Solve {l}{text{(i)}5/2(x-1)+3=2x-1}{text{(k)}z+z/2=z/4+5}

$$\left. \begin{array} { l } { \text { (i) } \frac { 5 } { 2 } ( x - 1 ) + 3 = 2 x - 1 } \\ { \text { (k) } z + \frac { z } { 2 } = \frac { z } { 4 } + 5 } \end{array} \right.$$

Solve for x, z

$$x=-3$$

$$z=4$$

Solution Steps

Consider the first equation. Use the distributive property to multiply \frac{5}{2} by x-1.

$$\frac{5}{2}x-\frac{5}{2}+3=2x-1$$

Add -\frac{5}{2} and 3 to get \frac{1}{2}.

$$\frac{5}{2}x+\frac{1}{2}=2x-1$$

Subtract 2x from both sides.

$$\frac{5}{2}x+\frac{1}{2}-2x=-1$$

Combine \frac{5}{2}x and -2x to get \frac{1}{2}x.

$$\frac{1}{2}x+\frac{1}{2}=-1$$

Subtract \frac{1}{2} from both sides.

$$\frac{1}{2}x=-1-\frac{1}{2}$$

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

$$\frac{1}{2}x=-\frac{3}{2}$$

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$x=-\frac{3}{2}\times 2$$

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

$$x=-3$$

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$4z+2z=z+20$$

Combine 4z and 2z to get 6z.

$$6z=z+20$$

Subtract z from both sides.

$$6z-z=20$$

Combine 6z and -z to get 5z.

$$5z=20$$

Divide both sides by 5.

$$z=\frac{20}{5}$$

Divide 20 by 5 to get 4.

$$z=4$$

The system is now solved.

$$x=-3 z=4$$