Solve 3=frac{55+n_{4}*4}{20+7}

$$3 = \frac { 55 + n _ { 4 } * 4 } { 20 + 7 }$$

Solve for n_4

$$n_{4} = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

Steps for Solving Linear Equation

Add 20 and 7 to get 27.

$$3=\frac{55+n_{4}\times 4}{27}$$

Divide each term of 55+n_{4}\times 4 by 27 to get \frac{55}{27}+\frac{4}{27}n_{4}.

$$3=\frac{55}{27}+\frac{4}{27}n_{4}$$

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

$$\frac{55}{27}+\frac{4}{27}n_{4}=3$$

Subtract \frac{55}{27} from both sides.

$$\frac{4}{27}n_{4}=3-\frac{55}{27}$$

Convert 3 to fraction \frac{81}{27}.

$$\frac{4}{27}n_{4}=\frac{81}{27}-\frac{55}{27}$$

Since \frac{81}{27} and \frac{55}{27} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4}{27}n_{4}=\frac{81-55}{27}$$

Subtract 55 from 81 to get 26.

$$\frac{4}{27}n_{4}=\frac{26}{27}$$

Multiply both sides by \frac{27}{4}, the reciprocal of \frac{4}{27}.

$$n_{4}=\frac{26}{27}\times \frac{27}{4}$$

Multiply \frac{26}{27} times \frac{27}{4} by multiplying numerator times numerator and denominator times denominator.

$$n_{4}=\frac{26\times 27}{27\times 4}$$

Cancel out 27 in both numerator and denominator.

$$n_{4}=\frac{26}{4}$$

Reduce the fraction \frac{26}{4} to lowest terms by extracting and canceling out 2.

$$n_{4}=\frac{13}{2}$$