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For a certain value of k, the system  3a + 4b = 7, 6a + 4b = k- 4b has infinitely many solutions (a,b). What is k?

Jun 5, 2020

#1
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Ok. If we look at this, it seems like the best substitute for a and b is one.(Simple). So, 10 = k-4. Wouldn't k be 6a+8b? Now, if we look at this, it says 6a+4b=6a+8b-4b. K doesn't have to be constant does it? But if yes, if both sides will output the same thing every time, I am pretty sure that counts as infinite solutions. Hope this helps!

Jun 5, 2020
#2
0

Ok. If we look at this, it seems like the best substitute for a and b is one.(Simple). So, 10 = k-4. Wouldn't k be 6a+8b? Now, if we look at this, it says 6a+4b=6a+8b-4b. K doesn't have to be constant does it? But if yes, if both sides will output the same thing every time, I am pretty sure that counts as infinite solutions. Hope this helps!

Jun 5, 2020
#3
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Ok. If we look at this, it seems like the best substitute for a and b is one.(Simple). So, 10 = k-4. Wouldn't k be 6a+8b? Now, if we look at this, it says 6a+4b=6a+8b-4b. K doesn't have to be constant does it? But if yes, if both sides will output the same thing every time, I am pretty sure that counts as infinite solutions. Hope this helps!

Jun 5, 2020
#4
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A system of two linear equations will have infinitely many solution if both equations reduce down to the same one.

3a + 4b  =  7

6a + 4b  =  k - 4b   --->   6a + 8b  =  k

Muliply the first equation by 2:  3a + 4b  =  7   --->   x 2   --->   6a + 8b  =  14

Compare that equation to the second equation:                       6a + 8b  =  k

They reduce to the same equation if  k = 14.

Jun 5, 2020