Solve {l}{(3x+1)^2/9x-5=x+1}{y=3x+2}{z=77}{a=x+6*2x-8}{text{Solvefor}b,c,dtext{where}}{b=y}{c=z}{d=a}

$$\left. \begin{array} { l } { \frac{{(3 x + 1)} ^ {2}}{9 x - 5} = x + 1 }\\ { y = 3 x + 2 }\\ { z = 77 }\\ { a = x + 6 \cdot 2 x - 8 }\\ { \text{Solve for } b,c,d \text{ where} } \\ { b = y }\\ { c = z }\\ { d = a } \end{array} \right.$$

Solve for x, y, z, a, b, c, d

$$b=-7$$

$$c=77$$

$$d=-47$$

Solution Steps

Consider the first equation. Variable x cannot be equal to \frac{5}{9} since division by zero is not defined. Multiply both sides of the equation by 9x-5.

$$\left(3x+1\right)^{2}=\left(9x-5\right)x+9x-5$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+6x+1=\left(9x-5\right)x+9x-5$$

Use the distributive property to multiply 9x-5 by x.

$$9x^{2}+6x+1=9x^{2}-5x+9x-5$$

Combine -5x and 9x to get 4x.

$$9x^{2}+6x+1=9x^{2}+4x-5$$

Subtract 9x^{2} from both sides.

$$9x^{2}+6x+1-9x^{2}=4x-5$$

Combine 9x^{2} and -9x^{2} to get 0.

$$6x+1=4x-5$$

Subtract 4x from both sides.

$$6x+1-4x=-5$$

Combine 6x and -4x to get 2x.

$$2x+1=-5$$

Subtract 1 from both sides.

$$2x=-5-1$$

Subtract 1 from -5 to get -6.

$$2x=-6$$

Divide both sides by 2.

$$x=\frac{-6}{2}$$

Divide -6 by 2 to get -3.

$$x=-3$$

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

$$y=3\left(-3\right)+2$$

Multiply 3 and -3 to get -9.

$$y=-9+2$$

Add -9 and 2 to get -7.

$$y=-7$$

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

$$a=-3+6\times 2\left(-3\right)-8$$

Multiply 6 and 2 to get 12.

$$a=-3+12\left(-3\right)-8$$

Multiply 12 and -3 to get -36.

$$a=-3-36-8$$

Subtract 36 from -3 to get -39.

$$a=-39-8$$

Subtract 8 from -39 to get -47.

$$a=-47$$

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

$$b=-7$$

Consider the equation (7). Insert the known values of variables into the equation.

$$d=-47$$

The system is now solved.

$$x=-3 y=-7 z=77 a=-47 b=-7 c=77 d=-47$$