Having difficulty with linear equations

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June 3rd 2012, 04:28 PM
astuart

Having difficulty with linear equations

Hello,

I've been out of school for over 6 years so these questions are probably relatively simple, but I'm just not grasping the concept here. I'm doing a bridging course for a university degree, and my high-school math has just gone out the window. I don't remember any of it.

Anyway, I'm being asked the following question.

If the line passing through points (1 , a) and (4, -2) is parallel to the line passing through the points (2, 8) and (-7, a+4), what is the value of a?

I don't even have any idea on where to begin to work this out :(.

Any help would be appreciated.

Cheers,
June 3rd 2012, 04:38 PM
skeeter

Re: Having difficulty with linear equations

Quote:

Originally Posted by astuart

Hello,

I've been out of school for over 6 years so these questions are probably relatively simple, but I'm just not grasping the concept here. I'm doing a bridging course for a university degree, and my high-school math has just gone out the window. I don't remember any of it.

Anyway, I'm being asked the following question.

If the line passing through points (1 , a) and (4, -2) is parallel to the line passing through the points (2, 8) and (-7, a+4), what is the value of a?

I don't even have any idea on where to begin to work this out :(.

Any help would be appreciated.

Cheers,

slope between any two points $(x_1,y_1)$ , $(x_2,y_2)$ is

$m = \frac{y_1-y_2}{x_1-x_2}$

note that parallel lines have the same slope.
June 3rd 2012, 04:50 PM
astuart

Re: Having difficulty with linear equations

Quote:

Originally Posted by skeeter

slope between any two points $(x_1,y_1)$ , $(x_2,y_2)$ is

$m = \frac{y_1-y_2}{x_1-x_2}$

note that parallel lines have the same slope.

Yep, I understand that, but I don't know how to solve for (a) because it's a variable so from what I can gather the equation would be

$m=\frac{(-2)-a}{4-1}$ .

This means that the other point would be

$m = \frac{(a+4)-8}{-7-2}$

However, from here, how do I solve for a?

A tip in the right direction is suitable, I'm really wanting to solve this myself and hopefully it just 'clicks'
June 3rd 2012, 04:56 PM
Plato

Re: Having difficulty with linear equations

Quote:

Originally Posted by astuart

Yep, I understand that, but I don't know how to solve for (a) because it's a variable so from what I can gather the equation would be

$m=\frac{(-2)-a}{4-1}$ .

This means that the other point would be

$m = \frac{(a+4)-8}{-7-2}$

However, from here, how do I solve for a?

A tip in the right direction is suitable, I'm really wanting to solve this myself and hopefully it just 'clicks'

$\frac{(-2)-a}{4-1}= \frac{(a+4)-8}{-7-2}$ solve for $a$
June 3rd 2012, 05:04 PM
skeeter

Re: Having difficulty with linear equations

so, I see you do know where to start ...

$\frac{-2-a}{4-1} = \frac{(a+4) - 8}{-7-2}$

$\frac{-2-a}{3} = \frac{a-4}{-9}$

cross multiply ...

$-9(-2-a) = 3(a-4)$

finish solving for $a$ .
June 3rd 2012, 05:11 PM
astuart

Re: Having difficulty with linear equations

Quote:

Originally Posted by Plato

$\frac{(-2)-a}{4-1}= \frac{(a+4)-8}{-7-2}$ solve for $a$

This is where I'm falling down. From what I think I understand, I need to isolate a. Isolating a on one of these equations would give me the answer to the other side, correct (due to them being equal)?
June 3rd 2012, 05:36 PM
astuart

Re: Having difficulty with linear equations

Quote:

Originally Posted by skeeter

so, I see you do know where to start ...

$\frac{-2-a}{4-1} = \frac{(a+4) - 8}{-7-2}$

$\frac{-2-a}{3} = \frac{a-4}{-9}$

cross multiply ...

$-9(-2-a) = 3(a-4)$

finish solving for $a$ .

Still not getting it :(. I can't believe It's not clicking - I think I need to re-visit some basic algebra.

Would the next step of the equation be

$-9(-2-a) = 3(a-4)$

$-18-9a=3a-12$

Is this heading in the right direction?
June 3rd 2012, 05:49 PM
Reckoner

Re: Having difficulty with linear equations

Quote:

Originally Posted by astuart

Is this heading in the right direction?

You got the idea, yes. But let's be careful with our distribution:

$-9(-2-a) = 3(a-4)$

$\Rightarrow(-9)(-2)+(-9)(-a) = 3a - 3(4)$

$\Rightarrow18+9a = 3a - 12$ .

Remember that any time you multiply two numbers with the same sign (both positive or both negative), the answer will be positive.
June 3rd 2012, 07:46 PM
astuart

Re: Having difficulty with linear equations

Quote:

Originally Posted by Reckoner

You got the idea, yes. But let's be careful with our distribution:

$-9(-2-a) = 3(a-4)$

$\Rightarrow(-9)(-2)+(-9)(-a) = 3a - 3(4)$

$\Rightarrow18+9a = 3a - 12$ .

Remember that any time you multiply two numbers with the same sign (both positive or both negative), the answer will be positive.

Right, so after having a break for an hour or so, it's a little bit clearer.

$\Rightarrow18+9a = 3a - 12$ . (subtract 3a)

$\Rightarrow18+6a = - 12$ . (divide by 6)

$\Rightarrow3+a = - 2$ . (subtract 3)

$\Rightarrow a = - 5$ .

It's pretty simple really. Thanks for everybody's patience. Great forum too!