Author Topic: 1,000,000 Question Thread :D (Read 44794 times) Tweet Share

kenhung123 · « **Reply #150 on:** December 08, 2009, 05:15:45 pm »

When solving simultaneous equations like this:
ax+by=a^2+2ab-b^2
bx+ay=a^2+b^2

Is it better to do elimination or substitution?
If I do substitution does it matter which equation and which variable I rearrange for? E.g. Use equation 2 to obtain y=.... then sub in equation 1 to get x=.... Because its possible to do this with 2 equations and with either x or y. I am wondering if the answers would be different.

TrueTears · « **Reply #151 on:** December 08, 2009, 05:19:34 pm »

At first glance I'd say elimination works pretty well, since $a^2$ can cancel or the $b^2$ can cancel.

And then you can see what you can do from there.

And no it doesn't matter which equation you rearrange.

kenhung123 · « **Reply #152 on:** December 08, 2009, 05:56:43 pm »

Thanks
I tried that and ended up with
$y=(b(a^2+2ab-b^2)-a(a^2+b^2))/(b-a)(b+a)$ How do I cancel this? lol

TrueTears · « **Reply #153 on:** December 08, 2009, 06:37:30 pm »

$ax+by=a^2+2ab-b^2 ...[1]$

$bx+ay=a^2+b^2 ...[2]$

$[1]+[2] \implies ax+by+bx+ay=2a^2+2ab$

$x(a+b)+y(a+b) = 2a^2+2ab$

$(x+y)(a+b) = 2a(a+b)$

$x+y=2a ...[3]$

$[2]-[1]$

$bx+ay - ax -by = 2b^2-2ab$

$x(b-a)+y(a-b) = 2b(b-a)$

$-x(a-b)+y(a-b) = 2b(b-a)$

$(y-x)(a-b) = -2b(a-b)$

$y-x=-2b$

$x-y=2b ..[4]$

So now we have:

$x+y=2a ...[3]$

$x-y=2b ...[4]$

Can you solve simultaneously for $[3]$ and $[4]$ now?

kenhung123 · « **Reply #154 on:** December 08, 2009, 09:29:21 pm »

Thank you

kenhung123 · « **Reply #155 on:** December 08, 2009, 09:30:24 pm »

How do I cancel $(a^3-b^3)/(a^2-b^2)$ ?

/0 · « **Reply #156 on:** December 08, 2009, 09:42:07 pm »

Another way

$ax+by=a^2+2ab-b^2$ ...[1]

$bx+ay = a^2+b^2$ ...[2]

$abx+b^2y = b(a^2+2ab-b^2)$ ...[1]*b

$abx+a^2y = a(a^2+b^2)$ ...[2]*a

$(a^2-b^2)y = a(a^2+b^2) - b(a^2+2ab-b^2)$ ...[2]*a-[1]*b

$(a^2-b^2)y =a^3+ab^2-a^2b-2ab^2+b^3$

$(a^2-b^2)y =a^3-a^2b-ab^2+b^3$

$(a^2-b^2)y =(a^3-ab^2)+(b^3-a^2b)$

$(a^2-b^2)y =a(a^2-b^2)+b(b^2-a^2)$

$(a^2-b^2)y =(a-b)(a^2-b^2)$

$y=a-b$

... $x=a+b$

Quote from: kenhung123 on December 08, 2009, 09:30:24 pm

How do I cancel $(a^3-b^3)/(a^2-b^2)$ ?

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

$(a^2-b^2) = (a-b)(a+b)$

$\frac{a^3-b^3}{a^2-b^2} = \frac{a^2+ab+b^2}{a+b} = \frac{a^2+2ab+b^2-ab}{a+b} = \frac{(a+b)^2-ab}{a+b} = a+b - \frac{ab}{a+b}$

kenhung123 · « **Reply #157 on:** December 08, 2009, 09:54:04 pm »

Quote from: /0 on December 08, 2009, 09:42:07 pm

Quote from: kenhung123 on December 08, 2009, 09:30:24 pm
How do I cancel $(a^3-b^3)/(a^2-b^2)$ ?

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

$(a^2-b^2) = (a-b)(a+b)$

$\frac{a^3-b^3}{a^2-b^2} = \frac{a^2+ab+b^2}{a+b} = \frac{a^2+2ab+b^2[b]-ab[/b]}{a+b} = \frac{(a+b)^2-ab}{a+b} = a+b - \frac{ab}{a+b}$

How did you get that -ab in your equation after 2nd equal sign?

TrueTears · « **Reply #158 on:** December 08, 2009, 09:56:19 pm »

kenhung have you been taught long division? You can use long division on $\frac{a^3-b^3}{a^2-b^2}$

/0 · « **Reply #159 on:** December 08, 2009, 10:01:31 pm »

Quote from: kenhung123 on December 08, 2009, 09:54:04 pm

Quote from: /0 on December 08, 2009, 09:42:07 pm

Quote from: kenhung123 on December 08, 2009, 09:30:24 pm
How do I cancel $(a^3-b^3)/(a^2-b^2)$ ?

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

$(a^2-b^2) = (a-b)(a+b)$

$\frac{a^3-b^3}{a^2-b^2} = \frac{a^2+ab+b^2}{a+b} = \frac{a^2+2ab+b^2[b]-ab[/b]}{a+b} = \frac{(a+b)^2-ab}{a+b} = a+b - \frac{ab}{a+b}$
How did you get that -ab in your equation after 2nd equal sign?

You could probably just leave it as $\frac{a^2+ab+b^2}{a+b}$ if that's what the question asked. I just decided to be a smartass and expand it further.

Oh and to get $\frac{a^2+2ab+b^2-ab}{a+b}$ I just did

$\frac{a^2+ab+b^2+(ab-ab)}{a+b}$ . Since $(ab-ab)$ is 0, I'm not actually changing anything, but this trick lets you create a perfect square since $(a+b)^2 = a^2+2ab+b^2$

kenhung123 · « **Reply #160 on:** December 08, 2009, 10:09:44 pm »

Oh I get it. Wow, how do you learn these tricks

TrueTears · « **Reply #161 on:** December 08, 2009, 10:11:13 pm »

Quote from: kenhung123 on December 08, 2009, 10:09:44 pm

Oh I get it. Wow, how do you learn these tricks

Wishful thinking + adding 0's + experience.

kenhung123 · « **Reply #162 on:** December 08, 2009, 10:21:42 pm »

For this questions: The length of a rectanle is 4cm more than the width. If the length were to be decreased by 5cm and the width decreased by 2cm, the perimeter would be 18cm. Calculate the dimensions of the rectangle.
Since:
$l=w+4$
Why can't I do: $2(l-5)+2(w+4-2)=18$ ?

kyzoo · « **Reply #163 on:** December 08, 2009, 10:40:52 pm »

Because it's

2(l-5) + 2(w-2) = 18

or

2(l-5) + 2(l-4-2) = 18

...

Since l = 4 + w

Therefore your equation "2(l-5) + 2(w+4-2) = 18" is really saying "2(l-5) + 2(l-2) = 18"

It says in the conditions that the width is reduced by 2cm, not the length.

Hope that helped, not sure if I explained clearly enough.

kenhung123 · « **Reply #164 on:** December 08, 2009, 10:58:18 pm »

Oh I understand now. Thanks for clarification

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