Solve {l}{(x+y)*2=56}{[(x+4)*(y-2)]-[x*y]=8}

$$\left\{ \begin{array} { l } { ( x + y ) \cdot 2 = 56 } \\ { [ ( x + 4 ) \cdot ( y - 2 ) ] - [ x \cdot y ] = 8 } \end{array} \right.$$

Solve for x, y

$$x=16$$

$$y=12$$

Steps Using Substitution

Consider the first equation. Divide both sides by 2.

$$x+y=\frac{56}{2}$$

Divide 56 by 2 to get 28.

$$x+y=28$$

Consider the second equation. Use the distributive property to multiply x+4 by y-2.

$$xy-2x+4y-8-xy=8$$

Combine xy and -xy to get 0.

$$-2x+4y-8=8$$

Add 8 to both sides.

$$-2x+4y=8+8$$

Add 8 and 8 to get 16.

$$-2x+4y=16$$

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.

$$x+y=28,-2x+4y=16$$

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

$$x+y=28$$

Subtract y from both sides of the equation.

$$x=-y+28$$

Substitute -y+28 for x in the other equation, -2x+4y=16.

$$-2\left(-y+28\right)+4y=16$$

Multiply -2 times -y+28.

$$2y-56+4y=16$$

Add 2y to 4y.

$$6y-56=16$$

Add 56 to both sides of the equation.

$$6y=72$$

Divide both sides by 6.

$$y=12$$

Substitute 12 for y in x=-y+28. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-12+28$$

Add 28 to -12.

$$x=16$$

The system is now solved.

$$x=16,y=12$$

Steps Using Matrices

Consider the first equation. Divide both sides by 2.

$$x+y=\frac{56}{2}$$

Divide 56 by 2 to get 28.

$$x+y=28$$

Consider the second equation. Use the distributive property to multiply x+4 by y-2.

$$xy-2x+4y-8-xy=8$$

Combine xy and -xy to get 0.

$$-2x+4y-8=8$$

Add 8 to both sides.

$$-2x+4y=8+8$$

Add 8 and 8 to get 16.

$$-2x+4y=16$$

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

$$x+y=28,-2x+4y=16$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\-2&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}28\\16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-2&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\-2&4\end{matrix}\right))\left(\begin{matrix}1&1\\-2&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-2&4\end{matrix}\right))\left(\begin{matrix}28\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-2&4\end{matrix}\right))\left(\begin{matrix}28\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-2&4\end{matrix}\right))\left(\begin{matrix}28\\16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-\left(-2\right)}&-\frac{1}{4-\left(-2\right)}\\-\frac{-2}{4-\left(-2\right)}&\frac{1}{4-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}28\\16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}&-\frac{1}{6}\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}28\\16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\times 28-\frac{1}{6}\times 16\\\frac{1}{3}\times 28+\frac{1}{6}\times 16\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=16,y=12$$

Steps Using Elimination

Consider the first equation. Divide both sides by 2.

$$x+y=\frac{56}{2}$$

Divide 56 by 2 to get 28.

$$x+y=28$$

Consider the second equation. Use the distributive property to multiply x+4 by y-2.

$$xy-2x+4y-8-xy=8$$

Combine xy and -xy to get 0.

$$-2x+4y-8=8$$

Add 8 to both sides.

$$-2x+4y=8+8$$

Add 8 and 8 to get 16.

$$-2x+4y=16$$

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.

$$x+y=28,-2x+4y=16$$

To make x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 1.

$$-2x-2y=-2\times 28,-2x+4y=16$$

Simplify.

$$-2x-2y=-56,-2x+4y=16$$

Subtract -2x+4y=16 from -2x-2y=-56 by subtracting like terms on each side of the equal sign.

$$-2x+2x-2y-4y=-56-16$$

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

$$-2y-4y=-56-16$$

Add -2y to -4y.

$$-6y=-56-16$$

Add -56 to -16.

$$-6y=-72$$

Divide both sides by -6.

$$y=12$$

Substitute 12 for y in -2x+4y=16. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x+4\times 12=16$$

Multiply 4 times 12.

$$-2x+48=16$$

Subtract 48 from both sides of the equation.

$$-2x=-32$$

Divide both sides by -2.

$$x=16$$

The system is now solved.

$$x=16,y=12$$