Halp! MX1 exam tomorrow! (1 Viewer)

ohexploitable · Feb 15, 2010

$x\; +\; y\; =\; 1,\; \; x^{3}\; +\; y^{3}\; =\; 19,\; \; Find\; \; x^{2}\; +\; y^{2}$

$a\; +\; b\; +c\; =\; 0,\; \; Prove\; \; (2a\; -\; b)^{3}\; +\; (2b\; -\; c)^{3}\; +\; (2c\; -\; a)^{3}\; = 3(2a\; -\; b)(2b\; -\; c)(2c\; -\; a)$

$Solve\; simultaneously,\; \; \frac{7}{x}\; -\; \frac{5}{y}\; =\; 3\; \; and\; \; \frac{2}{x}\; +\; \frac{25}{2y}\; =\; 12$

And all of Excercise 1l from question 7 onwards in the Cambridge 3 Unit textbook.

EDIT: Oh and this too plz...

A certain tank can be filled by two pipes in 80 minutes. The larger pipe by itself can fill the tank in 2 hours less than the smaller pipe by itself. How long does each pipe take to fill the tank on its own?

shaon0 · Feb 15, 2010

ohexploitable said:
$x\; +\; y\; =\; 1,\; \; x^{3}\; +\; y^{3}\; =\; 19,\; \; Find\; \; x^{2}\; +\; y^{2}$

$a\; +\; b\; +c\; =\; 0,\; \; Prove\; \; (2a\; -\; b)^{3}\; +\; (2b\; -\; c)^{3}\; +\; (2c\; -\; a)^{3}\; = 3(2a\; -\; b)(2b\; -\; c)(2c\; -\; a)$

$Solve\; simultaneously,\; \; \frac{7}{x}\; -\; \frac{5}{y}\; =\; 3\; \; and\; \; \frac{2}{x}\; +\; \frac{25}{2y}\; =\; 12$

And all of Excercise 1l from question 7 onwards in the Cambridge 3 Unit textbook.

Consider (x+y)^3:
(x+y)^3=(x+y)^2.(x+y)
=(x^2+2xy+y^2)(x+y)
=x^3+2x^2.y+xy^2+2xy^2+y^3
=(x^3+y^3)+2xy(x+y)+xy(x+y)
1=19+3xy
xy=-6

(x+y)^2=x^2+y^2+2xy
1=x^2+y^2-12
13=x^2+y^2

For the next one, just try to get each variable as a subject of the other two and sub in. It should workout.

Multiply both equations by 2xy and then divide both equations by one another. You should get a 65x=52y relationship between variables and sub this in to either equation and solve for y and x where y=1.25 and x=1.

Trebla · Feb 15, 2010

An alternative solution to the first question:

$x^3+y^3=19\\\\(x+y)(x^2-xy+y^2)=19\quad$ but $ x+y=1\\\\x^2+y^2=19+xy\\\\$But $\quad x+y=1\\\\\Rightarrow (x+y)^2=1\\\\x^2+y^2=1-2xy\\\\$Solve simultaneously $\\\\1-2xy=19+xy\\\\\Rightarrow xy = -6\\\\\therefore x^2+y^2=19-6=13$

For the second question, will use a little bit of manipulation and derive an existing formula:

$$Consider $\\\\x^3+y^3+z^3-3xyz\\\\=x^3+3x^2y+3xy^2+y^3+z^3-3xyz-3x^2y-3xy^2\\\\=(x+y)^3+z^3-3xy(x+y+z)\\\\=[(x+y)+z][(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)\\\\=(x+y+z)(x^2+y^2+z^2-xz-yz+2xy)-3xy(x+y+z)\\\\=(x+y+z)(x^2+y^2+z^2-xz-yz-xy)\\\\\therefore x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xz-yz-xy)\\\\$Let $x=2a-b, y=2b-c, z=2c-a\\\\$Note that $x+y+z=2a-b+2b-c+2c-a=a+b+c=0\\\\$Hence the RHS is zero, leaving $\\\\(2a-b)^3+(2b-c)^3+(2c-a)^3-3(2a-b)(2b-c)(2c-a)=0\\\\\therefore (2a-b)^3+(2b-c)^3+(2c-a)^3=3(2a-b)(2b-c)(2c-a)$

For the third question:

$\dfrac{7}{x}-\dfrac{5}{y}=3\Rightarrow\dfrac{1}{x}=\dfrac{1}{7}\left ( 3+\dfrac{5}{y} \right )\\\\\\$Also $\quad \dfrac{2}{x}+\dfrac{25}{2y}=12\\\\\Rightarrow \dfrac{2}{7}\left ( 3+\dfrac{5}{y} \right )+\dfrac{25}{2y}=12\\\\\dfrac{6}{7}+\dfrac{10}{7y}+\dfrac{25}{2y}=12\\\\\dfrac{195}{14y}=\dfrac{78}{7}\\\\\therefore y=\dfrac{5}{4}\\\\\Rightarrow \dfrac{1}{x}=\dfrac{1}{7}\left ( 3+4 \right )\\\\\therefore x=1$

life92 · Feb 15, 2010

For the last question, I've solved it before for someone else...
Here's the link.

http://community.boredofstudies.org/13/mathematics-extension-1/237865/need-help-question.html

Btw, good luck for your exam tomorrow !

Halp! MX1 exam tomorrow! (1 Viewer)

ohexploitable

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shaon0

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Trebla

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