Solve Simultaneous Equations to Find OP: a & c Given

[riddle]

Ok, I'm new here, and I'm not sure if this is the right place to ask. So anyway here's the question:

OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AP and MQ meet at P. If OA=a and OC=c, express OP in terms of a and c.

The question's easy enough... until the point I get stuck.

Here's what I've got so far:
OA = a, MO = -0.5a, OC = c, CQ = .75a, MQ= .25a+c, AC = c-a

OP = OA + x(AC) = 0.5(OA) + y(MQ)
=> OP = a + x(c-a) = 0.5a + y(c+0.25a)
=> a + x(c-a) = 0.5a + y(c+0.25a)

Aaand this is where I get stuck. I've got the textbook. It mentions "forming simultaneous euqations by equating the coefficients", but I've got no idea how that works. It's the first time I've ever heard of solving equations like that. I've done some looking up, but I'm still confused. Can someone please help? Thanks

P.S.: Please excuse the lack of arrows on top of the lines. There should be "$$\rightarrow$$" on top of them all, I just wasn't sure how to use Latex to do it.

 

[Mentallic]

If we have an equality say,$$ax+by+c=2x+3y+5$$

then we can deduce that a=2, b=3 and c=5. This is what it means to equate the coefficients. The coefficients of each variable, mainly a, b and c (the variables being x and y) have to be equal on both ends of the equation.

 

[riddle]

Well, I knew that. But what I don't get is how would you solve the equality that I came up with. My books got the answer with all the steps, and I've done exactly what's in the book, but then it goes on to solve the equality by "equating the coefficients".

 

[Mentallic]

Well then you have to get each side exactly into that form. If you have 2x+5x on one side, you would factorize to make it (2+5)x=7x.

 

[riddle]

bleh. ok, it's either the effects of sleep deprivation, an acceleration in my cognitive dysfunction, or a combination of both, but I can't see what they've done here. Well I can see what they've done, but not WHY theyve done it.
These are the equations that they make by equating the co-efficients.
Can you please explain what they've done here.

From:a + x(c-a) = 0.5a + y(c+0.25a)
To: 0.5 + 0.25x =1− y and x=y

 

[Mentallic]

Ok I'll get give your head a little break, too much thinking can hurt

Just expand everything, then factorize out all the a's, and then the c's.

So on the left side, a + x(c-a)=a+cx-ax=a(1-x)+cx

Now you can do the same for the right side and equate coefficients.

 

[riddle]

Ohhhhh.
Wow. I feel stupid now.
Oh and thanks for your help.

 

[Mentallic]

I'm guessing you slept? Haha see how much it helps

 

[riddle]

Haha, just a short nap.
I'm looking at some trigonometry now. Bearings. I've spent around an hour looking at the question not getting how to find this one angle, and it's supposed to be the easy part.
I'll post here asking for help when the frustrations pointing a gun at my head

 

[Mentallic]

If you're failing at the easy part, just post the question already!

 

[riddle]

I got the tough parts done, but I might as well post it. I think it's ok if I post it here, considering the question's in the same chapter as the first question I asked for help for.

An aeroplane flies from airport A to airport B 80 km away on a bearing of 070∘. From B the aeroplane flies to airport C, 60 km from B. Airport C is 90 km from A. Find the two possible directions for the course set by the aeroplane on the second stage of its journey.

I've made two triangles, triangle ABC₁ and ABC₂, and I've found the angle; Angle ABC₁ = Angle ABC₂ = arccos (19/96)
Now what I don't get is, how do I work out the rest of the angle, i.e, the angle made by north and BC.

Oh... I was going to edit the topic title, making it something more like "Solving vectors and bearings", but I didn't know I couldn't change the title

 

[Mentallic]

Look back at A and draw the NSEW lines from that point, then add the angles you know, now do the same for B. Notice any parallel lines?

 

[riddle]

Ooooh. *facepalm*
Thanks Mentallic.

 

[Mentallic]

forehead slap no problem, good luck with the rest.