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For a certain value of k, the system

 

3a + 4b = 7,

6a + 4b = k - 3a - 8b

 

has infinitely many solutions (a,b). What is k?

 Apr 4, 2021
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To have infinitely many solutions, we must have that the two equations are multiples of each other. Moving all variables to the left of the second equation gives $9a+12b=k$. Aha! The second equation is $3$ times the first! So we need $k$ to be $3(7)=\boxed{21}$

 Apr 4, 2021

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