\left\{ \begin{array} { l } { 3 x - 3 y = 100 } \\ { 5 x + 6 y = 460 } \end{array} \right.
Solve for x, y
x=60
y = \frac{80}{3} = 26\frac{2}{3} \approx 26.666666667
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3x-3y=100,5x+6y=460
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-3y=100
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=3y+100
Add 3y to both sides of the equation.
x=\frac{1}{3}\left(3y+100\right)
Divide both sides by 3.
x=y+\frac{100}{3}
Multiply \frac{1}{3} times 3y+100.
5\left(y+\frac{100}{3}\right)+6y=460
Substitute y+\frac{100}{3} for x in the other equation, 5x+6y=460.
5y+\frac{500}{3}+6y=460
Multiply 5 times y+\frac{100}{3}.
11y+\frac{500}{3}=460
Add 5y to 6y.
11y=\frac{880}{3}
Subtract \frac{500}{3} from both sides of the equation.
y=\frac{80}{3}
Divide both sides by 11.
x=\frac{80+100}{3}
Substitute \frac{80}{3} for y in x=y+\frac{100}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=60
Add \frac{100}{3} to \frac{80}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=60,y=\frac{80}{3}
The system is now solved.
3x-3y=100,5x+6y=460
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&-3\\5&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\460\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&-3\\5&6\end{matrix}\right))\left(\begin{matrix}3&-3\\5&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-3\\5&6\end{matrix}\right))\left(\begin{matrix}100\\460\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-3\\5&6\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&-3\\5&6\end{matrix}\right))\left(\begin{matrix}100\\460\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&-3\\5&6\end{matrix}\right))\left(\begin{matrix}100\\460\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{6}{3\times 6-\left(-3\times 5\right)}&-\frac{-3}{3\times 6-\left(-3\times 5\right)}\\-\frac{5}{3\times 6-\left(-3\times 5\right)}&\frac{3}{3\times 6-\left(-3\times 5\right)}\end{matrix}\right)\left(\begin{matrix}100\\460\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{2}{11}&\frac{1}{11}\\-\frac{5}{33}&\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}100\\460\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{11}\times 100+\frac{1}{11}\times 460\\-\frac{5}{33}\times 100+\frac{1}{11}\times 460\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\\frac{80}{3}\end{matrix}\right)
Do the arithmetic.
x=60,y=\frac{80}{3}
Extract the matrix elements x and y.
3x-3y=100,5x+6y=460
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 3x+5\left(-3\right)y=5\times 100,3\times 5x+3\times 6y=3\times 460
To make 3x 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 3.
15x-15y=500,15x+18y=1380
Simplify.
15x-15x-15y-18y=500-1380
Subtract 15x+18y=1380 from 15x-15y=500 by subtracting like terms on each side of the equal sign.
-15y-18y=500-1380
Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.
-33y=500-1380
Add -15y to -18y.
-33y=-880
Add 500 to -1380.
y=\frac{80}{3}
Divide both sides by -33.
5x+6\times \frac{80}{3}=460
Substitute \frac{80}{3} for y in 5x+6y=460. Because the resulting equation contains only one variable, you can solve for x directly.
5x+160=460
Multiply 6 times \frac{80}{3}.
5x=300
Subtract 160 from both sides of the equation.
x=60
Divide both sides by 5.
x=60,y=\frac{80}{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}