Solve 3/5(n-4)-1/3(2n-9)=1/4(n-1)-2

$$\frac { 3 } { 5 } ( n - 4 ) - \frac { 1 } { 3 } ( 2 n - 9 ) = \frac { 1 } { 4 } ( n - 1 ) - 2$$

Solve for n

$$n=9$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{3}{5} by n-4.

$$\frac{3}{5}n+\frac{3}{5}\left(-4\right)-\frac{1}{3}\left(2n-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Express \frac{3}{5}\left(-4\right) as a single fraction.

$$\frac{3}{5}n+\frac{3\left(-4\right)}{5}-\frac{1}{3}\left(2n-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Multiply 3 and -4 to get -12.

$$\frac{3}{5}n+\frac{-12}{5}-\frac{1}{3}\left(2n-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Fraction \frac{-12}{5} can be rewritten as -\frac{12}{5} by extracting the negative sign.

$$\frac{3}{5}n-\frac{12}{5}-\frac{1}{3}\left(2n-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Use the distributive property to multiply -\frac{1}{3} by 2n-9.

$$\frac{3}{5}n-\frac{12}{5}-\frac{1}{3}\times 2n-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Express -\frac{1}{3}\times 2 as a single fraction.

$$\frac{3}{5}n-\frac{12}{5}+\frac{-2}{3}n-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Fraction \frac{-2}{3} can be rewritten as -\frac{2}{3} by extracting the negative sign.

$$\frac{3}{5}n-\frac{12}{5}-\frac{2}{3}n-\frac{1}{3}\left(-9\right)=\frac{1}{4}\left(n-1\right)-2$$

Express -\frac{1}{3}\left(-9\right) as a single fraction.

$$\frac{3}{5}n-\frac{12}{5}-\frac{2}{3}n+\frac{-\left(-9\right)}{3}=\frac{1}{4}\left(n-1\right)-2$$

Multiply -1 and -9 to get 9.

$$\frac{3}{5}n-\frac{12}{5}-\frac{2}{3}n+\frac{9}{3}=\frac{1}{4}\left(n-1\right)-2$$

Divide 9 by 3 to get 3.

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

Combine \frac{3}{5}n and -\frac{2}{3}n to get -\frac{1}{15}n.

$$-\frac{1}{15}n-\frac{12}{5}+3=\frac{1}{4}\left(n-1\right)-2$$

Convert 3 to fraction \frac{15}{5}.

$$-\frac{1}{15}n-\frac{12}{5}+\frac{15}{5}=\frac{1}{4}\left(n-1\right)-2$$

Since -\frac{12}{5} and \frac{15}{5} have the same denominator, add them by adding their numerators.

$$-\frac{1}{15}n+\frac{-12+15}{5}=\frac{1}{4}\left(n-1\right)-2$$

Add -12 and 15 to get 3.

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}\left(n-1\right)-2$$

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

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}n+\frac{1}{4}\left(-1\right)-2$$

Multiply \frac{1}{4} and -1 to get -\frac{1}{4}.

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}n-\frac{1}{4}-2$$

Convert 2 to fraction \frac{8}{4}.

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}n-\frac{1}{4}-\frac{8}{4}$$

Since -\frac{1}{4} and \frac{8}{4} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}n+\frac{-1-8}{4}$$

Subtract 8 from -1 to get -9.

$$-\frac{1}{15}n+\frac{3}{5}=\frac{1}{4}n-\frac{9}{4}$$

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

$$-\frac{1}{15}n+\frac{3}{5}-\frac{1}{4}n=-\frac{9}{4}$$

Combine -\frac{1}{15}n and -\frac{1}{4}n to get -\frac{19}{60}n.

$$-\frac{19}{60}n+\frac{3}{5}=-\frac{9}{4}$$

Subtract \frac{3}{5} from both sides.

$$-\frac{19}{60}n=-\frac{9}{4}-\frac{3}{5}$$

Least common multiple of 4 and 5 is 20. Convert -\frac{9}{4} and \frac{3}{5} to fractions with denominator 20.

$$-\frac{19}{60}n=-\frac{45}{20}-\frac{12}{20}$$

Since -\frac{45}{20} and \frac{12}{20} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{19}{60}n=\frac{-45-12}{20}$$

Subtract 12 from -45 to get -57.

$$-\frac{19}{60}n=-\frac{57}{20}$$

Multiply both sides by -\frac{60}{19}, the reciprocal of -\frac{19}{60}.

$$n=-\frac{57}{20}\left(-\frac{60}{19}\right)$$

Multiply -\frac{57}{20} times -\frac{60}{19} by multiplying numerator times numerator and denominator times denominator.

$$n=\frac{-57\left(-60\right)}{20\times 19}$$

Do the multiplications in the fraction \frac{-57\left(-60\right)}{20\times 19}.

$$n=\frac{3420}{380}$$

Divide 3420 by 380 to get 9.

$$n=9$$