Question

(y+z)/4=(z+x)/3=(x+y)/3; x+y+z=27.discuss the nature of solutions of the following system of equation.​

Answer 1

Answer:

xyz

Step-by-step explanation:

abcdefghijklmnopqrstuvwxyz

Answer 2

Nature of solutions:

Solution.

Starting

We are given $\sf{x+y+z=27}$.

This gives three equations

• $\sf{x+y=27-z}$
• $\sf{y+z=27-x}$
• $\sf{z+x=27-y}$

From the above

$\sf{\dfrac{27-x}{4} =\dfrac{27-y}{3} =\dfrac{27-z}{3} }$

First, this gives three equations

• $\sf{3(27-x)=4(27-y)}$
• $\sf{27-y=27-z}$
• $\sf{4(27-z)=3(27-x)}$

These are the followings

• $\sf{3x-4y+27=0}$ ...(1)
• $\sf{y=z}$ ...(2)
• $\sf{4z-3x-27=0}$ ...(3)

Solving the Problem

After solving the equations, we get the solutions.

The solutions are

• $\sf{x=\dfrac{27}{5}}$, $\sf{y=\dfrac{54}{5}}$, $\sf{z=\dfrac{54}{5}}$.

For your information.

1. We only have to solve (1), (2), (3).

Because except them are already used to find them.

It is actually solving one equation.

It will result in infinitely many solutions.

2. Linear equations have three natures.

They are the following.

• Solution.
• No solution.
• Infinitely many solutions.