Solve {l}{y=2x*3+8sqrt{4}}{2y=2x+4}

$$\left\{ \begin{array} { l } { y = 2 x \cdot 3 + 8 \sqrt { 4 } } \\ { 2 y = 2 x + 4 } \end{array} \right.$$

Solve for y, x

$$x = -\frac{14}{5} = -2\frac{4}{5} = -2.8$$

$$y=-\frac{4}{5}=-0.8$$

Steps Using Substitution

Consider the first equation. Multiply 2 and 3 to get 6.

$$y=6x+8\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$y=6x+8\times 2$$

Multiply 8 and 2 to get 16.

$$y=6x+16$$

Substitute 6x+16 for y in the other equation, 2y-2x=4.

$$2\left(6x+16\right)-2x=4$$

Multiply 2 times 6x+16.

$$12x+32-2x=4$$

Add 12x to -2x.

$$10x+32=4$$

Subtract 32 from both sides of the equation.

$$10x=-28$$

Divide both sides by 10.

$$x=-\frac{14}{5}$$

Substitute -\frac{14}{5} for x in y=6x+16. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=6\left(-\frac{14}{5}\right)+16$$

Multiply 6 times -\frac{14}{5}.

$$y=-\frac{84}{5}+16$$

Add 16 to -\frac{84}{5}.

$$y=-\frac{4}{5}$$

The system is now solved.

$$y=-\frac{4}{5},x=-\frac{14}{5}$$

Steps Using Matrices

Consider the first equation. Multiply 2 and 3 to get 6.

$$y=6x+8\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$y=6x+8\times 2$$

Multiply 8 and 2 to get 16.

$$y=6x+16$$

Subtract 6x from both sides.

$$y-6x=16$$

Consider the second equation. Subtract 2x from both sides.

$$2y-2x=4$$

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

$$y-6x=16,2y-2x=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}16\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-6\\2&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right))\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right))\left(\begin{matrix}16\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right))\left(\begin{matrix}16\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-6\\2&-2\end{matrix}\right))\left(\begin{matrix}16\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-\left(-6\times 2\right)}&-\frac{-6}{-2-\left(-6\times 2\right)}\\-\frac{2}{-2-\left(-6\times 2\right)}&\frac{1}{-2-\left(-6\times 2\right)}\end{matrix}\right)\left(\begin{matrix}16\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{3}{5}\\-\frac{1}{5}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}16\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 16+\frac{3}{5}\times 4\\-\frac{1}{5}\times 16+\frac{1}{10}\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5}\\-\frac{14}{5}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-\frac{4}{5},x=-\frac{14}{5}$$

Steps Using Elimination

Consider the first equation. Multiply 2 and 3 to get 6.

$$y=6x+8\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$y=6x+8\times 2$$

Multiply 8 and 2 to get 16.

$$y=6x+16$$

Subtract 6x from both sides.

$$y-6x=16$$

Consider the second equation. Subtract 2x from both sides.

$$2y-2x=4$$

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.

$$y-6x=16,2y-2x=4$$

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.

$$2y+2\left(-6\right)x=2\times 16,2y-2x=4$$

Simplify.

$$2y-12x=32,2y-2x=4$$

Subtract 2y-2x=4 from 2y-12x=32 by subtracting like terms on each side of the equal sign.

$$2y-2y-12x+2x=32-4$$

Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.

$$-12x+2x=32-4$$

Add -12x to 2x.

$$-10x=32-4$$

Add 32 to -4.

$$-10x=28$$

Divide both sides by -10.

$$x=-\frac{14}{5}$$

Substitute -\frac{14}{5} for x in 2y-2x=4. Because the resulting equation contains only one variable, you can solve for y directly.

$$2y-2\left(-\frac{14}{5}\right)=4$$

Multiply -2 times -\frac{14}{5}.

$$2y+\frac{28}{5}=4$$

Subtract \frac{28}{5} from both sides of the equation.

$$2y=-\frac{8}{5}$$

Divide both sides by 2.

$$y=-\frac{4}{5}$$

The system is now solved.

$$y=-\frac{4}{5},x=-\frac{14}{5}$$