Use back substitution to solve the system of linear equations
x-2y+z=8
y+z=5
z=2
**I tried to solve it eventho im still confused/lost.
1=3 idk ugh
well, let's put z=2 into the 2nd
z=y + 2 = 5
y = 3
So now we know that z=2 and y = 3
now into the 1st:
x - 6 + 2 = 8
z = 12
So, where was your problem with this ?
Looks very very easy.
Thanks ...its easy for you because your good in math. IM not.
To solve the system of linear equations using back substitution, we start from the last equation and substitute its solution into the previous equations until we find all the variables.
Given system:
1) x - 2y + z = 8
2) y + z = 5
3) z = 2
From the third equation, we have the solution z = 2. Now, we substitute this value into the second equation:
2) y + 2 = 5
Solving this equation for y, we get y = 5 - 2 = 3. Now, we substitute the values of y = 3 and z = 2 into the first equation:
1) x - 2(3) + 2 = 8
x - 6 + 2 = 8
x - 4 = 8
Solving the final equation for x, we get x = 8 + 4 = 12.
Therefore, the solution to the system of linear equations is:
x = 12, y = 3, z = 2.