joshoyen
  • joshoyen
Solve this system of equation. 2x+4y+z=1 -4x-2y+2z=22 -2x-6y-4z=-14
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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triciaal
  • triciaal
one approach using eqn1 find the expression for z then substitute in eqn 2 using this expression for z substitute in eqb 3 as well you will now have simultaneous eqn with 2 variables (x and y) there are about 4 methods to choose from to solve. Can you solve 2-variable eqn? with those values of x and y substitute in the expression for z to get that value, notice that if you add eqn 1 and eqn 3 you can elimiinate x work with this and let me know if you need more
joshoyen
  • joshoyen
eqn1: z = -2x-4y+1
joshoyen
  • joshoyen
eqn2: -8x-10y=20

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joshoyen
  • joshoyen
im confused what to do next
triciaal
  • triciaal
|dw:1450328379478:dw|
triciaal
  • triciaal
gibe me a min let me find my glasses
joshoyen
  • joshoyen
oh alright
whpalmer4
  • whpalmer4
\[2x+4y+z=1\]\[ -4x-2y+2z=22\]\[ -2x-6y-4z=-14\] I would start by adding equations 1 & 3. This will give you an equation only in terms of \(y\) and \(z\). Next, multiply equation 1 by \(2\) and add it to equation 2. This will give you another equation only in terms of \(y\) and \(z\). Call the first equation you got by combining equations equation A, and the second one equation B. Multiply equation A by 3 and add it to equation B. You should now have an equation in only 1 variable. Solve it for the value of that variable. Then use that value to work backwards through the equations to get the other two.
triciaal
  • triciaal
@whpalmer4 agree did not use the fastest I have x = -5
joshoyen
  • joshoyen
wait so it would be -2y-3z=-13 for eq 1 and 3?
whpalmer4
  • whpalmer4
Yes, so far so good!
triciaal
  • triciaal
|dw:1450329709130:dw|
joshoyen
  • joshoyen
how did you get Z
whpalmer4
  • whpalmer4
Which of us are you asking?
triciaal
  • triciaal
|dw:1450329880455:dw|
joshoyen
  • joshoyen
triciaal
triciaal
  • triciaal
|dw:1450330038896:dw|
joshoyen
  • joshoyen
on my paper it says x = -5
whpalmer4
  • whpalmer4
no, \[-x =5\] means \[x = -5\]
whpalmer4
  • whpalmer4
she just forgot the - sign...
triciaal
  • triciaal
x = -5
joshoyen
  • joshoyen
i just don't get how to get Z
whpalmer4
  • whpalmer4
\[1: 2x+4y+z=1\]\[2: −4x−2y+2z=22\]\[3: −2x−6y−4z=−14\] Add equations 1 and 3 together: \[2x+(-2x)+4y+(-6y)+z+(-4z) = 1 + (-14)\]\[0x -2y -3z = -13\]\[A: -2y -3z = -13\] Next add 2*equation 1 + equation 2 together: \[2*2x + (-4x) + 2*4y + (-2y) + 2*z + 2z= 2*1 + 22\]\[0x + 6y + 4z = 24\]\[B: 6y + 4z = 24\] Now add 3*equation A + equation B together: \[3*(-2y) + 3(-3z) + 6y + 4z = 3*(-13) + 24\]\[0y - 5z = -15\]\[z = 3\] Plug \(z = 3\) into equation A or B and solve for \(y\): \[6y+4(3) = 24\]\[y=2\] Now plug \(z = 3,\ y=2\) into equation 1, 2 or 3 and solve for \(x\): \[2x+4(2)+(3)=1\]\[2x+11=1\]\[x=-5\] To verify this answer, you should put \((-5,2,3)\) into all 3 equations and verify that they work. That is a necessary condition for the solution to be correct.
joshoyen
  • joshoyen
Oooh I get it now, Thanks guys!
triciaal
  • triciaal
|dw:1450330730829:dw|
triciaal
  • triciaal
do you understand the general process?
joshoyen
  • joshoyen
Yep now I do, took me a while but i got it now
triciaal
  • triciaal
ok and as stated, always check the solution in the original. keep practicing!
joshoyen
  • joshoyen
will do

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