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I need some help with my kid's Algebra homework - Printable Version

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I need some help with my kid's Algebra homework by barnburner on 05-21-2007 at 01:39 AM

I need some help with my kid's Algebra homework, he is about to make me pull my hair out, can someone help me with this:

Slove each equation

2x / x-4    -2   = v4/x+5   Common Dom


3/ x^2+5x+6  -  7/x+3  = x-1/x+2    Common Dom

Simplify

x^2+2x-15/  X^2+9x+20  / (x-3)

x+5/x+2 +6 Common dom

X^2-1/X+1 - X^2+1/ X-1

Algebra was 30 years ago for me, can someone help????


RE: I need some help with my kid's Algebra homework by Nagamasa on 05-21-2007 at 03:46 AM

If you dont mind, I will solve once I get home and use equation wizard.

Naga :D


RE: I need some help with my kid's Algebra homework by Volv on 05-21-2007 at 11:49 AM

I've written all the answers which I got + working out (there shouldn't be any errors but I apologise if there are).
I would just like to mention that if your son/daughter wishes to do further mathematics in the future that he/she should really learn this topic and methods for solving, as it is something which is really important in pretty much all areas of mathematics.

quote:
Originally posted by barnburner
2x / x-4    -2   = v4/x+5   Common Dom
Multiply all terms by (x-4)(x+5) and you get:
2x(x+5) - 2(x-4)(x+5) = 4(x-4)

expand:
2x^2 + 10x - 2(x^2 + x - 20) = 4x - 16
2x^2 + 10x - 2x^2 - 2x + 40 = 4x - 16

Cancel out:
10x - 2x + 40 = 4x - 16

Move all x to one side and everything else to the other:
10x - 2x - 4x = -40 - 16
4x = -56

ANSWER:
x = -14


quote:
Originally posted by barnburner
3/ x^2+5x+6  -  7/x+3  = x-1/x+2    Common Dom
Factorise the denominator of 3/ x^2+5x+6 to get:
3/((x+3)(x+2)) - 7/(x+3) = (x-1)/(x+2)

Multiply everything by (x+3)(x+2):
3 - 7(x+2) = (x-1)(x+3)
3 - 7x - 14 = x^2 + 2x - 3

Move everything to one side:
0 = x^2 + 2x + 7x - 3 - 3 + 14
x^2 + 9x + 8 = 0

Factorise quadratic:
(x+1)(x+8) = 0

therefore the answer is:
x = -1 or x = -8

quote:
Originally posted by barnburner
Simplify

x^2+2x-15/  X^2+9x+20  / (x-3)
Factorise both quadratics:
((x+5)(x-3)) / ((x+4)(x+5)) / (x-3)

Now a/b/c = ac/b, so from the above we'll say
a = (x+5)(x-3)
b = (x+4)(x+5)
c = (x-3)
and by applying the rule we get:
a/b/c
= ac/b
= ((x+5)(x-3)(x-3)) / ((x+4)(x+5))

Cancel out (x+5)'s:
= ((x-3)(x-3)) / (x+4)

Simplify:
= (x-3)^2 / (x+4)

quote:
Originally posted by barnburner
x+5/x+2 +6 Common dom
Not sure what you meant, if you mean:
x+5/x+2x+6
Then there are no real solutions (as the bottom can't be factorised unless you're going to be using complex numbers - which I really don't think is the case at this stage).

If you actually meant Edit: Now that I look at it I know that you mean this (because of your space) - Sorry :p
(x+5)/(x+2)   + 6
Then I suppose you multiply everything by (x+2):
= x + 5 + 6(x+2)
= x + 5 + 6x + 12
= 7x + 17

quote:
Originally posted by barnburner
X^2-1/X+1 - X^2+1/ X-1
Factorise the top of the left fraction (using difference of two squares rule):
= ((x+1)(x-1))/(x+1) - (x^2+1)/(x-1)

Cancel out the (x+1)'s on the left:
= (x-1) - (x^2+1)/(x-1)

Then multiple everything by (x-1):
= (x-1)^2 - x^2 + 1

Expand:
= x^2 - 2x + 1 - x^2 + 1

Simplify:
= -2x + 2

Rearrange and Factorise (to make prettier):
= 2 - 2x
= 2(1 - x)
RE: I need some help with my kid's Algebra homework by MeEtc on 05-21-2007 at 11:50 AM

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