\left\{ \begin{array} { l } { 2 x + 3 y = 780 } \\ { 5 x + 4 y = 1320 } \end{array} \right.
Solve for x, y
x=120
y=180
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2x+3y=780,5x+4y=1320
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+3y=780
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-3y+780
Subtract 3y from both sides of the equation.
x=\frac{1}{2}\left(-3y+780\right)
Divide both sides by 2.
x=-\frac{3}{2}y+390
Multiply \frac{1}{2} times -3y+780.
5\left(-\frac{3}{2}y+390\right)+4y=1320
Substitute -\frac{3y}{2}+390 for x in the other equation, 5x+4y=1320.
-\frac{15}{2}y+1950+4y=1320
Multiply 5 times -\frac{3y}{2}+390.
-\frac{7}{2}y+1950=1320
Add -\frac{15y}{2} to 4y.
-\frac{7}{2}y=-630
Subtract 1950 from both sides of the equation.
y=180
Divide both sides of the equation by -\frac{7}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3}{2}\times 180+390
Substitute 180 for y in x=-\frac{3}{2}y+390. Because the resulting equation contains only one variable, you can solve for x directly.
x=-270+390
Multiply -\frac{3}{2} times 180.
x=120
Add 390 to -270.
x=120,y=180
The system is now solved.
2x+3y=780,5x+4y=1320
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}780\\1320\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&3\\5&4\end{matrix}\right))\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\5&4\end{matrix}\right))\left(\begin{matrix}780\\1320\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\5&4\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&3\\5&4\end{matrix}\right))\left(\begin{matrix}780\\1320\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&3\\5&4\end{matrix}\right))\left(\begin{matrix}780\\1320\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{4}{2\times 4-3\times 5}&-\frac{3}{2\times 4-3\times 5}\\-\frac{5}{2\times 4-3\times 5}&\frac{2}{2\times 4-3\times 5}\end{matrix}\right)\left(\begin{matrix}780\\1320\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{4}{7}&\frac{3}{7}\\\frac{5}{7}&-\frac{2}{7}\end{matrix}\right)\left(\begin{matrix}780\\1320\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{7}\times 780+\frac{3}{7}\times 1320\\\frac{5}{7}\times 780-\frac{2}{7}\times 1320\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\180\end{matrix}\right)
Do the arithmetic.
x=120,y=180
Extract the matrix elements x and y.
2x+3y=780,5x+4y=1320
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.
5\times 2x+5\times 3y=5\times 780,2\times 5x+2\times 4y=2\times 1320
To make 2x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 2.
10x+15y=3900,10x+8y=2640
Simplify.
10x-10x+15y-8y=3900-2640
Subtract 10x+8y=2640 from 10x+15y=3900 by subtracting like terms on each side of the equal sign.
15y-8y=3900-2640
Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.
7y=3900-2640
Add 15y to -8y.
7y=1260
Add 3900 to -2640.
y=180
Divide both sides by 7.
5x+4\times 180=1320
Substitute 180 for y in 5x+4y=1320. Because the resulting equation contains only one variable, you can solve for x directly.
5x+720=1320
Multiply 4 times 180.
5x=600
Subtract 720 from both sides of the equation.
x=120
Divide both sides by 5.
x=120,y=180
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
699 * 533
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
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}