Solve {l}{{(x-9)}!0-9{(5-6x)}={(4x-1)}2+5{(1+2x)}}{y=4x-{(2x+3)}{(3x-5)}-40-{(6x-1)}{(x-2)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(x - 9)} ! 0 - 9 {(5 - 6 x)} = {(4 x - 1)} 2 + 5 {(1 + 2 x)} }\\ { y = 4 x - {(2 x + 3)} {(3 x - 5)} - 40 - {(6 x - 1)} {(x - 2)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = -\frac{73}{3} = -24\frac{1}{3} \approx -24.333333333$$

Solution Steps

Consider the first equation. Use the distributive property to multiply -9 by 5-6x.

$$\left(x-9\right)!\times 0-45+54x=\left(4x-1\right)\times 2+5\left(1+2x\right)$$

Use the distributive property to multiply 4x-1 by 2.

$$\left(x-9\right)!\times 0-45+54x=8x-2+5\left(1+2x\right)$$

Use the distributive property to multiply 5 by 1+2x.

$$\left(x-9\right)!\times 0-45+54x=8x-2+5+10x$$

Add -2 and 5 to get 3.

$$\left(x-9\right)!\times 0-45+54x=8x+3+10x$$

Combine 8x and 10x to get 18x.

$$\left(x-9\right)!\times 0-45+54x=18x+3$$

Subtract 18x from both sides.

$$\left(x-9\right)!\times 0-45+54x-18x=3$$

Combine 54x and -18x to get 36x.

$$\left(x-9\right)!\times 0-45+36x=3$$

Add 45 to both sides.

$$\left(x-9\right)!\times 0+36x=3+45$$

Add 3 and 45 to get 48.

$$\left(x-9\right)!\times 0+36x=48$$

Reorder the terms.

$$36x=48$$

Divide both sides by 36.

$$x=\frac{48}{36}$$

Reduce the fraction \frac{48}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{4}{3}$$

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

$$y=4\times \frac{4}{3}-\left(2\times \frac{4}{3}+3\right)\left(3\times \frac{4}{3}-5\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Multiply 4 and \frac{4}{3} to get \frac{16}{3}.

$$y=\frac{16}{3}-\left(2\times \frac{4}{3}+3\right)\left(3\times \frac{4}{3}-5\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Multiply 2 and \frac{4}{3} to get \frac{8}{3}.

$$y=\frac{16}{3}-\left(\frac{8}{3}+3\right)\left(3\times \frac{4}{3}-5\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Add \frac{8}{3} and 3 to get \frac{17}{3}.

$$y=\frac{16}{3}-\frac{17}{3}\left(3\times \frac{4}{3}-5\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

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

$$y=\frac{16}{3}-\frac{17}{3}\left(4-5\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Subtract 5 from 4 to get -1.

$$y=\frac{16}{3}-\frac{17}{3}\left(-1\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Multiply \frac{17}{3} and -1 to get -\frac{17}{3}.

$$y=\frac{16}{3}-\left(-\frac{17}{3}\right)-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

The opposite of -\frac{17}{3} is \frac{17}{3}.

$$y=\frac{16}{3}+\frac{17}{3}-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Add \frac{16}{3} and \frac{17}{3} to get 11.

$$y=11-40-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Subtract 40 from 11 to get -29.

$$y=-29-\left(6\times \frac{4}{3}-1\right)\left(\frac{4}{3}-2\right)$$

Multiply 6 and \frac{4}{3} to get 8.

$$y=-29-\left(8-1\right)\left(\frac{4}{3}-2\right)$$

Subtract 1 from 8 to get 7.

$$y=-29-7\left(\frac{4}{3}-2\right)$$

Subtract 2 from \frac{4}{3} to get -\frac{2}{3}.

$$y=-29-7\left(-\frac{2}{3}\right)$$

Multiply 7 and -\frac{2}{3} to get -\frac{14}{3}.

$$y=-29-\left(-\frac{14}{3}\right)$$

The opposite of -\frac{14}{3} is \frac{14}{3}.

$$y=-29+\frac{14}{3}$$

Add -29 and \frac{14}{3} to get -\frac{73}{3}.

$$y=-\frac{73}{3}$$

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

$$z=-\frac{73}{3}$$

The system is now solved.

$$x=\frac{4}{3} y=-\frac{73}{3} z=-\frac{73}{3}$$