
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.