Solve for y, x
x=24.3
y=145.8
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y-6x=0
Consider the first equation. Subtract 6x from both sides.
x+2y=315.9
Consider the second equation. Combine y and y to get 2y.
y-6x=0,2y+x=315.9
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.
y-6x=0
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
y=6x
Add 6x to both sides of the equation.
2\times 6x+x=315.9
Substitute 6x for y in the other equation, 2y+x=315.9.
12x+x=315.9
Multiply 2 times 6x.
13x=315.9
Add 12x to x.
x=24.3
Divide both sides by 13.
y=6\times 24.3
Substitute 24.3 for x in y=6x. Because the resulting equation contains only one variable, you can solve for y directly.
y=145.8
Multiply 6 times 24.3.
y=145.8,x=24.3
The system is now solved.
y-6x=0
Consider the first equation. Subtract 6x from both sides.
x+2y=315.9
Consider the second equation. Combine y and y to get 2y.
y-6x=0,2y+x=315.9
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-6\\2&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\315.9\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-6\\2&1\end{matrix}\right))\left(\begin{matrix}1&-6\\2&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&1\end{matrix}\right))\left(\begin{matrix}0\\315.9\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-6\\2&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&1\end{matrix}\right))\left(\begin{matrix}0\\315.9\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&1\end{matrix}\right))\left(\begin{matrix}0\\315.9\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-6\times 2\right)}&-\frac{-6}{1-\left(-6\times 2\right)}\\-\frac{2}{1-\left(-6\times 2\right)}&\frac{1}{1-\left(-6\times 2\right)}\end{matrix}\right)\left(\begin{matrix}0\\315.9\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}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{13}&\frac{6}{13}\\-\frac{2}{13}&\frac{1}{13}\end{matrix}\right)\left(\begin{matrix}0\\315.9\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{6}{13}\times 315.9\\\frac{1}{13}\times 315.9\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{729}{5}\\\frac{243}{10}\end{matrix}\right)
Do the arithmetic.
y=\frac{729}{5},x=\frac{243}{10}
Extract the matrix elements y and x.
y-6x=0
Consider the first equation. Subtract 6x from both sides.
x+2y=315.9
Consider the second equation. Combine y and y to get 2y.
y-6x=0,2y+x=315.9
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.
2y+2\left(-6\right)x=0,2y+x=315.9
To make y and 2y equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.
2y-12x=0,2y+x=315.9
Simplify.
2y-2y-12x-x=-315.9
Subtract 2y+x=315.9 from 2y-12x=0 by subtracting like terms on each side of the equal sign.
-12x-x=-315.9
Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.
-13x=-315.9
Add -12x to -x.
x=\frac{243}{10}
Divide both sides by -13.
2y+\frac{243}{10}=315.9
Substitute \frac{243}{10} for x in 2y+x=315.9. Because the resulting equation contains only one variable, you can solve for y directly.
2y=\frac{1458}{5}
Subtract \frac{243}{10} from both sides of the equation.
y=\frac{729}{5}
Divide both sides by 2.
y=\frac{729}{5},x=\frac{243}{10}
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}