# Algebra and Fractions

In yesterday’s post, we solved $\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}$. We saw that when we first multiplied both side of the equation by 2, we had to multiply the entire left side. Let’s now look at this equation: $\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{12}$.

Again, it appears that the first thing to do in our quest to get by itself is to multiply both sides by 2. But this time, the whole left side of the equation is divided by 2, not just x like it was in yesterday’s equation. So,

${2}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{12}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{2}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{24}$

Now let me stop here to explain a bit more of what’s happening. I put the implied “1” below the “2” so you can see what the next step should be. Remember that when multiplying two fractions together, you multiply the numerators together and the denominators together. But realise that when multiplying the numerators, the “2” is multiplying the whole expression “3 + 3x” so I will need to use brackets to indicate that:

$\frac{2}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{24}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{{1}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}$

Now we can see that we can cancel the “2”s, leaving the stuff in the brackets:

$\begin{array}{l} {\frac{{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{{1}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}}\\ {\Longrightarrow\hspace{0.33em}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{24}} \end{array}$

Now we can subtract the 3 and the other methods we have covered to solve the equation:

3 + 3x -3 = 24 – 3 ⇒ 3x = 21

$\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{21}{3}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{7}$
Posted on Categories Algebra, Pre-VCE