Solve {l}{4x-1/9{(25x-9{(-1+8x)})}=15}{y={(9-x-3)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 4 x - \frac{1}{9} {(25 x - 9 {(-1 + 8 x)})} = 15 }\\ { y = {(9 - x - 3)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{354}{83} = 4\frac{22}{83} \approx 4.265060241$$

Solution Steps

Consider the first equation. Use the distributive property to multiply -9 by -1+8x.

$$4x-\frac{1}{9}\left(25x+9-72x\right)=15$$

Combine 25x and -72x to get -47x.

$$4x-\frac{1}{9}\left(-47x+9\right)=15$$

Use the distributive property to multiply -\frac{1}{9} by -47x+9.

$$4x+\frac{47}{9}x-1=15$$

Combine 4x and \frac{47}{9}x to get \frac{83}{9}x.

$$\frac{83}{9}x-1=15$$

Add 1 to both sides.

$$\frac{83}{9}x=15+1$$

Add 15 and 1 to get 16.

$$\frac{83}{9}x=16$$

Multiply both sides by \frac{9}{83}, the reciprocal of \frac{83}{9}.

$$x=16\times \frac{9}{83}$$

Multiply 16 and \frac{9}{83} to get \frac{144}{83}.

$$x=\frac{144}{83}$$

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

$$y=9-\frac{144}{83}-3$$

Subtract \frac{144}{83} from 9 to get \frac{603}{83}.

$$y=\frac{603}{83}-3$$

Subtract 3 from \frac{603}{83} to get \frac{354}{83}.

$$y=\frac{354}{83}$$

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

$$z=\frac{354}{83}$$

The system is now solved.

$$x=\frac{144}{83} y=\frac{354}{83} z=\frac{354}{83}$$