-1\right)-1}\\-\frac{1}{2\left(-1\right)-1}&\frac{2}{2\left(-1\right)-1}\end{matrix}\right)\left(\begin{matrix}500\\300\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{1}{3}&\frac{1}{3}\\\frac{1}{3}&-\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}500\\300\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 500+\frac{1}{3}\times 300\\\frac{1}{3}\times 500-\frac{2}{3}\times 300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{800}{3}\\-\frac{100}{3}\end{matrix}\right)
Do the arithmetic.
x=\frac{800}{3},y=-\frac{100}{3}
Extract the matrix elements x and y.
2x+y=500,x-y=300
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.
2x+y=500,2x+2\left(-1\right)y=2\times 300
To make 2x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 2.
2x+y=500,2x-2y=600
Simplify.
2x-2x+y+2y=500-600
Subtract 2x-2y=600 from 2x+y=500 by subtracting like terms on each side of the equal sign.
y+2y=500-600
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
3y=500-600
Add y to 2y.
3y=-100
Add 500 to -600.
y=-\frac{100}{3}
Divide both sides by 3.
x-\left(-\frac{100}{3}\right)=300
Substitute -\frac{100}{3} for y in x-y=300. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{800}{3}
Subtract \frac{100}{3} from both sides of the equation.
x=\frac{800}{3},y=-\frac{100}{3}
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}