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

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

Evaluate

$$\frac{xy}{12}-\frac{4z}{3}+y+3x+\frac{1}{2}$$

Solution Steps

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

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

The opposite of -1 is 1.

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

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

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

Since \frac{4}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

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

Add 4 and 1 to get 5.

$$x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\left(-2x\right)-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

The opposite of -2x is 2x.

$$x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy+2x-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

Combine x and 2x to get 3x.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

To find the opposite of \frac{2}{3}xy+\frac{3}{4}, find the opposite of each term.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\frac{2}{3}xy-\frac{3}{4}$$

Combine \frac{3}{4}xy and -\frac{2}{3}xy to get \frac{1}{12}xy.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{1}{12}xy-\frac{3}{4}$$

Since \frac{5}{4} and \frac{3}{4} have the same denominator, subtract them by subtracting their numerators.

$$3x+y-\frac{4}{3}z+\frac{5-3}{4}+\frac{1}{12}xy$$

Subtract 3 from 5 to get 2.

$$3x+y-\frac{4}{3}z+\frac{2}{4}+\frac{1}{12}xy$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$3x+y-\frac{4}{3}z+\frac{1}{2}+\frac{1}{12}xy$$

Expand

$$\frac{xy}{12}-\frac{4z}{3}+y+3x+\frac{1}{2}$$

Solution Steps

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

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

The opposite of -1 is 1.

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

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

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

Since \frac{4}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

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

Add 4 and 1 to get 5.

$$x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\left(-2x\right)-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

The opposite of -2x is 2x.

$$x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy+2x-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

Combine x and 2x to get 3x.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\left(\frac{2}{3}xy+\frac{3}{4}\right)$$

To find the opposite of \frac{2}{3}xy+\frac{3}{4}, find the opposite of each term.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{3}{4}xy-\frac{2}{3}xy-\frac{3}{4}$$

Combine \frac{3}{4}xy and -\frac{2}{3}xy to get \frac{1}{12}xy.

$$3x+y-\frac{4}{3}z+\frac{5}{4}+\frac{1}{12}xy-\frac{3}{4}$$

Since \frac{5}{4} and \frac{3}{4} have the same denominator, subtract them by subtracting their numerators.

$$3x+y-\frac{4}{3}z+\frac{5-3}{4}+\frac{1}{12}xy$$

Subtract 3 from 5 to get 2.

$$3x+y-\frac{4}{3}z+\frac{2}{4}+\frac{1}{12}xy$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$3x+y-\frac{4}{3}z+\frac{1}{2}+\frac{1}{12}xy$$