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Quote:

$$\begin{aligned}3a+10b+5c&=0\\4a+6b+2c&=0\end{aligned}$$ solves to $$\frac{a}{30-20}=\frac{b}{6-20}=\frac{c}{-18+40}$$ so that $$\frac{a}{5}=\frac{b}{-7}=\frac{c}{11}$$

Can someone explain in deep how this equations are solved in this way?

First I thought it to do with using Cramer's rule but we are all aware that Cramer's rule work only for square matrix. And we can see this are only 2 equations with will not make matrix square.

So in my last question I was tempted to do only using Cramer's rule and now it has nothing to do with previous question!

I wish someone will find helping more important

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    You need to take a better photo or transcript the whole exercise here, Im unable to understand what is in the photo or what you are asking. – 2017-02-06
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    @Masacroso iam interested to know how this denominators are achieved (I.e., of a,b and c) – 2017-02-06
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    Why the urgency? – 2017-02-06
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    Instead of (or as well as) saying "thank you very much" to all responders, you can click on the double-arrow at the top left of a question to upvote those you especially like. – 2017-02-06

3 Answers 3

5

You have two equations in three unknowns so in general there is infinity of solutions. Namely, for any $a\in\mathbb{R}$, $b=-\frac{7a}{5}$ and $c=\frac{11a}{5}$ solve the system.

To see this, rewrite the system treating $a$ as a parameter. That is, as $$\begin{aligned}10b+5c&=-3a\\6b+2c&=-4a\end{aligned}$$ Now you can apply the Cramer's rule. For example, $$b=\frac{\left|\begin{array}{cc}-3a&5\\-4a&2\end{array}\right|}{\left|\begin{array}{cc}10&5\\6&2\end{array}\right|}=-\frac{14a}{10}$$ With this, you can solve for $c$ yourself.

2

First, eliminate $c$: subtracting $5 \times(3)$ from $2\times(2)$ gives $$(6-20)a +(20-30)b = 0$$ or $$\frac{a}{30-20}=\frac{b}{6-20}$$ Eliminating $b$ in a similar way gives you $$\frac{a}{30-20}=\frac{c}{-18+40}$$ This is enough. But if you want you can eliminate $a$ to check that your algebra is correct. You should get $$\frac{b}{6-20}=\frac{c}{-18+40}$$

2

We can set up a system of equations:

$$ \begin{cases} \text{a}\cdot\left(3-0\right)+\text{b}\cdot\left(9+1\right)+\text{c}\cdot\left(4+1\right)=0\\ \\ \text{a}\cdot\left(4-0\right)+\text{b}\cdot\left(5+1\right)+\text{c}\cdot\left(1+1\right)=0\\ \end{cases}\tag1 $$

This simplifies to:

$$ \begin{cases} 3\text{a}+10\text{b}+5\text{c}=0\\ \\ 4\text{a}+6\text{b}+2\text{c}=0\\ \end{cases}\tag2 $$

And so we get:

$$6\text{a}+20\text{b}+\left(-20\text{a}-30\text{b}\right)=0\space\Longleftrightarrow\space\text{b}=-\frac{7\text{a}}{5}\tag3$$

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    @user412463 You're welcome, I'm glad that I could help! – 2017-02-06