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Solve the following system of linear equations:
x + 3y − 2z = 1
5x + 16y − 5z = −5
x + 7y + 19z= −41
i got x = 32.78
y= -471/41
z=55/-41

but those are all wrong please help

2006-10-25 10:21:49 · 4 answers · asked by Diggler AKA The Cab Driver 1 in Science & Mathematics Mathematics

4 answers

x = -3
y = 0
z = -2

2006-10-25 10:52:34 · answer #1 · answered by titanium007 4 · 1 0

Start with #1 and solve for x:
x = -3y + 2z + 1

Substitute into #2:
5(-3y + 2z) + 16y - 5z = -5

-15y + 10z + 16y - 5z = -5
y + 5z = -5
y = -5 - 5z

Now substitute into your equation for x to get it in terms of z:

x = -3y + 2z + 1
x = -3(-5 - 5z) + 2z + 1
x = 15 - 15z + 2z + 1
x = -13z + 16

Finally plug this and the equation for y into the last equation:
x + 7y + 19z= −41
(-13z + 16) + 7(-5 - 5z) + 19z = -41
-13z + 16 -35 - 35z + 19z = -41
-29z - 19 = -41
-29z = -22
z = 22/29

Next plug these into your equations for y and x:
y = -5 - 5(22/29)
y = -5 -110/29
y = (-145 - 110)/29
y = -255/29

x = -13z + 16
x = -13(22/29) + 16
x = (-13)(22) / 29 + (16)(29) / 29
x = (-286 + 464) / 29
x = 178 / 29

Summary:
x = 178 / 29 = 6 2/29
y = -255 / 29 = -8 23/29
z = 22/29

2006-10-25 10:23:57 · answer #2 · answered by Puzzling 7 · 0 1

A, it quite isn't any answer do you may hear additionally the data? Sorry i'm incorrect, it is nice if the final -11 is +11 (my mistake i did no longer be conscious the minus sign) So now i'm with D :) you comprehend all equations have x - 2y And z satisfies the three equations as a result it is sturdy for being the arbitrary parameter. Edited: I even wrote the respond on the resource area ... i think of i could have a relax ;) could I upload that for the time of any new layout I made (electronics circuits in communications and administration), I succeed ending it after approximately one hundred severe blunders. as quickly as my instructor informed me: it quite is not suitable how many circumstances you fail in case you finally end up with the wonderful element. Edited: nicely achieved, gianlino

2016-10-16 10:02:06 · answer #3 · answered by Anonymous · 0 0

(3)-(1)
4y+21z=-42
(2)-(1)*5
y+5z=-10
so4y+20z=-40
4y+21z=-42
z=-2
y-10=-10
y=0
x+4=1
x=-3
check
equation(1) -3+4=1 checks
equation (2) -15+10=-5 checks
equation (3) -3-38=-41 checks
so the solution set {-3,0,-2)

2006-10-25 11:02:04 · answer #4 · answered by raj 7 · 1 0

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