Solve {l}{4x-1=0}{y={(4x-1)}{(x+5)}-{(4x-1)}{(2x+3)}}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { 4 x - 1 = 0 }\\ { y = {(4 x - 1)} {(x + 5)} - {(4 x - 1)} {(2 x + 3)} }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

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

$$c=0$$

Solution Steps

Consider the first equation. Add 1 to both sides. Anything plus zero gives itself.

$$4x=1$$

Divide both sides by 4.

$$x=\frac{1}{4}$$

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

$$y=\left(4\times \frac{1}{4}-1\right)\left(\frac{1}{4}+5\right)-\left(4\times \frac{1}{4}-1\right)\left(2\times \frac{1}{4}+3\right)$$

Multiply 4 and \frac{1}{4} to get 1.

$$y=\left(1-1\right)\left(\frac{1}{4}+5\right)-\left(4\times \frac{1}{4}-1\right)\left(2\times \frac{1}{4}+3\right)$$

Subtract 1 from 1 to get 0.

$$y=0\left(\frac{1}{4}+5\right)-\left(4\times \frac{1}{4}-1\right)\left(2\times \frac{1}{4}+3\right)$$

Add \frac{1}{4} and 5 to get \frac{21}{4}.

$$y=0\times \frac{21}{4}-\left(4\times \frac{1}{4}-1\right)\left(2\times \frac{1}{4}+3\right)$$

Multiply 0 and \frac{21}{4} to get 0.

$$y=0-\left(4\times \frac{1}{4}-1\right)\left(2\times \frac{1}{4}+3\right)$$

Multiply 4 and \frac{1}{4} to get 1.

$$y=0-\left(1-1\right)\left(2\times \frac{1}{4}+3\right)$$

Subtract 1 from 1 to get 0.

$$y=0-0\left(2\times \frac{1}{4}+3\right)$$

Multiply 2 and \frac{1}{4} to get \frac{1}{2}.

$$y=0-0\left(\frac{1}{2}+3\right)$$

Add \frac{1}{2} and 3 to get \frac{7}{2}.

$$y=0-0\times \frac{7}{2}$$

Multiply 0 and \frac{7}{2} to get 0.

$$y=0-0$$

Subtracting 0 from itself leaves 0.

$$y=0$$

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

$$z=0$$

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

$$a=0$$

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

$$b=0$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=0$$

The system is now solved.

$$x=\frac{1}{4} y=0 z=0 a=0 b=0 c=0$$