SOLUTION: Good Morning: I am still working with back substitution to solve the system of linear equations. Can you help me? {3/4x+2y+2/3z=-3 {-4/5y-4/3z=1 {z=3 Thanks so much!

Question 385955: Good Morning:
I am still working with back substitution to solve the system of linear equations. Can you help me?
{3/4x+2y+2/3z=-3
{-4/5y-4/3z=1
{z=3
Thanks so much!
Denise
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
x + 2y + z = -3
y - z = 1
z = 3
:
Replace z with 3 in the 1st two equations
x + 2y + (3) = -3
y - (3) = 1
Fortunately, this will cause two of fractions to cancel out
x + 2y + 2 = -3
y - 4 = 1
:
x + 2y = -3 -2
y = 1 + 4
:
x + 2y = -5
y = +5
:
Multiply the above equation by -5, get rid of the denominator change the signs, results:
4y = -5(5)
y =
y = -6.25
:
Find x, substitute for y and z
x + 2y + z = -3
y=-6.25, z=3
x + 2(-6.25) + (3) = -3
x -12.5 + 2 = -3
x -10.5 = - 3
x = -3 + 10.5
x = 7.5
Multiply by 4
3x = 4(7.5)
3x = 30
x = 10
:
Summarize: x=10; y=-6.25, z=3

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