Solve {l}{10x-10y=0}{6x+6y=20x+0}

$$\left. \begin{array} { l } { 10 x - 10 y = 0 } \\ { 6 x + 6 y = 20 x + 0 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the second equation. Anything plus zero gives itself.

$$6x+6y=20x$$

Subtract 20x from both sides.

$$6x+6y-20x=0$$

Combine 6x and -20x to get -14x.

$$-14x+6y=0$$

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.

$$10x-10y=0,-14x+6y=0$$

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

$$10x-10y=0$$

Add 10y to both sides of the equation.

$$10x=10y$$

Divide both sides by 10.

$$x=\frac{1}{10}\times 10y$$

Multiply \frac{1}{10} times 10y.

$$x=y$$

Substitute y for x in the other equation, -14x+6y=0.

$$-14y+6y=0$$

Add -14y to 6y.

$$-8y=0$$

Divide both sides by -8.

$$y=0$$

Substitute 0 for y in x=y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,y=0$$

Steps Using Matrices

Consider the second equation. Anything plus zero gives itself.

$$6x+6y=20x$$

Subtract 20x from both sides.

$$6x+6y-20x=0$$

Combine 6x and -20x to get -14x.

$$-14x+6y=0$$

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

$$10x-10y=0,-14x+6y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&-10\\-14&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right))\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-10\\-14&6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{10\times 6-\left(-10\left(-14\right)\right)}&-\frac{-10}{10\times 6-\left(-10\left(-14\right)\right)}\\-\frac{-14}{10\times 6-\left(-10\left(-14\right)\right)}&\frac{10}{10\times 6-\left(-10\left(-14\right)\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{40}&-\frac{1}{8}\\-\frac{7}{40}&-\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=0$$

Steps Using Elimination

Consider the second equation. Anything plus zero gives itself.

$$6x+6y=20x$$

Subtract 20x from both sides.

$$6x+6y-20x=0$$

Combine 6x and -20x to get -14x.

$$-14x+6y=0$$

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.

$$10x-10y=0,-14x+6y=0$$

To make 10x and -14x equal, multiply all terms on each side of the first equation by -14 and all terms on each side of the second by 10.

$$-14\times 10x-14\left(-10\right)y=0,10\left(-14\right)x+10\times 6y=0$$

Simplify.

$$-140x+140y=0,-140x+60y=0$$

Subtract -140x+60y=0 from -140x+140y=0 by subtracting like terms on each side of the equal sign.

$$-140x+140x+140y-60y=0$$

Add -140x to 140x. Terms -140x and 140x cancel out, leaving an equation with only one variable that can be solved.

$$140y-60y=0$$

Add 140y to -60y.

$$80y=0$$

Divide both sides by 80.

$$y=0$$

Substitute 0 for y in -14x+6y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-14x=0$$

Divide both sides by -14.

$$x=0$$

The system is now solved.

$$x=0,y=0$$