Couple of preliminary extension 1 questions. (1 Viewer)

Starrzz · Feb 21, 2010

So I got a quiz recently, and I had almost no clue with some questions on what to do and how the answer was such.

Help would be appreciated. Step by step, if possible >.<

$$1.Simplify:$
${\frac{2}{x^2-9}}-{\frac{1}{x-3}}-{\frac{4}{x+3}$

$$2.Solve simultaneously for x:$
$3x-7\leq 2x-3\leq 5+x$

$$3.Factorise fully:$
${\frac{1}{ab^3}}-{\frac{1}{a^3b}}$

$$4.Factorise fully:$
$x^4a^4+b^4-x^4b^4-a^4$

$$5.Simplify:$
$\frac{ax-bx+ay-by}{2bx-2ay+2by-2ax}$

$$6.Simplify:$
${\frac{x^3-1}{x^3}}\times{\frac{2x^2+2x}{x^2-1}}$

$$7.Simplify:$
${\frac{c^2-1}{c^2+c-2}}\times{\frac{c^2-4}{c^2+2}}+{\frac{2-c}{c}$

$$8.$
$If$ x={\frac{1}{c-1}},y={\frac{1}{c^2-1}},z={\frac{1}{c^3-1}} $and$ w=c^2+c+1,$rewrite in its simplest form, the expression${\frac{x}{y}}+wz,$in terms of c only.$

fullonoob · Feb 21, 2010

Starrzz said:
So I got a quiz recently, and I had almost no clue with some questions on what to do and how the answer was such.

Help would be appreciated.

$$1.Simplify:$
${\frac{2}{x^2-9}}-{\frac{1}{x-3}}-{\frac{4}{x+3}$

$$2.Solve simultaneously for x:$
$3x-7\leq 2x-3\leq 5+x+x$

$$3.Factorise fully:$
${\frac{1}{ab^3}}-{\frac{1}{a^3b}}$

$$4.Factorise fully:$
$x^4a^4+b^4-x^4b^4-a^4$

$$5.Simplify:$
$\frac{ax-bx+ay-by}{2bx-2ay+2by-2ax}$

$$6.Simplify:$
${\frac{x^3-1}{x^3}}\times{\frac{2x^2+2x}{x^2-1}}$

$$7.Simplify:$
${\frac{c^2-1}{c^2+c-2}}\times{\frac{c^2-4}{c^2+2}}+{\frac{2-c}{c}$

$$8.$
$If$ x={\frac{1}{c-1}},y={\frac{1}{c^2-1}},z={\frac{1}{c^3-1}} $and$ w=c^2+c+1,$rewrite in its simplest form, the expression${\frac{x}{y}}+wz,$in terms of c only.$

whih one do you nee help on he most?
first one is comon denomintor with one fraction, then obtain one whole fraction.Aftewards -the oter.
2) is triple simultaneous. simultaneous with one of the eqations to obtain a few pronuemerals then simultaneous with the other.
keyboard batts running out....zomg

Starrzz · Feb 21, 2010

fullonoob said:
whih one do you nee help on he most?
first one is comon denomintor with one fraction, then obtain one whole fraction.Aftewards -the oter.
2) is triple simultaneous. simultaneous with one of the eqations to obtain a few pronuemerals then simultaneous with the other.
keyboard batts running out....zomg

All the questions I've posted are the ones I need help with.

ohexploitable · Feb 21, 2010

ohexploitable · Feb 21, 2010

madsam · Feb 21, 2010

For 2,

You do the lhs and rhs independantly, 3x-7< 2x - 3 and 2x - 3<5+x

And you just solve them, you end with x<4 and x<8
so both criteria are met when x<4

frenzal_dude · Feb 22, 2010

isn't the answer +ve 1/2?
This is what I got:

$\frac{ax-bx+ay-by}{2bx-2ay+2by-2ax} = \frac{x(a-b)+y(a-b)}{2(bx-ay+by-ax)}=\frac{(x+y)(a-b)}{2[b(x+y)-a(x+y)]}=\frac{(x+y)(a-b)}{2[(b-a)(x+y)]}=\frac{a-b}{2(b-a)}=\frac{-1(-a+b)}{-2(b-a)}=\frac{-(b-a)}{-2(b-a)}=\frac{1}{2}$

frenzal_dude · Feb 22, 2010

Question 6:
$\frac{x^{3}-1}{x^{3}}*\frac{2x^{2}+2x}{x^{2}-1}=\frac{(x-1)(x^{2}+x+1)(2x)(x+1)}{x^{3}(x-1)(x+1)}=\frac{(x^{2}+x+1)(2x)}{x^{3}}=\frac{2(x^{2}+x+1)}{x^{2}}$

Question 7:
$\frac{c^{2}-1}{c^{2}+c-2}*\frac{c^{2}-4}{c^{2}+2}+\frac{2-c}{c}=\frac{(c-1)(c+1)(c-2)(c+2)}{(c+2)(c-1)(c^{2}+2)}+\frac{2-c}{c}=\frac{(c+1)(c-2)}{c^{2}+2}+\frac{2-c}{c}$

Question 8:
$x=\frac{1}{c-1} \cdots y=\frac{1}{(c-1)(c+1)}\cdots z=\frac{1}{(c-1)(c^{2}+c+1)}\therefore \frac{x}{y}+wz = \frac{\frac{1}{c-1}}{\frac{1}{(c-1)(c+1)}}+\frac{c^{2}+c+1}{(c-1)(c^{2}+c+1)}=\frac{(c-1)(c+1)}{c-1}+\frac{1}{c-1}=c+1+\frac{1}{c-1}$

harrowed2 · Feb 22, 2010

frenzal_dude said:
isn't the answer +ve 1/2?
This is what I got:

$\frac{ax-bx+ay-by}{2bx-2ay+2by-2ax} = \frac{x(a-b)+y(a-b)}{2(bx-ay+by-ax)}=\frac{(x+y)(a-b)}{2[b(x+y)-a(x+y)]}=\frac{(x+y)(a-b)}{2[(b-a)(x+y)]}=\frac{a-b}{2(b-a)}=\frac{-1(-a+b)}{-2(b-a)}=\frac{-(b-a)}{-2(b-a)}=\frac{1}{2}$

The answer is negative.
$\frac{ax-bx+ay-by}{2bx-2ay+2by-2ax}=\frac{ax-bx+ay-by}{-2(ax-bx+ay-by)}=\frac{-1}{\: 2}$

frenzal_dude · Feb 22, 2010

Yeh you're right, the question is actually a lot more simpler than what it looks like.

Couple of preliminary extension 1 questions. (1 Viewer)

Starrzz

Member

fullonoob

fail engrish? unpossible!

Starrzz

Member

ohexploitable

hey

ohexploitable

hey

madsam

God among men

frenzal_dude

UTS Student

frenzal_dude

UTS Student

harrowed2

Member

frenzal_dude

UTS Student

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