# Simultaneous equations with two unknowns

The question I've been given is $$\begin{array}{c|c}t&v\\\hline3&38\\12&200\end{array}$$ Modelling equation is $$v=k\sqrt{t-a}$$ Calculate $a$ and $k$.

I tried to solve like:

$$38 = k \sqrt{12-a}$$ $$200 = k \sqrt{3-a}$$

$$162 = k \sqrt{12-a} - k \sqrt{3-a}$$

$$\frac{162}k = \sqrt{12-a} - \sqrt{3-a}$$

$$(162/k)^2 = 12-a - 3 + a$$

$$(162/k)^2 = 9$$

$$162/k = 3$$

$$162/3 = k$$

$$k = 54$$

WHICH IS GIVEN AS WRONG. $k$ should be $65.45$

• mathjax resources to learn how to typeset maths. – Siong Thye Goh Jul 22 '18 at 4:54
• You swapped $3$ and $12$ ! – Yves Daoust Jul 23 '18 at 11:50

## 4 Answers

It seems to me that in general this question would be easier if you were to square both sides:

$$v^2=k^2(t-a)$$

With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have

$$1444 = 3k^2-ak^2 \\ 40000 = 12k^2-ak^2$$

Subtracting the former from the latter gives us

$$9k^2 = 38556 \\ k^2 = 4284 k = \sqrt{4284} \approx 65.452$$

Then substituting $k^2 = 4284$ back into either of the above yields

$$a = 12-\frac{40000}{4284} = 3-\frac{1444}{4284} \approx 2.663$$

Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.

• Thanks, that got it – DJx Jul 22 '18 at 5:59

First mistake:

• When time is $3$, the volume is $38$.

• When time is $12$, the volume is $200$.

• You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.

Second mistake:

• In general, $(\sqrt{a} - \sqrt{b})^2 \ne a - b$.

Guide:

• You might like to divide the equations rather than subtract the equations.
• Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that – DJx Jul 22 '18 at 5:59

Using your data, you have in fact $$38 = k \sqrt{3-a}\tag 1$$ $$200 = k \sqrt{12-a}\tag 2$$ Make the ratio $$\frac{38}{200}=\frac{\sqrt{3-a} } {\sqrt{12-a} }$$ Square both sides to get $$\frac{361}{10000}=\frac{3-a}{12-a}\implies a=\frac{2852}{1071}$$ Plug in $(1)$ to get $k=6 \sqrt{119}\approx 65.4523$.

Rewrite the model as

$$k^2t-k^2a=v^2$$

and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.

$$\begin{cases}3k^2-k^2a&=38^2,\\12k^2-k^2a&=200^2.\end{cases}$$

Then solving the system

$$k^2= 4284,$$ $$k^2a=11408.$$

The rest is easy.