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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!
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
thank you so much! would you be willing to explain this one? x-y=2 -2x+2y=5
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.
okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?
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.
here is one... 2x-y+5z=24 y+2z=4 z=6
for using back-substitution
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
you are incredible thank you!
You're welcome
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
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
May i ask you what the R stands for
R stands for the row So R1 means "row 1", R2 means "row 2", and R3 means "row 3"
okay so how did you determine subtract row 3- 3 by row1
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
okay can you tell me what you get for this one then... 6y+4z=-18 3x+3y = 9 2x -z =12
using the same method of row reduction or some other method?
row reduction :)
alright thanks, one sec
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
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? :)
sure whats the problem
same thing so solving by row operations :)... its x-2y+5z=2 4x -z=0
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.
A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.
so the answer is consistent and dependent or x = 79z - 30 sorry I am a little confused haha
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.
That makes sense! Thank you again! I appreciate all your help!
You're welcome, glad it's all clicking.