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,

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 $\displaystyle (x_1,y_1)$ , $\displaystyle (x_2,y_2)$ is

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

note that parallel lines have the same slope.

Re: Having difficulty with linear equations

Quote:

Originally Posted by

**skeeter** slope between any two points $\displaystyle (x_1,y_1)$ , $\displaystyle (x_2,y_2)$ is

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

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

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

This means that the other point would be

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

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'

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

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

This means that the other point would be

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

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'

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

Re: Having difficulty with linear equations

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

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

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

cross multiply ...

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

finish solving for $\displaystyle a$.

Re: Having difficulty with linear equations

Quote:

Originally Posted by

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

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)?

Re: Having difficulty with linear equations

Quote:

Originally Posted by

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

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

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

cross multiply ...

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

finish solving for $\displaystyle a$.

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

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

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

Is this heading in the right direction?

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:

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

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

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

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

Re: Having difficulty with linear equations

Quote:

Originally Posted by

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

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

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

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

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.

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

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

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

$\displaystyle \Rightarrow a = - 5$.

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