# Question ffde9

Jun 20, 2017

#### Answer:

See the explanation

#### Explanation:

It all depends on the correct interpretation of the question. I have given two interpretations to demonstrate some mathematical processes. Even if they are not the correct solutions $\textcolor{m a \ge n t a}{\text{the methods are important.}}$

You state $m + \frac{2}{3} + \frac{1}{4} m - 1$ and you use the word 'solve'. This implies that you wish to determine the value of $m$.

As there is no equals sign it is not possible to 'solve' for $m$

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$\textcolor{b l u e}{\text{Taking the work on from your part 'solution'}}$

$\textcolor{g r e e n}{12 m + 8 = 3 m - 1}$

subtract $\textcolor{red}{3 m}$ from both sides

$\textcolor{g r e e n}{12 m \textcolor{red}{- 3 m} + 8 \text{ "=" } 3 m \textcolor{red}{- 3 m} - 1}$

$\textcolor{g r e e n}{\text{ "9m" "color(white)(.)+8" "=" "0" } - 1}$

Subtract $\textcolor{red}{8}$ from both sides

color(green)(9m+8color(red)(-8)" "=" "-1color(red)(-8)#

$\textcolor{g r e e n}{9 m \text{ "+0" "=" } - 9}$

Divide both sides by $\textcolor{red}{9}$

$\textcolor{g r e e n}{\frac{9}{\textcolor{red}{9}} m \text{ "=" } \frac{- 9}{\textcolor{red}{9}}}$

but $\frac{9}{9} = 1 \mathmr{and} \frac{- 9}{9} = - 1$

$\textcolor{g r e e n}{1 m = - 1}$

but writing $1 m$ is bad practice so write this as just $m$

$m = - 1$
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$\textcolor{b l u e}{\text{Assuming that we have an expression and not an equation}}$

An expression does not have an equals sign in it.

Simplifying the expression$\to m + \frac{2}{3} + \frac{1}{4} m - 1$
Note that $m$ is the same as $1 m$ which is also the same as $\frac{4}{4} m$

$m + \frac{2}{3} + \frac{1}{4} m - 1 \text{ "->" } \frac{4}{4} m + \frac{2}{3} + \frac{1}{4} m - 1$

$\text{ "->" } \frac{4}{4} m + \frac{1}{4} m + \frac{2}{3} - 1$

$\text{ "->" } \frac{5}{4} m + \frac{2}{3} - 1$

Note that $- 1$ is the same as $- \frac{3}{3}$

$\frac{5}{4} m + \frac{2}{3} - 1 \text{ "->" } \frac{5}{4} m + \frac{2}{3} - \frac{3}{3}$

$\text{ "->" "5/4m-1/3 larr" Simplified}$