Solve x+y-(4/3z-1)+3/4xy+3/4-(-2x)]-(2/3xy+frac{3}{4}-frac{14}{3}z)-[2z+(frac{4}{3}z+1)+frac{1}{12}xy)

$$x + y - ( \frac { 4 } { 3 } z - 1 ) + \frac { 3 } { 4 } x y + \frac { 3 } { 4 } - ( - 2 x ) ] - ( \frac { 2 } { 3 } x y + \frac { 3 } { 4 } - \frac { 14 } { 3 } z ) - [ 2 z + ( \frac { 4 } { 3 } z + 1 ) + \frac { 1 } { 12 } x y )$$

Evaluate

$$\frac{3xy}{4}-\frac{4z}{3}+y+3x+\frac{7}{4}$$

Solution Steps

To find the opposite of \frac{4}{3}z-1, find the opposite of each term.

$$x+y-\frac{4}{3}z-\left(-1\right)+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

The opposite of -1 is 1.

$$x+y-\frac{4}{3}z+1+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

Convert 1 to fraction \frac{4}{4}.

$$x+y-\frac{4}{3}z+\frac{4}{4}+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

Since \frac{4}{4} and \frac{3}{4} have the same denominator, add them by adding their numerators.

$$x+y-\frac{4}{3}z+\frac{4+3}{4}+\frac{3}{4}xy-\left(-2x\right)$$

Add 4 and 3 to get 7.

$$x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy-\left(-2x\right)$$

The opposite of -2x is 2x.

$$x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy+2x$$

Combine x and 2x to get 3x.

$$3x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy$$

Expand

$$\frac{3xy}{4}-\frac{4z}{3}+y+3x+\frac{7}{4}$$

Solution Steps

To find the opposite of \frac{4}{3}z-1, find the opposite of each term.

$$x+y-\frac{4}{3}z-\left(-1\right)+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

The opposite of -1 is 1.

$$x+y-\frac{4}{3}z+1+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

Convert 1 to fraction \frac{4}{4}.

$$x+y-\frac{4}{3}z+\frac{4}{4}+\frac{3}{4}xy+\frac{3}{4}-\left(-2x\right)$$

Since \frac{4}{4} and \frac{3}{4} have the same denominator, add them by adding their numerators.

$$x+y-\frac{4}{3}z+\frac{4+3}{4}+\frac{3}{4}xy-\left(-2x\right)$$

Add 4 and 3 to get 7.

$$x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy-\left(-2x\right)$$

The opposite of -2x is 2x.

$$x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy+2x$$

Combine x and 2x to get 3x.

$$3x+y-\frac{4}{3}z+\frac{7}{4}+\frac{3}{4}xy$$