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Solve x(y+z)=44, y(z+x)=50 and z(x+y)=54. Find the value of x,y&z​

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Explanation:

To solve the given system of equations:

1. x(y+z) = 44

2. y(z+x) = 50

3. z(x+y) = 54

We can use substitution or elimination method. Let's use the substitution method.

From equation 1, we have y + z = 44/x.

Rearranging equation 2 gives us y = 50/(z+x), and equation 3 gives us z = 54/(x+y).

Now, substitute these expressions for y and z in equation 1:

(y + z) = 44/x

(50/(z+x)) + (54/(x+y)) = 44/x

To simplify, let's cross-multiply:

50(x+y) + 54(z+x) = 44(z+x)

Expanding both sides:

50x + 50y + 54z + 54x = 44z + 44x

Collecting like terms:

50x + 54x - 44x + 50y + 54z = 44z

Now, simplify further:

60x + 50y + 10z = 44z

Rearranging terms:

60x + 50y = 34z

Divide both sides by 2:

30x + 25y = 17z

Now, we have three equations:

1. y + z = 44/x

2. 30x + 25y = 17z

3. z = 54/(x+y)

To find the values of x, y, and z, we need further information or equations. The given system is not sufficient to solve for the values of x, y, and z.

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