Solve {l}{3x{(1+12x)}-{(6x-1)}{(6x+1)}=2}{y=5x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 3 x {(1 + 12 x)} - {(6 x - 1)} {(6 x + 1)} = 2 }\\ { y = 5 x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 3x by 1+12x.

$$3x+36x^{2}-\left(6x-1\right)\left(6x+1\right)=2$$

Consider \left(6x-1\right)\left(6x+1\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 1.

$$3x+36x^{2}-\left(\left(6x\right)^{2}-1\right)=2$$

Expand \left(6x\right)^{2}.

$$3x+36x^{2}-\left(6^{2}x^{2}-1\right)=2$$

Calculate 6 to the power of 2 and get 36.

$$3x+36x^{2}-\left(36x^{2}-1\right)=2$$

To find the opposite of 36x^{2}-1, find the opposite of each term.

$$3x+36x^{2}-36x^{2}+1=2$$

Combine 36x^{2} and -36x^{2} to get 0.

$$3x+1=2$$

Subtract 1 from both sides.

$$3x=2-1$$

Subtract 1 from 2 to get 1.

$$3x=1$$

Divide both sides by 3.

$$x=\frac{1}{3}$$

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

$$y=5\times \frac{1}{3}$$

Multiply 5 and \frac{1}{3} to get \frac{5}{3}.

$$y=\frac{5}{3}$$

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

$$z=\frac{5}{3}$$

The system is now solved.

$$x=\frac{1}{3} y=\frac{5}{3} z=\frac{5}{3}$$