\left\{ \begin{array} { l } { 3 x + y = 450 } \\ { 2 x + 3 y = 650 } \end{array} \right.
Solve for x, y
x=100
y=150
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3x+y=450,2x+3y=650
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+y=450
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-y+450
Subtract y from both sides of the equation.
x=\frac{1}{3}\left(-y+450\right)
Divide both sides by 3.
x=-\frac{1}{3}y+150
Multiply \frac{1}{3} times -y+450.
2\left(-\frac{1}{3}y+150\right)+3y=650
Substitute -\frac{y}{3}+150 for x in the other equation, 2x+3y=650.
-\frac{2}{3}y+300+3y=650
Multiply 2 times -\frac{y}{3}+150.
\frac{7}{3}y+300=650
Add -\frac{2y}{3} to 3y.
\frac{7}{3}y=350
Subtract 300 from both sides of the equation.
y=150
Divide both sides of the equation by \frac{7}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{3}\times 150+150
Substitute 150 for y in x=-\frac{1}{3}y+150. Because the resulting equation contains only one variable, you can solve for x directly.
x=-50+150
Multiply -\frac{1}{3} times 150.
x=100
Add 150 to -50.
x=100,y=150
The system is now solved.
3x+y=450,2x+3y=650
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&1\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}450\\650\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&1\\2&3\end{matrix}\right))\left(\begin{matrix}3&1\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&1\\2&3\end{matrix}\right))\left(\begin{matrix}450\\650\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&1\\2&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&1\\2&3\end{matrix}\right))\left(\begin{matrix}450\\650\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&1\\2&3\end{matrix}\right))\left(\begin{matrix}450\\650\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}&-\frac{1}{3\times 3-2}\\-\frac{2}{3\times 3-2}&\frac{3}{3\times 3-2}\end{matrix}\right)\left(\begin{matrix}450\\650\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{3}{7}&-\frac{1}{7}\\-\frac{2}{7}&\frac{3}{7}\end{matrix}\right)\left(\begin{matrix}450\\650\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7}\times 450-\frac{1}{7}\times 650\\-\frac{2}{7}\times 450+\frac{3}{7}\times 650\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\150\end{matrix}\right)
Do the arithmetic.
x=100,y=150
Extract the matrix elements x and y.
3x+y=450,2x+3y=650
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.
2\times 3x+2y=2\times 450,3\times 2x+3\times 3y=3\times 650
To make 3x 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 3.
6x+2y=900,6x+9y=1950
Simplify.
6x-6x+2y-9y=900-1950
Subtract 6x+9y=1950 from 6x+2y=900 by subtracting like terms on each side of the equal sign.
2y-9y=900-1950
Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.
-7y=900-1950
Add 2y to -9y.
-7y=-1050
Add 900 to -1950.
y=150
Divide both sides by -7.
2x+3\times 150=650
Substitute 150 for y in 2x+3y=650. Because the resulting equation contains only one variable, you can solve for x directly.
2x+450=650
Multiply 3 times 150.
2x=200
Subtract 450 from both sides of the equation.
x=100
Divide both sides by 2.
x=100,y=150
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
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}