Author Topic: yr 11 questions (Read 5209 times) Tweet Share

Hielly · « **on:** February 08, 2009, 05:58:19 pm »

hey

Need help on 2 questions

1)Find x.
$ax+by=a^2+2ab-b^2$
$bx+ay=a^2+b^2$

2)Find x and y
3(x-a)-2(y+a)=5-4a
2(x+a)+3(y-a)=4a-1

Thanks

Flaming_Arrow · « **Reply #1 on:** February 08, 2009, 06:21:36 pm »

$ax+by=a^2+2ab-b^2$

$bx+ay=a^2+b^2$

times the first equation by a and the second by b

$a^2x+aby=a^3+2a^2b-ab^2$

$b^2x+aby=a^2b+b^3$

now minus the first from the 2nd

$a^2x-b^2x= a^3-a^2b+2a^2b-b^3-ab^2$

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

factorise it

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

$x = (a+b)$

Flaming_Arrow · « **Reply #2 on:** February 08, 2009, 06:26:13 pm »

$3(x-a)-2(y+a)=5-4a$

$2(x+a)+3(y-a)=4a-1$

expand

$3x-3a-2y-2a=5-4a$

$2x+2a+3y-3a=4a-1$

times the first by 3 and the second by 2

$9x-9a-6y-6a=15-12a$

$4x+4a+6y-6a=8a-2$

add them

$13x-5a-12a=13-4a$

$13x-17a=13-4a$

$13x=13+13a$

$x=1+a$

Hielly · « **Reply #3 on:** February 08, 2009, 06:26:30 pm »

Hey flaming_arrow, the answer is x=a+b, there isn't a square root.

1st question

Flaming_Arrow · « **Reply #4 on:** February 08, 2009, 06:29:23 pm »

oh whoops typo
fixed.

Hielly · « **Reply #5 on:** February 08, 2009, 06:45:47 pm »

$a^3+a^2b-b^3-ab^2$

when factorized i got this, in my previous attempt

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

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

$x = (a-b)$

i believe i screwed up there, wondering why this is wrong?

Flaming_Arrow · « **Reply #6 on:** February 08, 2009, 06:51:59 pm »

Hielly · « **Reply #7 on:** February 08, 2009, 06:53:34 pm »

ohhhh silly me, i forgot about (a-b)^2 which equals to (a-b)(a+b)

thanks!!

Hielly · « **Reply #8 on:** February 08, 2009, 07:21:14 pm »

okay im trying to solve for y first...

$3(x-a)-2(y+a)=5-4a$

$2(x+a)+3(y-a)=4a-1$

expand

$3x-3a-2y-2a=5-4a$

$2x+2a+3y-3a=4a-1$

times the first by 2 and the second by 3....

$6x-6a-4y-4a=10-8a$

$6x+6a+9y-9a=12a-3$

add them

$-12a-13y+5a=10-12a-8a-3$

$-13y=7-20a-5a+12a$

$-13y=7-13a$

$y=\frac {7-13a}{-13}$

hrm, what did i do wrong here?

Flaming_Arrow · « **Reply #9 on:** February 08, 2009, 07:24:44 pm »

i think you mean minus them

anyway

10--3 = 13

Hielly · « **Reply #10 on:** February 08, 2009, 07:27:16 pm »

oh so thats how it is, i thought it was like 10-12a and then the 12a takes the minus, so then its 10+ something

Flaming_Arrow · « **Reply #11 on:** February 08, 2009, 07:30:26 pm »

you can do it like this

$(10-8a)-(12a-3) = 13-20a$

Mao · « **Reply #12 on:** February 08, 2009, 10:45:46 pm »

Quote from: Hielly on February 08, 2009, 06:53:34 pm

ohhhh silly me, i forgot about (a-b)^2 which equals to (a-b)(a+b)

thanks!!

no

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

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

Hielly · « **Reply #13 on:** February 10, 2009, 04:34:56 pm »

okay guys this is the same one as before but trying to find y.
equations are
ax+by=p
bx-ay=q

atm im at q(aq-bp)(p+b)/ap(-a^2-b^2)-b^3q

Flaming_Arrow · « **Reply #14 on:** February 10, 2009, 04:37:37 pm »

$ax+by=p$

$bx-ay=q$

times first one by b and second by a

$abx+b^2y=bp$

$abx-a^2y=aq$

take away

$b^2y+a^2y=bp-aq$

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

$\implies y = \frac{bp-aq}{a^2+b^2}$

Search the forums now!

We have moved!

Author Topic: yr 11 questions (Read 5209 times) Tweet Share

Hielly

yr 11 questions

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Mao

Re: quick 2 questions on simulanteous equations

Hielly

Re: quick 2 questions on simulanteous equations

Flaming_Arrow

Re: quick 2 questions on simulanteous equations

Recent Posts

User:
Password: