Solve {l}{6y-18x+4=0}{6x-18y+4=0}

$$\left\{ \begin{array} { l } { 6 y - 18 x + 4 = 0 } \\ { 6 x - 18 y + 4 = 0 } \end{array} \right.$$

Solve for y, x

$$x=\frac{1}{3}\approx 0.333333333$$

$$y=\frac{1}{3}\approx 0.333333333$$

Steps Using Substitution

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.

$$6y-18x+4=0,-18y+6x+4=0$$

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

$$6y-18x+4=0$$

Subtract 4 from both sides of the equation.

$$6y-18x=-4$$

Add 18x to both sides of the equation.

$$6y=18x-4$$

Divide both sides by 6.

$$y=\frac{1}{6}\left(18x-4\right)$$

Multiply \frac{1}{6} times 18x-4.

$$y=3x-\frac{2}{3}$$

Substitute 3x-\frac{2}{3} for y in the other equation, -18y+6x+4=0.

$$-18\left(3x-\frac{2}{3}\right)+6x+4=0$$

Multiply -18 times 3x-\frac{2}{3}.

$$-54x+12+6x+4=0$$

Add -54x to 6x.

$$-48x+12+4=0$$

Add 12 to 4.

$$-48x+16=0$$

Subtract 16 from both sides of the equation.

$$-48x=-16$$

Divide both sides by -48.

$$x=\frac{1}{3}$$

Substitute \frac{1}{3} for x in y=3x-\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=3\times \frac{1}{3}-\frac{2}{3}$$

Multiply 3 times \frac{1}{3}.

$$y=1-\frac{2}{3}$$

Add -\frac{2}{3} to 1.

$$y=\frac{1}{3}$$

The system is now solved.

$$y=\frac{1}{3},x=\frac{1}{3}$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$6y-18x+4=0,-18y+6x+4=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&-18\\-18&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right))\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right))\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right))\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&-18\\-18&6\end{matrix}\right))\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{6}{6\times 6-\left(-18\left(-18\right)\right)}&-\frac{-18}{6\times 6-\left(-18\left(-18\right)\right)}\\-\frac{-18}{6\times 6-\left(-18\left(-18\right)\right)}&\frac{6}{6\times 6-\left(-18\left(-18\right)\right)}\end{matrix}\right)\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{48}&-\frac{1}{16}\\-\frac{1}{16}&-\frac{1}{48}\end{matrix}\right)\left(\begin{matrix}-4\\-4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{48}\left(-4\right)-\frac{1}{16}\left(-4\right)\\-\frac{1}{16}\left(-4\right)-\frac{1}{48}\left(-4\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\\\frac{1}{3}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{1}{3},x=\frac{1}{3}$$

Steps Using Elimination

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.

$$6y-18x+4=0,-18y+6x+4=0$$

To make 6y and -18y equal, multiply all terms on each side of the first equation by -18 and all terms on each side of the second by 6.

$$-18\times 6y-18\left(-18\right)x-18\times 4=0,6\left(-18\right)y+6\times 6x+6\times 4=0$$

Simplify.

$$-108y+324x-72=0,-108y+36x+24=0$$

Subtract -108y+36x+24=0 from -108y+324x-72=0 by subtracting like terms on each side of the equal sign.

$$-108y+108y+324x-36x-72-24=0$$

Add -108y to 108y. Terms -108y and 108y cancel out, leaving an equation with only one variable that can be solved.

$$324x-36x-72-24=0$$

Add 324x to -36x.

$$288x-72-24=0$$

Add -72 to -24.

$$288x-96=0$$

Add 96 to both sides of the equation.

$$288x=96$$

Divide both sides by 288.

$$x=\frac{1}{3}$$

Substitute \frac{1}{3} for x in -18y+6x+4=0. Because the resulting equation contains only one variable, you can solve for y directly.

$$-18y+6\times \frac{1}{3}+4=0$$

Multiply 6 times \frac{1}{3}.

$$-18y+2+4=0$$

Add 2 to 4.

$$-18y+6=0$$

Subtract 6 from both sides of the equation.

$$-18y=-6$$

Divide both sides by -18.

$$y=\frac{1}{3}$$

The system is now solved.

$$y=\frac{1}{3},x=\frac{1}{3}$$