Solve {l}{-7{(x+5)}=3}{5y+4=3x}{z=5y}{a=0}{text{Solvefor}b,ctext{where}}{b=z}{c=a}

$$\left. \begin{array} { l } { -7 {(x + 5)} = 3 }\\ { 5 y + 4 = 3 x }\\ { z = 5 y }\\ { a = 0 }\\ { \text{Solve for } b,c \text{ where} } \\ { b = z }\\ { c = a } \end{array} \right.$$

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

$$b = -\frac{142}{7} = -20\frac{2}{7} \approx -20.285714286$$

$$c=0$$

Solution Steps

Consider the first equation. Divide both sides by -7.

$$x+5=-\frac{3}{7}$$

Subtract 5 from both sides.

$$x=-\frac{3}{7}-5$$

Subtract 5 from -\frac{3}{7} to get -\frac{38}{7}.

$$x=-\frac{38}{7}$$

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

$$5y+4=3\left(-\frac{38}{7}\right)$$

Multiply 3 and -\frac{38}{7} to get -\frac{114}{7}.

$$5y+4=-\frac{114}{7}$$

Subtract 4 from both sides.

$$5y=-\frac{114}{7}-4$$

Subtract 4 from -\frac{114}{7} to get -\frac{142}{7}.

$$5y=-\frac{142}{7}$$

Divide both sides by 5.

$$y=\frac{-\frac{142}{7}}{5}$$

Express \frac{-\frac{142}{7}}{5} as a single fraction.

$$y=\frac{-142}{7\times 5}$$

Multiply 7 and 5 to get 35.

$$y=\frac{-142}{35}$$

Fraction \frac{-142}{35} can be rewritten as -\frac{142}{35} by extracting the negative sign.

$$y=-\frac{142}{35}$$

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

$$z=5\left(-\frac{142}{35}\right)$$

Multiply 5 and -\frac{142}{35} to get -\frac{142}{7}.

$$z=-\frac{142}{7}$$

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

$$b=-\frac{142}{7}$$

The system is now solved.

$$x=-\frac{38}{7} y=-\frac{142}{35} z=-\frac{142}{7} a=0 b=-\frac{142}{7} c=0$$