Question

n%3D%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="\frac{3}{4}(m+n)-\frac{1}{4}(m-n);m=\frac{1}{2},n=\frac{1}{3}" alt="\frac{3}{4}(m+n)-\frac{1}{4}(m-n);m=\frac{1}{2},n=\frac{1}{3}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\frac{3}{4}(m+n)-\frac{1}{4}(m-n) = \frac{14}{24}$

Step-by-step explanation:

Given

$\frac{3}{4}(m+n)-\frac{1}{4}(m-n);m=\frac{1}{2},n=\frac{1}{3}$

Required

Solve

To do this, we simply substitute values of m and n in the given expression

$\frac{3}{4}(\frac{1}{2}+\frac{1}{3})-\frac{1}{4}(\frac{1}{2}-\frac{1}{3})$

Evaluate the brackets

$\frac{3}{4}(\frac{3 + 2}{6})-\frac{1}{4}(\frac{3-2}{6})$

$\frac{3}{4}(\frac{5}{6})-\frac{1}{4}(\frac{1}{6})$

Open Brackets

$\frac{15}{24} - \frac{1}{24}$

Take LCM

$\frac{15 - 1}{24}$

$\frac{14}{24}$

Hence:

$\frac{3}{4}(m+n)-\frac{1}{4}(m-n) = \frac{14}{24}$