Solve {l}{5/3-4x=4/7x-1}{y={(2x-1)}-4{(3-4x)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{5}{3 - 4 x} = \frac{4}{7 x - 1} }\\ { y = {(2 x - 1)} - 4 {(3 - 4 x)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=-7$$

Solution Steps

Consider the first equation. Variable x cannot be equal to any of the values \frac{1}{7},\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-3\right)\left(7x-1\right), the least common multiple of 3-4x,7x-1.

$$\left(1-7x\right)\times 5=\left(4x-3\right)\times 4$$

Use the distributive property to multiply 1-7x by 5.

$$5-35x=\left(4x-3\right)\times 4$$

Use the distributive property to multiply 4x-3 by 4.

$$5-35x=16x-12$$

Subtract 16x from both sides.

$$5-35x-16x=-12$$

Combine -35x and -16x to get -51x.

$$5-51x=-12$$

Subtract 5 from both sides.

$$-51x=-12-5$$

Subtract 5 from -12 to get -17.

$$-51x=-17$$

Divide both sides by -51.

$$x=\frac{-17}{-51}$$

Reduce the fraction \frac{-17}{-51} to lowest terms by extracting and canceling out -17.

$$x=\frac{1}{3}$$

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

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

Multiply 2 and \frac{1}{3} to get \frac{2}{3}.

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

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

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

Multiply -4 and \frac{1}{3} to get -\frac{4}{3}.

$$y=-\frac{1}{3}-4\left(3-\frac{4}{3}\right)$$

Subtract \frac{4}{3} from 3 to get \frac{5}{3}.

$$y=-\frac{1}{3}-4\times \frac{5}{3}$$

Multiply -4 and \frac{5}{3} to get -\frac{20}{3}.

$$y=-\frac{1}{3}-\frac{20}{3}$$

Subtract \frac{20}{3} from -\frac{1}{3} to get -7.

$$y=-7$$

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

$$z=-7$$

The system is now solved.

$$x=\frac{1}{3} y=-7 z=-7$$