Question

if u =3x+2y-z,v=x-2y+z and w=x(x+2y-z) show that they are functionally related and find the relation​

Answer 1

$\textbf{Given:}$

$\textsf{u=3x+2y-z}$

$\textsf{v=x-2y+z}$

$\textsf{w=x(x+2y-z)}$

$\textbf{To find:}$

$\textsf{The relation connecting u,v and w}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\textsf{u=3x+2y-z}$.............(1)

$\textsf{v=x-2y+z}$...............(2)

$\textsf{w=x(x+2y-z)}$..........(3)

$\mathsf{Adding\;(1)\;and\;(2),}$

$\mathsf{u+v=4x}$

$\implies\boxed{\mathsf{x=\dfrac{u+v}{4}}}$...........(4)

$\mathsf{Now,\;(1)\,can\;be\;written\;as}$

$\mathsf{u=2x+(x+2y-z)}$

$\mathsf{u-2x=x+2y-z}$

$\mathsf{u-2(\dfrac{u+v}{4})=x+2y-z}$

$\mathsf{u-(\dfrac{u+v}{2})=x+2y-z}$

$\implies\boxed{\mathsf{\dfrac{u-v}{2}=x+2y-z}}$.............(5)

$\mathsf{Using,\;(4)\;and\;(5),\;w\;can\;be\;written\;as}$

$\mathsf{w=\dfrac{u+v}{4}\left(\dfrac{u-v}{2}\right)}$

$\mathsf{w=\dfrac{(u+v)(u-v)}{8}}$

$\implies\boxed{\mathsf{w=\dfrac{u^2-v^2}{8}}}$

$\textsf{This is the required relation.}$

$\textsf{Hence, u,v and w functionally related}$

Answer 2

SOLUTION

GIVEN

u =3x+2y-z,v=x-2y+z and w=x(x+2y-z)

TO SHOW

They are functionally related and find the relation

EVALUATION

Here it is given that

$\sf{u = 3x + 2y - z} \: \: \: ......(1)$

$\sf{v =x - 2y + z } \: \: \: ....(2)$

$\sf{ w = x(x + 2y - z)} \: \: \: ....(3)$

Equation (1) + Equation (2) gives

$\sf{u + v}$

$= \sf{(3x + 2y - z )+ (x - 2y + z)}$

$= \sf{4x}$

$\therefore \: \: \sf{u + v = 4x} \: \: \: \: \: ........(3)$

Now

$\sf{u - v}$

$= \sf{(3x + 2y - z ) - (x - 2y + z)}$

$= \sf{3x + 2y - z - x + 2y - z}$

$= \sf{2x + 4y -2 z }$

$= \sf{2(x + 2y - z ) }$

$\sf{ \therefore \: \: u - v = 2(x + 2y - z ) } \: \: \: ....(4)$

Equation (3) × Equation (4) gives

$\sf{(u + v)(u - v) = 4x .2 (x + 2y - z ) }$

$\implies \sf{ {u}^{2} - {v}^{2} = 8x (x + 2y - z ) }$

$\implies \sf{ {u}^{2} - {v}^{2} = 8w }$

Which shows that u, v, w are functionally related and the required relation is

$\boxed{ \: \: \sf{ {u}^{2} - {v}^{2} = 8w } \: \: }$

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