Solve {l}{(5x-2)^2={(5x-4)}{(5x+4)}-{(12x-28)}}{y=2*2*20}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(5 x - 2)} ^ {2} = {(5 x - 4)} {(5 x + 4)} - {(12 x - 28)} }\\ { y = 2 \cdot 2 \cdot 20 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=80$$

Solution Steps

Consider the first equation. Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4=\left(5x-4\right)\left(5x+4\right)-\left(12x-28\right)$$

Consider \left(5x-4\right)\left(5x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$25x^{2}-20x+4=\left(5x\right)^{2}-16-\left(12x-28\right)$$

Expand \left(5x\right)^{2}.

$$25x^{2}-20x+4=5^{2}x^{2}-16-\left(12x-28\right)$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-20x+4=25x^{2}-16-\left(12x-28\right)$$

To find the opposite of 12x-28, find the opposite of each term.

$$25x^{2}-20x+4=25x^{2}-16-12x+28$$

Add -16 and 28 to get 12.

$$25x^{2}-20x+4=25x^{2}+12-12x$$

Subtract 25x^{2} from both sides.

$$25x^{2}-20x+4-25x^{2}=12-12x$$

Combine 25x^{2} and -25x^{2} to get 0.

$$-20x+4=12-12x$$

Add 12x to both sides.

$$-20x+4+12x=12$$

Combine -20x and 12x to get -8x.

$$-8x+4=12$$

Subtract 4 from both sides.

$$-8x=12-4$$

Subtract 4 from 12 to get 8.

$$-8x=8$$

Divide both sides by -8.

$$x=\frac{8}{-8}$$

Divide 8 by -8 to get -1.

$$x=-1$$

Consider the second equation. Multiply 2 and 2 to get 4.

$$y=4\times 20$$

Multiply 4 and 20 to get 80.

$$y=80$$

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

$$z=80$$

The system is now solved.

$$x=-1 y=80 z=80$$