Question

Y+Z÷4=Z+X÷3=X+Y÷2,X+Y+Z=27 SOLVE ANF FIND THE NATURE OF SOLUTION?

Answer 1

Answer:

Solution is (x,y,z)=(15,9,3)

Step-by-step explanation:

I have applied elimination method to solve given system of equations.

Given:

$\frac{y+z}{4}=\frac{z+x}{3}=\frac{x+y}{2}=k(say)$

$\frac{y+z}{4}=k,\frac{z+x}{3}=k,\frac{x+y}{2}=k$

$\implies\:y+z=4k,x+z=3k,x+y=2k$

Adding all these equations, we get

$2x+2y+2z=9k$

$2(x+y+z)=9k$

$2(27)=9k$

$54=9k$

$k=6$

Now, we have

$y+z=24$..................(1)

$x+z=18$.................(2)

$x+y=12$..................(3)

$x+y+z=27$..............(4)

$(4)-(1)\implies\:x=3$

$(4)-(2)\implies\:y=9$

$(4)-(3)\implies\:z=15$

Solution is

(x,y,z)=(15,9,3)

Answer 2

Hope this helps ;)

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#mumukshu