# A couple of Questions

• Dec 10th 2012, 04:53 PM
bahamamath
A couple of Questions
Hey guys , i've pinpointed certain areas in the maths that i have weaknesses in so here are just a few question before next weeks final , i wanted to clarify wondering if you guys can solve for me asap |:0)

Now here's a Simultaneous Equation And Just a quick question ( when using the elimination method , which i use ) how can you tell when to add or subtract?

2x - 3y = 2
3x - 2y = -(7)

Also , i believe this is quadratics...

{ X^2 - 6x + 1 = 0 }

I am also trying to factorize

4x^2-5+1

and simplify lastly

4(x-1) -3(2x+1) = 3(x+4)

Anyhelp would be nice... These littler things tend to mess me up and i end up spending more time than i should on them.
• Dec 10th 2012, 05:19 PM
flor
Re: A couple of Questions
4(x-1)-3(2x+1)=3(x+4)
(4x-4)-(6x-3)=3x+12
4x-4-6x-3=3x+12
4x-6x-4-3=3x+12
-2x-7=3x+12
-7-12=3x-2x
-19=x
• Dec 10th 2012, 05:59 PM
topsquark
Re: A couple of Questions
Quote:

Originally Posted by flor
4(x-1)-3(2x+1)=3(x+4)

(4x - 4) - (6x + 3) = 3x +12

-Dan
• Dec 10th 2012, 06:04 PM
topsquark
Re: A couple of Questions
Quote:

Originally Posted by bahamamath
Now here's a Simultaneous Equation And Just a quick question ( when using the elimination method , which i use ) how can you tell when to add or subtract?

2x - 3y = 2
3x - 2y = -(7)

What you want to be able to do is to reduce the system to one variable. So let's try to get rid of the variable x. To do this we need to have the same coefficient for x in both equations. In this case, if we multiply the top equation by 3 and the bottom equation by 2 we get
6x - 9y = 6
6x - 4y = -14

If you subtract these two, the x terms disappear.

-Dan