Solve {l}{8{(2x-3)}=-2/3{(2x-5)}+0}{y=-{(frac{1*{(4)}+3}{4})}-0}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 8 {(2 x - 3)} = -\frac{2}{3} {(2 x - 5)} + 0 }\\ { y = -{(\frac{1 \cdot {(4)} + 3}{4})} - 0 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 8 by 2x-3.

$$16x-24=-\frac{2}{3}\left(2x-5\right)+0$$

Use the distributive property to multiply -\frac{2}{3} by 2x-5.

$$16x-24=-\frac{4}{3}x+\frac{10}{3}+0$$

Add \frac{10}{3} and 0 to get \frac{10}{3}.

$$16x-24=-\frac{4}{3}x+\frac{10}{3}$$

Add \frac{4}{3}x to both sides.

$$16x-24+\frac{4}{3}x=\frac{10}{3}$$

Combine 16x and \frac{4}{3}x to get \frac{52}{3}x.

$$\frac{52}{3}x-24=\frac{10}{3}$$

Add 24 to both sides.

$$\frac{52}{3}x=\frac{10}{3}+24$$

Add \frac{10}{3} and 24 to get \frac{82}{3}.

$$\frac{52}{3}x=\frac{82}{3}$$

Multiply both sides by \frac{3}{52}, the reciprocal of \frac{52}{3}.

$$x=\frac{82}{3}\times \frac{3}{52}$$

Multiply \frac{82}{3} and \frac{3}{52} to get \frac{41}{26}.

$$x=\frac{41}{26}$$

Consider the second equation. Multiply 1 and 4 to get 4.

$$y=-\frac{4+3}{4}-0$$

Add 4 and 3 to get 7.

$$y=-\frac{7}{4}-0$$

Subtract 0 from -\frac{7}{4} to get -\frac{7}{4}.

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

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

$$z=-\frac{7}{4}$$

The system is now solved.

$$x=\frac{41}{26} y=-\frac{7}{4} z=-\frac{7}{4}$$