Solve {l}{7{(x-1)}=4}{y=5x-1}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 7 {(x - 1)} = 4 }\\ { y = 5 x - 1 }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b = \frac{48}{7} = 6\frac{6}{7} \approx 6.857142857$$

Solution Steps

Consider the first equation. Divide both sides by 7.

$$x-1=\frac{4}{7}$$

Add 1 to both sides.

$$x=\frac{4}{7}+1$$

Add \frac{4}{7} and 1 to get \frac{11}{7}.

$$x=\frac{11}{7}$$

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

$$y=5\times \frac{11}{7}-1$$

Multiply 5 and \frac{11}{7} to get \frac{55}{7}.

$$y=\frac{55}{7}-1$$

Subtract 1 from \frac{55}{7} to get \frac{48}{7}.

$$y=\frac{48}{7}$$

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

$$z=\frac{48}{7}$$

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

$$a=\frac{48}{7}$$

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

$$b=\frac{48}{7}$$

The system is now solved.

$$x=\frac{11}{7} y=\frac{48}{7} z=\frac{48}{7} a=\frac{48}{7} b=\frac{48}{7}$$