Solve for y, x
x=21
y=3
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9y-6-x=0
Consider the first equation. Subtract x from both sides.
9y-x=6
Add 6 to both sides. Anything plus zero gives itself.
9y-x=6,-8y+2x=18
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.
9y-x=6
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
9y=x+6
Add x to both sides of the equation.
y=\frac{1}{9}\left(x+6\right)
Divide both sides by 9.
y=\frac{1}{9}x+\frac{2}{3}
Multiply \frac{1}{9} times x+6.
-8\left(\frac{1}{9}x+\frac{2}{3}\right)+2x=18
Substitute \frac{x}{9}+\frac{2}{3} for y in the other equation, -8y+2x=18.
-\frac{8}{9}x-\frac{16}{3}+2x=18
Multiply -8 times \frac{x}{9}+\frac{2}{3}.
\frac{10}{9}x-\frac{16}{3}=18
Add -\frac{8x}{9} to 2x.
\frac{10}{9}x=\frac{70}{3}
Add \frac{16}{3} to both sides of the equation.
x=21
Divide both sides of the equation by \frac{10}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.
y=\frac{1}{9}\times 21+\frac{2}{3}
Substitute 21 for x in y=\frac{1}{9}x+\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for y directly.
y=\frac{7+2}{3}
Multiply \frac{1}{9} times 21.
y=3
Add \frac{2}{3} to \frac{7}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
y=3,x=21
The system is now solved.
9y-6-x=0
Consider the first equation. Subtract x from both sides.
9y-x=6
Add 6 to both sides. Anything plus zero gives itself.
9y-x=6,-8y+2x=18
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}9&-1\\-8&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}6\\18\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}9&-1\\-8&2\end{matrix}\right))\left(\begin{matrix}9&-1\\-8&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}9&-1\\-8&2\end{matrix}\right))\left(\begin{matrix}6\\18\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&-1\\-8&2\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}9&-1\\-8&2\end{matrix}\right))\left(\begin{matrix}6\\18\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}9&-1\\-8&2\end{matrix}\right))\left(\begin{matrix}6\\18\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{2}{9\times 2-\left(-\left(-8\right)\right)}&-\frac{-1}{9\times 2-\left(-\left(-8\right)\right)}\\-\frac{-8}{9\times 2-\left(-\left(-8\right)\right)}&\frac{9}{9\times 2-\left(-\left(-8\right)\right)}\end{matrix}\right)\left(\begin{matrix}6\\18\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}{5}&\frac{1}{10}\\\frac{4}{5}&\frac{9}{10}\end{matrix}\right)\left(\begin{matrix}6\\18\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 6+\frac{1}{10}\times 18\\\frac{4}{5}\times 6+\frac{9}{10}\times 18\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}3\\21\end{matrix}\right)
Do the arithmetic.
y=3,x=21
Extract the matrix elements y and x.
9y-6-x=0
Consider the first equation. Subtract x from both sides.
9y-x=6
Add 6 to both sides. Anything plus zero gives itself.
9y-x=6,-8y+2x=18
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.
-8\times 9y-8\left(-1\right)x=-8\times 6,9\left(-8\right)y+9\times 2x=9\times 18
To make 9y and -8y equal, multiply all terms on each side of the first equation by -8 and all terms on each side of the second by 9.
-72y+8x=-48,-72y+18x=162
Simplify.
-72y+72y+8x-18x=-48-162
Subtract -72y+18x=162 from -72y+8x=-48 by subtracting like terms on each side of the equal sign.
8x-18x=-48-162
Add -72y to 72y. Terms -72y and 72y cancel out, leaving an equation with only one variable that can be solved.
-10x=-48-162
Add 8x to -18x.
-10x=-210
Add -48 to -162.
x=21
Divide both sides by -10.
-8y+2\times 21=18
Substitute 21 for x in -8y+2x=18. Because the resulting equation contains only one variable, you can solve for y directly.
-8y+42=18
Multiply 2 times 21.
-8y=-24
Subtract 42 from both sides of the equation.
y=3
Divide both sides by -8.
y=3,x=21
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}