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

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

Solve for x, y, z

$$z=10$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 3 by x+2.

$$3x+6=4\left(x-1\right)$$

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

$$3x+6=4x-4$$

Subtract 4x from both sides.

$$3x+6-4x=-4$$

Combine 3x and -4x to get -x.

$$-x+6=-4$$

Subtract 6 from both sides.

$$-x=-4-6$$

Subtract 6 from -4 to get -10.

$$-x=-10$$

Divide both sides by -1.

$$x=\frac{-10}{-1}$$

Fraction \frac{-10}{-1} can be simplified to 10 by removing the negative sign from both the numerator and the denominator.

$$x=10$$

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

$$y=10$$

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

$$z=10$$

The system is now solved.

$$x=10 y=10 z=10$$