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anonymous
 4 years ago
solving by elimination
1)
2xy=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!
anonymous
 4 years ago
solving by elimination 1) 2xy=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!

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jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.12xy=3 4x+3y=21 3*(2xy)=3*3 4x+3y=21 6x3y=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 2xy = 3 2(3)y = 3 6  y = 3 y = 36 y = 3 y = 3 So x = 3 and y = 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you so much! would you be willing to explain this one? xy=2 2x+2y=5

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1xy=2 2x+2y=5 2(xy)=2*2 2x+2y=5 2x2y=4 2x+2y=5  0x+0y = 9 0 = 9 ... which is FALSE So there are no solutions. This means that the system is inconsistent.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Solving by substitution: xy=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
 4 years ago
Best ResponseYou've already chosen the best response.0here is one... 2xy+5z=24 y+2z=4 z=6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for using backsubstitution

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1y+2z=4 y+2(6)=4 y + 12 = 4 y = 412 y = 8  2xy+5z = 24 2x  (8) + 5(6) = 24 2x + 8 + 30 = 24 2x + 38 = 24 2x = 2438 2x = 14 x = 14/2 x = 7 So the solutions are x = 7, y = 8 and z = 6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are incredible thank you!

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1You're welcome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey are you still on here? can you help me with solve using rowoperations... x+y+z=6 2xy+z=3 3x z=0

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1x+y+z=6 2xy+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 = 31 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
 4 years ago
Best ResponseYou've already chosen the best response.0May i ask you what the R stands for

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1R stands for the row So R1 means "row 1", R2 means "row 2", and R3 means "row 3"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so how did you determine subtract row 3 3 by row1

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1When 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 = 33 = 0 This basically "eliminates" the x variable in equation 3 which gets me one step closer to solving for x, y, and z

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay can you tell me what you get for this one then... 6y+4z=18 3x+3y = 9 2x z =12

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1using the same method of row reduction or some other method?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1alright thanks, one sec

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.16y+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
 4 years ago
Best ResponseYou've already chosen the best response.0THANK 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
 4 years ago
Best ResponseYou've already chosen the best response.1sure whats the problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same thing so solving by row operations :)... its x2y+5z=2 4x z=0

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1x2y+5z=2 4x z=0 1 2 5 2 4 0 1 0 1 2 5 2 0 8 21 8 R2  4*R1 x2y+5z = 2 8y21z=8  8y21z=8 8y = 8 + 21z y = (21z  8)/8  x2y+5z = 2 x2( (21z  8)/8 )+5z = 2 x  4(21z  8) + 5z = 2 x  84z + 32 + 5z = 2 x  79z + 32 = 2 x  79z = 232 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
 4 years ago
Best ResponseYou've already chosen the best response.1A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer is consistent and dependent or x = 79z  30 sorry I am a little confused haha

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Well both aspects make up the total answer. The idea here is that there are an infinite number of points where the two equations x2y+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 = 79s30, 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
 4 years ago
Best ResponseYou've already chosen the best response.0That makes sense! Thank you again! I appreciate all your help!

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1You're welcome, glad it's all clicking.
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