Please help me solve this.If a=x+y(3p-1) , b=x+y(3q-1) , c=x+y(3r-1) , find: i) the value of a(q-r)+b(r-p)+c(p-q) ii) prove that a-b/p-q = b-c/q-r = c-a/r-p

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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i) You need to replace the expressions of` a,b,c` in the expression you need to evaluate, such that:

`a=x+y(3p-1) , b=x+y(3q-1) , c=x+y(3r-1)`

`a(q-r)+b(r-p)+c(p-q) = (x+y(3p-1))(q-r)+(x+y(3q-1))(r-p)+(x+y(3r-1))(p-q)`

`a(q-r)+b(r-p)+c(p-q) = x(q - r) + x(r - p) + x(p - q) + y(3p-1)(q - r) + y(3q - 1)(r - p) + y(3r - 1)(p - q)`

`a(q-r)+b(r-p)+c(p-q) = xq - xr + xr - xp + xp - xq + y(3pq - 3pr - q + r + 3qr - 3pq - r + p + 3pr - 3qr - p + q)`

Reducing duplicate terms yields:

`a(q-r)+b(r-p)+c(p-q) = 0 + y*0`

`a(q-r)+b(r-p)+c(p-q) =0`

Hence, evaluating the expression `a(q-r)+b(r-p)+c(p-q),` under the given conditions, yields `a(q-r)+b(r-p)+c(p-q) = 0` .

ii) You need to prove that the equality `(a-b)/(p-q) = (b-c)/(q-r) = (c-a)/(r-p)` holds, for the given a,b,c, hence, you need to replace the expressions of a,b,c into the equality you need to test, such that:

`(x+y(3p-1)-x-y(3q-1))/(p-q) = (x+y(3q-1)-x-y(3r-1))/(q-r) = (x+y(3r-1)-x-y(3p-1))/(r-p)`

`(y(3p-1)-y(3q-1))/(p-q) = (y(3q-1)-y(3r-1))/(q-r) = (y(3r-1)-y(3p-1))/(r-p)`

Factoring out y, yields:

`(y(3p - 1 - 3q + 1))/(p-q) = (y(3q - 1 - 3r + 1))/(q-r) = (y(3r - 1 - 3p + 1))/(r-p)`

Reducing duplicate factors and terms yields:

`(3p - 3q)/(p-q) = (3q - 3r)/(q-r) = (3r - 3p)/(r-p)`

Factoring out 3, yields:

`(3(p - q))/(p-q) = (3(q - r))/(q-r) = (3(r - p))/(r-p)`

Reducing duplicate factors yields:

`1 = 1 = 1`

Hence, testing if the given equality holds, under the given conditions, yields `(a-b)/(p-q) = (b-c)/(q-r) = (c-a)/(r-p) ` holds.

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vaaruni | High School Teacher | (Level 1) Salutatorian

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Given  a = x + y(3p - 1)  -------(1)

          b = x + y(3q - 1)  -------(2)

          c = x + y(3r - 1)   -------(3)

( I ) Part : a(q-r) + b(r-p) + c(p-q)

 subtracting equation (3) from  equation (2)  :

  b = x + y(3q - 1)

  c = x + y(3r -1)                                                                            -      -   -                                                                              

 -----------------------

  b - c = 3yq - y - 3yr + y = 3yq - 3yr  = 3yq - 3yr

  b- c = 3y(q - r) 

  q - r = (b - c) / 3y      ---------(4)

similarly subtracting equation (1)  from equation (3)

 c = x + y(3r - 1)  

 a = x + y(3p - 1)                                                                           -     -   -

---------------------

 c - a = 3yr - y - 3yp + y = 3yr - 3yp

 c - a = 3y(r - p)

 r - p = (c - a)/3y   --------(5)

now subtracting equation (2) from equation (1)

 a = x + y(3p - 1)

 b = x + y(3q - 1)                                                                           -     -   -  

 ---------------------

 a - b = 3yp - y - 3yq + 3 = 3yp - 3yq

 a - b = 3(p - q)

 p - q = (a - b)/3y  -----------(6)

 Now  taking require expression :

 a(q-r) + b(r-p) + c(p-q) = a((b-c)/3y) + b((c-a)/3y) + c((a-b)/3y)

                                   = (ab - ac)/3y + (bc - ac)/3y + (ac - bc)/3y

                                   =(ab - ac - bc - ac - ac - bc)/3y

                                   = 0/3y = 0

Hence , a(q-r) + b(r-p) + c(p-q) = 0   Answer     

( II )  Part  ---->   To Prove : (a-b)/(p-q) = (b-c)/(q-r) = (c-a(/(r-p)

Taking : (a-b)/(p-q) = (a-b) / ((a-b)/3y)  [ substituting vllue of (p- q)                                                              from eqn. (6) ]

                              = (3y)(a-b) / (a-b)     

             (a-b)/(p-q)  = 3y    --------(A)

Taking  (b-c)/(q-r) = (b-c) / ((b-c)/3y) [substituting value of (q- r)                                                             from  eqn. (5) ]

            (b-c)/(q-r) = (3y)(b-c)/(b-c)                                               

            (b-c)/(q-r) = 3y   ----------(B)

Last taking (c-a)/(r-p)= (c-a) / ((c-a)/3y) [substituting value of (r- p)                                                               from eqn. (4) ]

            (c-a)/(r-p) = (3y)(r-p)/(r-p)

            (c-a)/(r-p) = 3y  ----------(C)

comparing equations (A), (B) and (C)  we conclude that

  (a - b)/(p - q) = (b - c)/(q - r) = (c - a)/(r - p) [each eqyal = 3y] 

 Proved      

          

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chandramohan | High School Teacher | (Level 1) eNoter

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i) a(q-r)+b(r-p)+c(p-q) = aq-ar+br-bp+cp-cq

= aq-cq+br-ar+cp-bp = q(a-c)+p(c-b)+r(b-a)

now a-c = [x+y(3p-1)]-[x+y(3r-1)]

= y(3p-3r)

now q(a-c) = qy(3p-3r)

similarly

       p(c-b)  =  py(3r-3q)

       r(b-a)   = ry(3q-3p)

Add above three

LHS = y(3pq-3qr+3pr-3pq+3qr-3pr)

       = y(0) = 0

ii)a-b = 3y(p-q)=>a-b/p-q = 3y

  b-c = 3y (q-r)=>b-c/q-r = 3y

  c-a = 3y(r-p)=> c-a/r-p = 3y

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