We have to solve the following system of equation sfor x, y and z

y + z - x = 8 ...(1)

5x = 4y ...(2)

x/4 = z/7 ...(3)

From (2) y = (5/4)x and from (3) z = (7/4)x

substitute these in (1)

(5/4)x + (7/4)x - x = 8

=> x(5/4 + 7/4 - 4/4) = 8

=> x*2 = 8

=> x = 4

y = 5

z = 7

**The solution is x = 4, y = 5 and z = 7**

y+z-x=8.......(a)

5x=4y...........(b)

x/4=z/7

or 7x=4z...........(c)

addind eq b and c

5x+7x=4y+4z

12x=4(y+z)

12x/4=y+z

3x=y+z

using value of y+z in aq a

3x -x=8

2x=8

x=4

eq b can be written as

5*4=4y

y=5

now eq c may be written as

7*4=4z

z=7

We'll recall the fact that the number of variables to be determined has to be equal to the number of relations provided.

We notice that we have to determine 3 variables and the problems gives us three conditions, therefore the variables can be determined.

We'll express the y and z terms with respect to x.

5x = 4y <=> y = 5x/4

x/4 = z/7 <=> z = 7x/4

We'll replace y and z into the relation y+z-x = 8, such as:

5x/4 + 7x/4 - x = 8 <=> 12x/4 - x = 8

3x - x = 8

2x = 8 => x = 4

We'll determine y and z:

x = 4 => y = 5*4/4 => y = 5

x = 4 => z = 7*4/4 => z = 7

**The variables x , y and z are: x=4 , y=5 , z=7.**