\left\{ \begin{array} { l } { 3 x + 2 y = 1020 } \\ { 4 x + 3 y = 1440 } \end{array} \right.
Solve for x, y
x=180
y=240
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3x+2y=1020,4x+3y=1440
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.
3x+2y=1020
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-2y+1020
Subtract 2y from both sides of the equation.
x=\frac{1}{3}\left(-2y+1020\right)
Divide both sides by 3.
x=-\frac{2}{3}y+340
Multiply \frac{1}{3} times -2y+1020.
4\left(-\frac{2}{3}y+340\right)+3y=1440
Substitute -\frac{2y}{3}+340 for x in the other equation, 4x+3y=1440.
-\frac{8}{3}y+1360+3y=1440
Multiply 4 times -\frac{2y}{3}+340.
\frac{1}{3}y+1360=1440
Add -\frac{8y}{3} to 3y.
\frac{1}{3}y=80
Subtract 1360 from both sides of the equation.
y=240
Multiply both sides by 3.
x=-\frac{2}{3}\times 240+340
Substitute 240 for y in x=-\frac{2}{3}y+340. Because the resulting equation contains only one variable, you can solve for x directly.
x=-160+340
Multiply -\frac{2}{3} times 240.
x=180
Add 340 to -160.
x=180,y=240
The system is now solved.
3x+2y=1020,4x+3y=1440
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1020\\1440\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\4&3\end{matrix}\right))\left(\begin{matrix}3&2\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\4&3\end{matrix}\right))\left(\begin{matrix}1020\\1440\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\4&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}3&2\\4&3\end{matrix}\right))\left(\begin{matrix}1020\\1440\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}3&2\\4&3\end{matrix}\right))\left(\begin{matrix}1020\\1440\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}{3\times 3-2\times 4}&-\frac{2}{3\times 3-2\times 4}\\-\frac{4}{3\times 3-2\times 4}&\frac{3}{3\times 3-2\times 4}\end{matrix}\right)\left(\begin{matrix}1020\\1440\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&-2\\-4&3\end{matrix}\right)\left(\begin{matrix}1020\\1440\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\times 1020-2\times 1440\\-4\times 1020+3\times 1440\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}180\\240\end{matrix}\right)
Do the arithmetic.
x=180,y=240
Extract the matrix elements x and y.
3x+2y=1020,4x+3y=1440
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.
4\times 3x+4\times 2y=4\times 1020,3\times 4x+3\times 3y=3\times 1440
To make 3x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 3.
12x+8y=4080,12x+9y=4320
Simplify.
12x-12x+8y-9y=4080-4320
Subtract 12x+9y=4320 from 12x+8y=4080 by subtracting like terms on each side of the equal sign.
8y-9y=4080-4320
Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.
-y=4080-4320
Add 8y to -9y.
-y=-240
Add 4080 to -4320.
y=240
Divide both sides by -1.
4x+3\times 240=1440
Substitute 240 for y in 4x+3y=1440. Because the resulting equation contains only one variable, you can solve for x directly.
4x+720=1440
Multiply 3 times 240.
4x=720
Subtract 720 from both sides of the equation.
x=180
Divide both sides by 4.
x=180,y=240
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}