Solve {l}{2x-2y/6=4}{15x+15y-5/3=0}

$$\left. \begin{array} { l } { \frac { 2 x - 2 y } { 6 } = 4 } \\ { \frac { 15 x + 15 y - 5 } { 3 } = 0 } \end{array} \right.$$

Solve for x, y

$$x = \frac{37}{6} = 6\frac{1}{6} \approx 6.166666667$$

$$y = -\frac{35}{6} = -5\frac{5}{6} \approx -5.833333333$$

Steps Using Substitution

Consider the first equation. Multiply both sides by 6.

$$2x-2y=4\times 6$$

Multiply 4 and 6 to get 24.

$$2x-2y=24$$

Consider the second equation. Multiply both sides by 3. Anything times zero gives zero.

$$15x+15y-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$15x+15y=5$$

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.

$$2x-2y=24,15x+15y=5$$

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

$$2x-2y=24$$

Add 2y to both sides of the equation.

$$2x=2y+24$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(2y+24\right)$$

Multiply \frac{1}{2} times 24+2y.

$$x=y+12$$

Substitute y+12 for x in the other equation, 15x+15y=5.

$$15\left(y+12\right)+15y=5$$

Multiply 15 times y+12.

$$15y+180+15y=5$$

Add 15y to 15y.

$$30y+180=5$$

Subtract 180 from both sides of the equation.

$$30y=-175$$

Divide both sides by 30.

$$y=-\frac{35}{6}$$

Substitute -\frac{35}{6} for y in x=y+12. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{35}{6}+12$$

Add 12 to -\frac{35}{6}.

$$x=\frac{37}{6}$$

The system is now solved.

$$x=\frac{37}{6},y=-\frac{35}{6}$$

Steps Using Matrices

Consider the first equation. Multiply both sides by 6.

$$2x-2y=4\times 6$$

Multiply 4 and 6 to get 24.

$$2x-2y=24$$

Consider the second equation. Multiply both sides by 3. Anything times zero gives zero.

$$15x+15y-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$15x+15y=5$$

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

$$2x-2y=24,15x+15y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-2\\15&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}24\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-2\\15&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-2\\15&15\end{matrix}\right))\left(\begin{matrix}2&-2\\15&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\15&15\end{matrix}\right))\left(\begin{matrix}24\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\15&15\end{matrix}\right))\left(\begin{matrix}24\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\15&15\end{matrix}\right))\left(\begin{matrix}24\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{2\times 15-\left(-2\times 15\right)}&-\frac{-2}{2\times 15-\left(-2\times 15\right)}\\-\frac{15}{2\times 15-\left(-2\times 15\right)}&\frac{2}{2\times 15-\left(-2\times 15\right)}\end{matrix}\right)\left(\begin{matrix}24\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}&\frac{1}{30}\\-\frac{1}{4}&\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}24\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\times 24+\frac{1}{30}\times 5\\-\frac{1}{4}\times 24+\frac{1}{30}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{37}{6}\\-\frac{35}{6}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{37}{6},y=-\frac{35}{6}$$

Steps Using Elimination

Consider the first equation. Multiply both sides by 6.

$$2x-2y=4\times 6$$

Multiply 4 and 6 to get 24.

$$2x-2y=24$$

Consider the second equation. Multiply both sides by 3. Anything times zero gives zero.

$$15x+15y-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$15x+15y=5$$

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.

$$2x-2y=24,15x+15y=5$$

To make 2x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 2.

$$15\times 2x+15\left(-2\right)y=15\times 24,2\times 15x+2\times 15y=2\times 5$$

Simplify.

$$30x-30y=360,30x+30y=10$$

Subtract 30x+30y=10 from 30x-30y=360 by subtracting like terms on each side of the equal sign.

$$30x-30x-30y-30y=360-10$$

Add 30x to -30x. Terms 30x and -30x cancel out, leaving an equation with only one variable that can be solved.

$$-30y-30y=360-10$$

Add -30y to -30y.

$$-60y=360-10$$

Add 360 to -10.

$$-60y=350$$

Divide both sides by -60.

$$y=-\frac{35}{6}$$

Substitute -\frac{35}{6} for y in 15x+15y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x+15\left(-\frac{35}{6}\right)=5$$

Multiply 15 times -\frac{35}{6}.

$$15x-\frac{175}{2}=5$$

Add \frac{175}{2} to both sides of the equation.

$$15x=\frac{185}{2}$$

Divide both sides by 15.

$$x=\frac{37}{6}$$

The system is now solved.

$$x=\frac{37}{6},y=-\frac{35}{6}$$