Question

if x+y=a y+z=b and x+z =c then average of x,y,z is

Answer 1

$x + y = a \\ y + z = b \\ x + z = c \\ add \: all \: the \: equations \\ 2(x + y + z) = a + b + c \\ x + y + z = \frac{a + b + c}{2} \\ average \: of \: x \: y \: and \: z \\ \frac{x + y + z}{3} \\ \frac{a + b + c}{6}$

Answer 2

Answer:

The average of x, y, z is $\frac{a-b+c}{6}$.

Step-by-step explanation:

Step 1 of 2

Consider the given conditions as follows:

x + y = a ____ (1)

y + z = b ____ (2)

x + z = c ____ (3)

From (1), we have

y = a - x ____ (4)

From (3), we have

z = c - x _____ (5)

Now, substitute the values of y and z in the equation (2) as follows:

⇒ (a - x) + (c - x) = b

Solve for x.

⇒ a - x + c - x - b = 0

⇒ a +b + c - 2x = 0

⇒ -2x = -(a + b + c)

⇒ x = $\frac{a+b+c}{2}$

Substitute the value of x in the equation (4) as (5), we get

y = a - $\frac{a+b+c}{2}$

  = $\frac{a-b-c}{2}$

z = c - $\frac{a+b+c}{2}$

  = $\frac{c-a-b}{2}$

Step 2 of 2

Find the average of x, y, z.

Average = $\frac{\frac{a+b+c}{2} +\frac{a-b-c}{2} +\frac{c-a-b}{y} }{3}$

              = $\frac{a+b+c +a-b-c +c-a-b}{6}$

              = $\frac{a-b+c}{6}$

Therefore, the average of x, y, z is $\frac{a-b+c}{6}$.

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