Another maths prob....

Forums: Science And Math
Email this Topic • Print this Page

I've been working on this problem for ages and Liessa helped me to come up with a solution. However, after inserting the numerical values given, it didn't work out. I've rewritten the equations and gone over this problem several times but I still can't figure out why. Can anyone help?

1) The sum of three numbers is S. The third one is K less than H times the sum of the first two. The first plus K times the second plus L times the third is P. What are the three numbers?

NOTE: the part which I didn't get was the value of K since the given value of K is different for each part. The 'K less than' part is K = 5 but the 'K times the second' part is K = 3. So how can you apply two different values of K into the same equation Confused

Values to be inserted: S = 7, K = 5, H = 2, K = 3, L = 2, P = 15

Anyway, I managed to derive three equations using x, y and z as values:

equation 1) x + y + z = S
equation 2) z = h( x + y ) - K
equation 3) x + Ky + Lz = P

Oh and can anyone help me manipulate this equation in terms of x? I hate working wit square roots so I'm convinced I got the wrong answer Rolling Eyes

2 * [ ( (square root of 3) / 4 ) * ( 4 / square root of 3x ) ^2 ] + 3 * ( 4 / square root of 3x ) x

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 534 • Replies: 4

No top replies

Simplify
After substituting values, you get:

1) x+y+z=7
2) 2x+2y-z=5
3) x+3y+2z=15

Removing x can be done this way

4=3-1) 2y + z = 8
5=Twice3-2) 4y + 5z = 25

Removing y can be done this way

6=5-Twice4) 3z=9

Therefore: z=3

From equation four, if z=3, y=2.5
From equation one, x=1.5

0 Replies

oh no no, what I mean by is generating a formula of x, y and z so that if you plug in those given values, you get the required answer. For example the solutions I got were:

x = (S+K) / (H+1) - (S+K)(LH+1) / (H+1)(1-K) - (P+LK) / (1-K)

y = (S+K)(LH+1) / (H+1)(1-K) - (P+LK) / (1-K)

z = S - (S+K)/(H+1)

0 Replies

Looks like the correct approach
You would use the same technique I used when applying the numbers you provided.

-Rewrite (2) as hx+hy-z=K
-Add (1) and (2) together and the z term drops out leaving an equation in x and y. (4) (H+1)x + (H+1)y = S + K1
-Multiply (2) by L and subtract it from (3). Once again Z drops out and now you have two equations in x and y. (5) (LH+1)x + (LH+K2)y = LK1+P
-Solve (4) and (5) for x.

Since you have two different K values, I called them K1 and K2. I got:

x = (LK1 + P)/(1-K2) - (S+K1)(LH+K2)/(H+1)/(1-K2)

I'll leave the other two for you. I did check that one by plugging in the example numbers you provided and it seemed to work out OK.

0 Replies

ahh thank you!! i got it now!! Thanks again Very Happy

0 Replies

Another maths prob....

Related Topics

Quick Links

My Account

able2know