\left\{ \begin{array} { l } { x + 2 y = 300 } \\ { 2 x + y = 270 } \end{array} \right.
Solve for x, y
x=80
y=110
Graph
Share
Copied to clipboard
x+2y=300,2x+y=270
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+2y=300
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-2y+300
Subtract 2y from both sides of the equation.
2\left(-2y+300\right)+y=270
Substitute -2y+300 for x in the other equation, 2x+y=270.
-4y+600+y=270
Multiply 2 times -2y+300.
-3y+600=270
Add -4y to y.
-3y=-330
Subtract 600 from both sides of the equation.
y=110
Divide both sides by -3.
x=-2\times 110+300
Substitute 110 for y in x=-2y+300. Because the resulting equation contains only one variable, you can solve for x directly.
x=-220+300
Multiply -2 times 110.
x=80
Add 300 to -220.
x=80,y=110
The system is now solved.
x+2y=300,2x+y=270
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\270\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}300\\270\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\2&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}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}300\\270\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&2\\2&1\end{matrix}\right))\left(\begin{matrix}300\\270\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}{1-2\times 2}&-\frac{2}{1-2\times 2}\\-\frac{2}{1-2\times 2}&\frac{1}{1-2\times 2}\end{matrix}\right)\left(\begin{matrix}300\\270\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{2}{3}\\\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}300\\270\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 300+\frac{2}{3}\times 270\\\frac{2}{3}\times 300-\frac{1}{3}\times 270\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\110\end{matrix}\right)
Do the arithmetic.
x=80,y=110
Extract the matrix elements x and y.
x+2y=300,2x+y=270
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+2\times 2y=2\times 300,2x+y=270
To make x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.
2x+4y=600,2x+y=270
Simplify.
2x-2x+4y-y=600-270
Subtract 2x+y=270 from 2x+4y=600 by subtracting like terms on each side of the equal sign.
4y-y=600-270
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
3y=600-270
Add 4y to -y.
3y=330
Add 600 to -270.
y=110
Divide both sides by 3.
2x+110=270
Substitute 110 for y in 2x+y=270. Because the resulting equation contains only one variable, you can solve for x directly.
2x=160
Subtract 110 from both sides of the equation.
x=80
Divide both sides by 2.
x=80,y=110
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}