algebra

2a+3b-4c=7,
a-b+2c=6

a) find an ordered triple of numbers (a,b,c) that satisfies both equations
b) can you find a second ordered triple that satisfies both equations
c) solve for b in terms of a, and solve for c in terms of a. how many solutions does this system have?

i dont understand how i can solve it because there are 3 variables but only 2 equations?

for c) i got
b=6+2c+a
i dont know how to transform the c in a term? please help

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  1. let's do c) first:

    first equation as is ---> 2a + 3b - 4c = 7
    double the second --> 2a - 2b + 4c = 12
    now add them:
    4a + b = 19
    b = 19 - 4a

    now you can pick any value for a that you want,
    e.g. a = 2
    then b = 19-8 = 11
    sub those values back into the original second equation:
    a-b+2c=6
    2-11+2c = 6
    2c = 15
    c = 7.5 , so a solution would be (2, 11, 7.5)

    how about a=0 , then
    b = 19-0 = 19
    in a-b+2c=6
    0 - 19 + 2c = 6
    c = 12.5

    etc
    There will be an infinite number of solutions.

    suppose you are given x + y = 10
    you are given 2 variables but only one solution, so there would be an infinite number of solutions, namely all the points on the given line.

    Your question has the same concept, except each of the given equations could represent the equation of a plane in 3D.
    So you are really looking at the intersection of two planes. Think of the case where a wall meets the ceiling, there would be an infinite number of points on that line of intersection.

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  2. thank you very much! i understand now

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