Solve {l}{5{(x+2)}+3{(2x-3)}=48+4{(x+2)}+7}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 5 {(x + 2)} + 3 {(2 x - 3)} = 48 + 4 {(x + 2)} + 7 }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{62}{7} = 8\frac{6}{7} \approx 8.857142857$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 5 by x+2.

$$5x+10+3\left(2x-3\right)=48+4\left(x+2\right)+7$$

Use the distributive property to multiply 3 by 2x-3.

$$5x+10+6x-9=48+4\left(x+2\right)+7$$

Combine 5x and 6x to get 11x.

$$11x+10-9=48+4\left(x+2\right)+7$$

Subtract 9 from 10 to get 1.

$$11x+1=48+4\left(x+2\right)+7$$

Use the distributive property to multiply 4 by x+2.

$$11x+1=48+4x+8+7$$

Add 48 and 8 to get 56.

$$11x+1=56+4x+7$$

Add 56 and 7 to get 63.

$$11x+1=63+4x$$

Subtract 4x from both sides.

$$11x+1-4x=63$$

Combine 11x and -4x to get 7x.

$$7x+1=63$$

Subtract 1 from both sides.

$$7x=63-1$$

Subtract 1 from 63 to get 62.

$$7x=62$$

Divide both sides by 7.

$$x=\frac{62}{7}$$

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

$$y=\frac{62}{7}$$

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

$$z=\frac{62}{7}$$

The system is now solved.

$$x=\frac{62}{7} y=\frac{62}{7} z=\frac{62}{7}$$