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

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

Solve for x, y, z

$$z=3$$

Solution Steps

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$-3\left(4+3\right)-0\times 8\left(2x-3\right)=-8\left(2x-5\right)+0$$

Add 4 and 3 to get 7.

$$-3\times 7-0\times 8\left(2x-3\right)=-8\left(2x-5\right)+0$$

Multiply -3 and 7 to get -21.

$$-21-0\times 8\left(2x-3\right)=-8\left(2x-5\right)+0$$

Multiply 0 and 8 to get 0.

$$-21-0\left(2x-3\right)=-8\left(2x-5\right)+0$$

Anything times zero gives zero.

$$-21-0=-8\left(2x-5\right)+0$$

Subtract 0 from -21 to get -21.

$$-21=-8\left(2x-5\right)+0$$

Use the distributive property to multiply -8 by 2x-5.

$$-21=-16x+40+0$$

Add 40 and 0 to get 40.

$$-21=-16x+40$$

Swap sides so that all variable terms are on the left hand side.

$$-16x+40=-21$$

Subtract 40 from both sides.

$$-16x=-21-40$$

Subtract 40 from -21 to get -61.

$$-16x=-61$$

Divide both sides by -16.

$$x=\frac{-61}{-16}$$

Fraction \frac{-61}{-16} can be simplified to \frac{61}{16} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{61}{16}$$

The system is now solved.

$$x=\frac{61}{16} y=3 z=3$$