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X = 2(Y + Z)/7                  

 21X = 6Y + 6Z       eqn  (1)

Y = 3(X + Z)/8

 16Y = 6X + 6Z         eqn (2)

eqn(2) – eqn (1)  gives

21X – 16Y = 6Y – 6X

27X = 22Y

Y = 27X/22           eqn(3)

Substitute (3) into (1):

21X = 6(27X)/22 + 6Z

6Z = (231X-81X)/11

Z = 150X/66 = 50X/22

X: Y : Z = 1 : 27/22 :  50/22 = 22  : 27 : 50

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Appreciate the help 👍

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Hmm… I guess there’s no way to solve this except by equations 🙁

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Actually, the method they learnt in primary school is to identify either the part, total or difference that remains unchanged (as per picture).

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Nice! I really like this solution.

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Very good solution!

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This method will be easier for most students to understand! Tq 

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Are they doing 3 unknowns at P6 now? Wow. Anyway here's a solution.

CHANGE FRACTIONS TO INTEGERS
  X = 2/7 (Y + Z) => 7X = 2Y + 2Z ...... (Eq1)
  Y = 3/8 (X + Z) => 8Y = 3X + 3Z ...... (Eq2)

THINKING PROCESS
If we eliminate one unknown, we have an equation in 2 unknowns.
That gives us a ratio. So we have to decide what to eliminate.
  For X: we can do 3 * Eq1 and 7 * Eq2 to get 21X
  For Y: we can do 4 * Eq1 and 1 * Eq2 to get 8Y
  For Z: we can do 3 * Eq1 and 2 * Eq2 to get 6Z
Seems easier to get rid of Y as it involves only 1 equation.

MECHANICAL ALGEBRA MANIPULATION
4 * Eq1: 28X = 8Y + 8Z
             = (3X + 3Z) + 8Z (substitute Eq2)
             = 3X + 11Z
So we have 25X = 11Z and X:Z = 11:25.

We let X=11u and Z=25u. We substitute them into Eq2:
      8Y = 3X + 3Z
         = 3(X + Z)
         = 3(36u)
         = (3 x 9 x 4)u 
So     Y = (27 x 4/8)u = (27/2)u

Therefore X:Y:Z = 11u : (27/2)u : 25u = 22:27:50.

CHECK
Let X=22, Y=27 and Z=50 (we can do 22u, 27u or 50u; it is the same).
X = 2/7 x (27 + 50) = 2/7 x 77 = 22
Y = 3/8 x (22 + 50) = 3/8 x 72 = 27
Both relations check out.
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Tq for taking time to do the detailed working. 👍

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