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!
Algebra ->
Coordinate Systems and Linear Equations
-> 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!
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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