Solve {l}{2^2/5(2x+y)=2y}{2y=3y+x-2(y-2)}

$$\left\{ \begin{array} { l } { \frac { 2 ^ { 2 } } { 5 } ( 2 x + y ) = 2 y } \\ { 2 y = 3 y + x - 2 ( y - 2 ) } \end{array} \right.$$

Solve for x, y

$$x=12$$

$$y=16$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 5.

$$2^{2}\left(2x+y\right)=10y$$

Calculate 2 to the power of 2 and get 4.

$$4\left(2x+y\right)=10y$$

Use the distributive property to multiply 4 by 2x+y.

$$8x+4y=10y$$

Subtract 10y from both sides.

$$8x+4y-10y=0$$

Combine 4y and -10y to get -6y.

$$8x-6y=0$$

Consider the second equation. Use the distributive property to multiply -2 by y-2.

$$2y=3y+x-2y+4$$

Combine 3y and -2y to get y.

$$2y=y+x+4$$

Subtract y from both sides.

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

Combine 2y and -y to get y.

$$y=x+4$$

Subtract x from both sides.

$$y-x=4$$

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.

$$8x-6y=0,-x+y=4$$

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

$$8x-6y=0$$

Add 6y to both sides of the equation.

$$8x=6y$$

Divide both sides by 8.

$$x=\frac{1}{8}\times 6y$$

Multiply \frac{1}{8} times 6y.

$$x=\frac{3}{4}y$$

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

$$-\frac{3}{4}y+y=4$$

Add -\frac{3y}{4} to y.

$$\frac{1}{4}y=4$$

Multiply both sides by 4.

$$y=16$$

Substitute 16 for y in x=\frac{3}{4}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{4}\times 16$$

Multiply \frac{3}{4} times 16.

$$x=12$$

The system is now solved.

$$x=12,y=16$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 5.

$$2^{2}\left(2x+y\right)=10y$$

Calculate 2 to the power of 2 and get 4.

$$4\left(2x+y\right)=10y$$

Use the distributive property to multiply 4 by 2x+y.

$$8x+4y=10y$$

Subtract 10y from both sides.

$$8x+4y-10y=0$$

Combine 4y and -10y to get -6y.

$$8x-6y=0$$

Consider the second equation. Use the distributive property to multiply -2 by y-2.

$$2y=3y+x-2y+4$$

Combine 3y and -2y to get y.

$$2y=y+x+4$$

Subtract y from both sides.

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

Combine 2y and -y to get y.

$$y=x+4$$

Subtract x from both sides.

$$y-x=4$$

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

$$8x-6y=0,-x+y=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&-6\\-1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right))\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right))\left(\begin{matrix}0\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right))\left(\begin{matrix}0\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-6\\-1&1\end{matrix}\right))\left(\begin{matrix}0\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8-\left(-6\left(-1\right)\right)}&-\frac{-6}{8-\left(-6\left(-1\right)\right)}\\-\frac{-1}{8-\left(-6\left(-1\right)\right)}&\frac{8}{8-\left(-6\left(-1\right)\right)}\end{matrix}\right)\left(\begin{matrix}0\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&3\\\frac{1}{2}&4\end{matrix}\right)\left(\begin{matrix}0\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\times 4\\4\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\16\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12,y=16$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 5.

$$2^{2}\left(2x+y\right)=10y$$

Calculate 2 to the power of 2 and get 4.

$$4\left(2x+y\right)=10y$$

Use the distributive property to multiply 4 by 2x+y.

$$8x+4y=10y$$

Subtract 10y from both sides.

$$8x+4y-10y=0$$

Combine 4y and -10y to get -6y.

$$8x-6y=0$$

Consider the second equation. Use the distributive property to multiply -2 by y-2.

$$2y=3y+x-2y+4$$

Combine 3y and -2y to get y.

$$2y=y+x+4$$

Subtract y from both sides.

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

Combine 2y and -y to get y.

$$y=x+4$$

Subtract x from both sides.

$$y-x=4$$

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.

$$8x-6y=0,-x+y=4$$

To make 8x and -x equal, multiply all terms on each side of the first equation by -1 and all terms on each side of the second by 8.

$$-8x-\left(-6y\right)=0,8\left(-1\right)x+8y=8\times 4$$

Simplify.

$$-8x+6y=0,-8x+8y=32$$

Subtract -8x+8y=32 from -8x+6y=0 by subtracting like terms on each side of the equal sign.

$$-8x+8x+6y-8y=-32$$

Add -8x to 8x. Terms -8x and 8x cancel out, leaving an equation with only one variable that can be solved.

$$6y-8y=-32$$

Add 6y to -8y.

$$-2y=-32$$

Divide both sides by -2.

$$y=16$$

Substitute 16 for y in -x+y=4. Because the resulting equation contains only one variable, you can solve for x directly.

$$-x+16=4$$

Subtract 16 from both sides of the equation.

$$-x=-12$$

Divide both sides by -1.

$$x=12$$

The system is now solved.

$$x=12,y=16$$