# Thread: Some problems that ave been giving me trouble

1. ## Some problems that ave been giving me trouble

16-X
------ = 2+x (thats a fraction lol)
5

Find the equation of the line joining the points (0,5) and (2,11)

Calculate

4 and 1 fifth (multiplied by) 1 and 1 seventh

(not sure how to type that out)

Nuno borrowed £400 for 9 months and paid £36 interest. What was the equivalent rate of interest per annum?

Also if you could show me how you got the answer it would help greatly.

thanks in advance for any help

2. Originally Posted by Xplicit
16-X
------ = 2+x (thats a fraction lol)
5
Just so you know, there is a difference between the variables X and x.

$\displaystyle \frac{16 - x}{5} = 2 + x$

$\displaystyle \frac{16 - x}{5} \cdot 5 = (2 + x) \cdot 5$

$\displaystyle 16 - x = 10 + 5x$

Can you solve this now?

Originally Posted by Xplicit
Find the equation of the line joining the points (0,5) and (2,11)
Given two points $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$, the slope of the line going through them may be calculated as
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

To get the rest of the equation for the line you can use the slope-intercept form:
$\displaystyle y = mx + b$

Plug in either point and the slope you just calculated and solve for b.

Originally Posted by Xplicit
Calculate
4 and 1 fifth (multiplied by) 1 and 1 seventh
$\displaystyle \left ( 4~\frac{1}{5} \right ) \cdot \left ( 1~\frac{1}{7} \right )$

I am personally against numbers written in this form, though I fully understand why they are useful. But I find that they rather confuse things by the time a student reaches their first Algebra course.

Anyway, I would convert these numbers to improper fractions:
$\displaystyle \left ( 4~\frac{1}{5} \right ) \cdot \left ( 1~\frac{1}{7} \right ) = \frac{21}{5} \cdot \frac{8}{7}$

You should be able to finish this now.

Someone else is going to have to help you with the last problem.

-Dan

3. Originally Posted by Xplicit
Nuno borrowed £400 for 9 months and paid £36 interest. What was the equivalent rate of interest per annum?
This is the formula for simple interest.

$\displaystyle Interest = Capital \ \times \ \frac{Rate}{100} \ \times \frac{Time (Months)}{12 \ Months}$

Thus:

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{t}{12}$

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{9}{12}$

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{3}{4}$

$\displaystyle 36 = 300 \times \frac{r}{100}$

$\displaystyle 36 = 3r$

$\displaystyle r = 12$