For a certain value of k, the system 3a + 4b = 7, 6a + 4b = k- 4b has infinitely many solutions (a,b). What is k?
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!
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!
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!
+23245 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.
