Solve {l}{{(x+3)}{(3x-1)}-{(x+2)}{(2x-1)}=(x+2)^2}{y=3{(x-1)}{(x-2)}-2{(x-2)}{(x-3)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(x + 3)} {(3 x - 1)} - {(x + 2)} {(2 x - 1)} = {(x + 2)} ^ {2} }\\ { y = 3 {(x - 1)} {(x - 2)} - 2 {(x - 2)} {(x - 3)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=24$$

Solution Steps

Consider the first equation. Use the distributive property to multiply x+3 by 3x-1 and combine like terms.

$$3x^{2}+8x-3-\left(x+2\right)\left(2x-1\right)=\left(x+2\right)^{2}$$

Use the distributive property to multiply x+2 by 2x-1 and combine like terms.

$$3x^{2}+8x-3-\left(2x^{2}+3x-2\right)=\left(x+2\right)^{2}$$

To find the opposite of 2x^{2}+3x-2, find the opposite of each term.

$$3x^{2}+8x-3-2x^{2}-3x+2=\left(x+2\right)^{2}$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}+8x-3-3x+2=\left(x+2\right)^{2}$$

Combine 8x and -3x to get 5x.

$$x^{2}+5x-3+2=\left(x+2\right)^{2}$$

Add -3 and 2 to get -1.

$$x^{2}+5x-1=\left(x+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+5x-1=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$x^{2}+5x-1-x^{2}=4x+4$$

Combine x^{2} and -x^{2} to get 0.

$$5x-1=4x+4$$

Subtract 4x from both sides.

$$5x-1-4x=4$$

Combine 5x and -4x to get x.

$$x-1=4$$

Add 1 to both sides.

$$x=4+1$$

Add 4 and 1 to get 5.

$$x=5$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=3\left(5-1\right)\left(5-2\right)-2\left(5-2\right)\left(5-3\right)$$

Subtract 1 from 5 to get 4.

$$y=3\times 4\left(5-2\right)-2\left(5-2\right)\left(5-3\right)$$

Multiply 3 and 4 to get 12.

$$y=12\left(5-2\right)-2\left(5-2\right)\left(5-3\right)$$

Subtract 2 from 5 to get 3.

$$y=12\times 3-2\left(5-2\right)\left(5-3\right)$$

Multiply 12 and 3 to get 36.

$$y=36-2\left(5-2\right)\left(5-3\right)$$

Subtract 2 from 5 to get 3.

$$y=36-2\times 3\left(5-3\right)$$

Multiply 2 and 3 to get 6.

$$y=36-6\left(5-3\right)$$

Subtract 3 from 5 to get 2.

$$y=36-6\times 2$$

Multiply 6 and 2 to get 12.

$$y=36-12$$

Subtract 12 from 36 to get 24.

$$y=24$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=24$$

The system is now solved.

$$x=5 y=24 z=24$$