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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: astroderp on February 14, 2014, 04:49:22 pm

Title: algebraic techniques application help
Post by: astroderp on February 14, 2014, 04:49:22 pm

This is from a chapter on algebraic techniques, and I assume you need to use simultaneous equations to solve it.
The questions are pretty long, so I've taken a photo of each. I hope you'll be able to read them, thanks.

sorry for the quality :(

Spoiler

(http://i.imgur.com/vnQHLMK.jpg)
(http://i.imgur.com/cEiJ8wH.jpg)
(http://i.imgur.com/Jy2NPza.jpg)

Title: Re: algebraic techniques application help
Post by: RKTR on February 14, 2014, 05:30:38 pm

1. Poise and Ivy  w1=1.20n +16
    friendly triffid   w2=0.40n+32
2. w1=w2
    1.20n+16=0.40n+32
      0.80n=16
        n=20
3. 1.20+16 = 1.20(20)+16
                  =$40
4. maybe average number of customers going to the shops?

Part B
1. a+b=21   
     b+c=24
     c+d=32
     d+e=37
     e+f=31
     f+g=25
     g+a=26
 
    using 1st two, we know c=a+3
    using 3 n 4 , we know e=c+5 =a+8
    using 5 n 6 ,we know e=g+6 
    a+8=g+6
    a=g-2
    sub into 7th , 2g-2=26  , g=14 , a=12
    then b=9,c=15,d=17,e=20,f=11

Part C
let a=lightest , b=2nd lightest and so on

a+b=16 a+c=18 a+d=19 a+e=20  b+c=21 b+d=22 b+e=23 c+d=24 c+e=26 d+e=27

using 1st two, we know c=b+2 ,using 2nd and third d=c+1=b+3 

  d+e=27
  b+3+e=27 ,b+e=24,e=a+8

using a+e=20 
        2a+8=20
         a=6
then b=10 , c=12,d=13,e=14

                       
                                             

Title: Re: algebraic techniques application help
Post by: astroderp on February 14, 2014, 05:34:04 pm

Quote from: RKTR on February 14, 2014, 05:30:38 pm

1. Poise and Ivy  w1=1.20n +16
    friendly triffid   w2=0.40n+32
2. w1=w2
    1.20n+16=0.40n+32
      0.80n=16
        n=20
3. 1.20+16 = 1.20(20)+16
                  =$40
4. maybe average number of customers going to the shops?

Part B
1. a+b=21   
     b+c=24
     c+d=32
     d+e=37
     e+f=31
     f+g=25
     g+a=26
 
    using 1st two, we know c=a+3
    using 3 n 4 , we know e=c+5 =a+8
    using 5 n 6 ,we know e=g+6 
    a+8=g+6
    a=g-2
    sub into 7th , 2g-2=26  , g=14 , a=12
    then b=9,c=15,d=17,e=20,f=11

Part C
let a=lightest , b=2nd lightest and so on

a+b=16 a+c=18 a+d=19 a+e=20  b+c=21 b+d=22 b+e=23 c+d=24 c+e=26 d+e=27

using 1st two, we know c=b+2 ,using 2nd and third d=c+1=b+3 

  d+e=27
  b+3+e=27 ,b+e=24,e=a+8

using a+e=20 
        2a+8=20
         a=6
then b=10 , c=12,d=13,e=14

                       
                                           

Thank you. Thank you so much for the quick response.