I need someone to help me with this problems: 6/v=3/11, 16j=320, 13x=15-2x.

Monday, March 16, 2015

I need someone to help me with this problems: 6/v=3/11, 16j=320, 13x=15-2x.

Hello!

As I understand, there are three unconnected equations, each with its own variable. Let's solve them one by one.

1) $\displaystyle\frac{{6}}{{v}}=\frac{{3}}{{11}}.$

This is a proportion and we can use a rule for solving it, but it is probably better to solve it step by step. First, multiply both sides by $\displaystyle{v}$ (it cannot be zero, so such a multiplication gives an equivalent equation):

$\displaystyle{6}=\frac{{3}}{{11}}\cdot{v}.$

Then multiply by $\displaystyle{11}:$ $\displaystyle{6}\cdot{11}={3}\cdot{v}.$

And finally divide by $\displaystyle{3}$ and compute: $\displaystyle{v}=\frac{{{6}\cdot{11}}}{{3}}={2}\cdot{11}={22}.$

2) $\displaystyle{16}{j}={320}.$

Only one step is here, divide both sides by $\displaystyle{16}$ and note that $\displaystyle{320}={32}\cdot{10}={2}\cdot{16}\cdot{10}:$

$\displaystyle{j}=\frac{{320}}{{16}}=\frac{{{2}\cdot{16}\cdot{10}}}{{16}}={2}\cdot{10}={20}.$

3) $\displaystyle{13}{x}={15}-{2}{x}.$

Two steps are here: first, add $\displaystyle{2}{x}$ to both sides and obtain $\displaystyle{13}{x}+{2}{x}={15},$ or $\displaystyle{15}{x}={15}.$ Second, divide both sides by $\displaystyle{15}$ and obviously $\displaystyle{x}={1}.$

Ninenotes

Monday, March 16, 2015

I need someone to help me with this problems: 6/v=3/11, 16j=320, 13x=15-2x.

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