Solve z+1/4=1/4n^2+2/3+3-[1/4(n-1)^2+frac{2}{3}(n-1)+3]

$$z + \frac { 1 } { 4 } = \frac { 1 } { 4 } n ^ { 2 } + \frac { 2 } { 3 } + 3 - [ \frac { 1 } { 4 } ( n - 1 ) ^ { 2 } + \frac { 2 } { 3 } ( n - 1 ) + 3 ]$$

Solve for n

$$n=5-6z$$

Steps for Solving Linear Equation

Add \frac{2}{3} and 3 to get \frac{11}{3}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{1}{4} by n^{2}-2n+1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{2}{3} by n-1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Combine -\frac{1}{2}n and \frac{2}{3}n to get \frac{1}{6}n.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Subtract \frac{2}{3} from \frac{1}{4} to get -\frac{5}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Add -\frac{5}{12} and 3 to get \frac{31}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

To find the opposite of \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, find the opposite of each term.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Combine \frac{1}{4}n^{2} and -\frac{1}{4}n^{2} to get 0.

$$z+\frac{1}{4}=\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Subtract \frac{31}{12} from \frac{11}{3} to get \frac{13}{12}.

$$z+\frac{1}{4}=\frac{13}{12}-\frac{1}{6}n$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{13}{12}-\frac{1}{6}n=z+\frac{1}{4}$$

Subtract \frac{13}{12} from both sides.

$$-\frac{1}{6}n=z+\frac{1}{4}-\frac{13}{12}$$

Subtract \frac{13}{12} from \frac{1}{4} to get -\frac{5}{6}.

$$-\frac{1}{6}n=z-\frac{5}{6}$$

Multiply both sides by -6.

$$\frac{-\frac{1}{6}n}{-\frac{1}{6}}=\frac{z-\frac{5}{6}}{-\frac{1}{6}}$$

Dividing by -\frac{1}{6} undoes the multiplication by -\frac{1}{6}.

$$n=\frac{z-\frac{5}{6}}{-\frac{1}{6}}$$

Divide z-\frac{5}{6} by -\frac{1}{6} by multiplying z-\frac{5}{6} by the reciprocal of -\frac{1}{6}.

$$n=5-6z$$

Solve for z

$$z=\frac{5-n}{6}$$

Solution Steps

Add \frac{2}{3} and 3 to get \frac{11}{3}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{1}{4} by n^{2}-2n+1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{2}{3} by n-1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Combine -\frac{1}{2}n and \frac{2}{3}n to get \frac{1}{6}n.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Subtract \frac{2}{3} from \frac{1}{4} to get -\frac{5}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Add -\frac{5}{12} and 3 to get \frac{31}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

To find the opposite of \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, find the opposite of each term.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Combine \frac{1}{4}n^{2} and -\frac{1}{4}n^{2} to get 0.

$$z+\frac{1}{4}=\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Subtract \frac{31}{12} from \frac{11}{3} to get \frac{13}{12}.

$$z+\frac{1}{4}=\frac{13}{12}-\frac{1}{6}n$$

Subtract \frac{1}{4} from both sides.

$$z=\frac{13}{12}-\frac{1}{6}n-\frac{1}{4}$$

Subtract \frac{1}{4} from \frac{13}{12} to get \frac{5}{6}.

$$z=\frac{5}{6}-\frac{1}{6}n$$