anonymous
  • anonymous
solving by elimination 1) 2x-y=3 4x+3y=21 when i solve it i get x=1.2 and y= -.6 but when i check it back in to the equation it is wrong will someone help me!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
2x-y=3 4x+3y=21 3*(2x-y)=3*3 4x+3y=21 6x-3y=9 4x+3y=21 ----------- 10x = 30 10x = 30 x = 30/10 x = 3 Now that we know that x = 3, we can use it to find y 2x-y = 3 2(3)-y = 3 6 - y = 3 -y = 3-6 -y = -3 y = 3 So x = 3 and y = 3
anonymous
  • anonymous
thank you so much! would you be willing to explain this one? x-y=2 -2x+2y=5
jim_thompson5910
  • jim_thompson5910
x-y=2 -2x+2y=5 2(x-y)=2*2 -2x+2y=5 2x-2y=4 -2x+2y=5 ----------- 0x+0y = 9 0 = 9 ... which is FALSE So there are no solutions. This means that the system is inconsistent.

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anonymous
  • anonymous
okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?
jim_thompson5910
  • jim_thompson5910
Solving by substitution: x-y=2 x = 2 + y ------------------------------------------ -2x+2y = 5 -2(2+y) + 2y = 5 -4 - 2y + 2y = 5 -4 = 5 ... which is FALSE So again, the system is inconsistent and there are no solutions.
anonymous
  • anonymous
here is one... 2x-y+5z=24 y+2z=4 z=6
anonymous
  • anonymous
for using back-substitution
jim_thompson5910
  • jim_thompson5910
y+2z=4 y+2(6)=4 y + 12 = 4 y = 4-12 y = -8 ------------------------------------------------------- 2x-y+5z = 24 2x - (-8) + 5(6) = 24 2x + 8 + 30 = 24 2x + 38 = 24 2x = 24-38 2x = -14 x = -14/2 x = -7 So the solutions are x = -7, y = -8 and z = 6
anonymous
  • anonymous
you are incredible thank you!
jim_thompson5910
  • jim_thompson5910
You're welcome
anonymous
  • anonymous
hey are you still on here? can you help me with solve using row-operations... x+y+z=6 2x-y+z=3 3x -z=0
jim_thompson5910
  • jim_thompson5910
x+y+z=6 2x-y+z=3 3x -z=0 1 1 1 6 2 -1 1 3 3 0 -1 0 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 1 1 1 6 0 -3 -1 -9 R2 - 2R1 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 (-1/3)*R2 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 0 0 -3 -9 R3 + 3R2 1 1 1 6 0 1 1/3 3 0 0 1 3 (-1/3)*R3 x+y+z = 6 y+(1/3)z = 3 z = 3 ============================================= ============================================= y+(1/3)z = 3 y+(1/3)(3) = 3 y + 1 = 3 y = 3-1 y = 2 ---------------- x+y+z = 6 x+2+3 = 6 x + 5 = 6 x = 1 So the solutions are x = 1, y = 2 and z = 3
anonymous
  • anonymous
May i ask you what the R stands for
jim_thompson5910
  • jim_thompson5910
R stands for the row So R1 means "row 1", R2 means "row 2", and R3 means "row 3"
anonymous
  • anonymous
okay so how did you determine subtract row 3- 3 by row1
jim_thompson5910
  • jim_thompson5910
When going from 1 1 1 6 2 -1 1 3 3 0 -1 0 to 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 I did this to make that lower left corner value go from 3 to 0 Notice that 3 - 3*1 = 3-3 = 0 This basically "eliminates" the x variable in equation 3 which gets me one step closer to solving for x, y, and z
anonymous
  • anonymous
okay can you tell me what you get for this one then... 6y+4z=-18 3x+3y = 9 2x -z =12
jim_thompson5910
  • jim_thompson5910
using the same method of row reduction or some other method?
anonymous
  • anonymous
row reduction :)
jim_thompson5910
  • jim_thompson5910
alright thanks, one sec
jim_thompson5910
  • jim_thompson5910
6y+4z=-18 3x+3y = 9 2x -z =12 0 6 4 -18 3 3 0 9 2 0 -1 12 0 6 4 -18 1 1 0 3 (1/3)*R2 2 0 -1 12 1 1 0 3 R1 <--> R2 0 6 4 -18 2 0 -1 12 1 1 0 3 0 6 4 -18 0 -2 -1 6 R3 - 2*R1 1 1 0 3 0 -2 -1 6 R2 <--> R3 0 6 4 -18 1 1 0 3 0 1 1/2 -3 (-1/2)*R2 0 6 4 -18 1 1 0 3 0 1 1/2 -3 0 0 1 0 R3 - 6*R2 x+y = 3 y + (1/2)z = -3 z = 0 ------------------------------------------------------- y + (1/2)z = -3 y + (1/2)(0) = -3 y = -3 -------------------- x+y = 3 x + (-3) = 3 x - 3 = 3 x = 3+3 x = 6 So the solutions are x = 6, y = -3, and z = 0
anonymous
  • anonymous
THANK YOU! You don't know how much you have helped me! there is only one more I can't seem to get right do you have time for one more? :)
jim_thompson5910
  • jim_thompson5910
sure whats the problem
anonymous
  • anonymous
same thing so solving by row operations :)... its x-2y+5z=2 4x -z=0
jim_thompson5910
  • jim_thompson5910
x-2y+5z=2 4x -z=0 1 -2 5 2 4 0 -1 0 1 -2 5 2 0 8 -21 -8 R2 - 4*R1 x-2y+5z = 2 8y-21z=-8 ------------------------------------------------------- 8y-21z=-8 8y = -8 + 21z y = (21z - 8)/8 ----------------------- x-2y+5z = 2 x-2( (21z - 8)/8 )+5z = 2 x - 4(21z - 8) + 5z = 2 x - 84z + 32 + 5z = 2 x - 79z + 32 = 2 x - 79z = 2-32 x - 79z = -30 x = 79z - 30 Now if you let z = s, where s is any number, then the solutions are x = 79s - 30, y = (21s - 8)/8, z = s This means that there are an infinite number of solutions. So the system is consistent and dependent.
jim_thompson5910
  • jim_thompson5910
A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.
anonymous
  • anonymous
so the answer is consistent and dependent or x = 79z - 30 sorry I am a little confused haha
jim_thompson5910
  • jim_thompson5910
Well both aspects make up the total answer. The idea here is that there are an infinite number of points where the two equations x-2y+5z=2 and 4x -z=0 intersect So this makes the system dependent (since one equation technically "depends" on the other -- it changes as the other one does) Because at least one solution exists, this makes the system consistent. The answers of x = 79s-30, etc... are the more explicit forms of the solutions and are more specific than saying "there are an infinite number of solutions" This is because you can't just say that something random like (1,2,3) is a solution even though I did say "there are an infinite number of solutions". All of the solutions fit a very specific algebraic form.
anonymous
  • anonymous
That makes sense! Thank you again! I appreciate all your help!
jim_thompson5910
  • jim_thompson5910
You're welcome, glad it's all clicking.

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