Question

Solve using elimination method:

x-2y+3z =4

2x-y+z = -1

4x + y + z = 5

What is z in the solution?

Answer 1

Using first equation, take x on left side and remaining terms on right side, we get

x = 4 +2y -3z

This gives us equation 2nd and 3rd as

2(4 +2y -3z) -y +z = -1.........4th equation and 4(4 +2y -3z) +y+z = 5.......5th equation

now, write equation 4th in terms of y

8 +4y -6z -y+z =-1

or y = (5z-9)/3

put this value into equation 5th to find the value of z

we get

4(4 +2((5z-9)/3) -3z) +((5z-9)/3)+z = 5

Solving this equation, we get

4(((z-18)/3)+4)+((5z-9)/3)+z = 5

this gives

4(((z-18)/3)+4)+((5z-9 +3z)/3 = 5........by LCM

(4z-72)/3 + 16 + ((5z-9 +3z)/3 = 5

taking LCM again, we get

(4z-72 + (16*3) +5z-9 +3z)/3 = 5

on solving, we get

(12z-33)/3 = 5

or 4z-11 =5

adding 11 on each side

4z = 11+5

this gives

4z = 16

or z = 16/4 = 4

Therefore, required value of z is 4