In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true. In other words, it is where the two graphs intersect, what they have in common. So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system
Answer:
Step-by-step explanation:
In this question, you would solve for "a".
Solve:
K = 4a + 9ab
Since we have our "a" on the same side, we can factor it out from the variables:
K = a(9b + 4)
To get "a" by itself, we would have to divide both sides by 9b + 4:
K/9b+ 4 = a
Your answer would be K/9b+ 4 = a
It would look like this:
Answer:No it's not an exponential equation in which f(x)is not qualified to an exponential number. no number from 1 - 20 can be used and get subtracted by 2 to give a square number.
eg-
4×2-2≠ square number
Step-by-step explanation:
Answer:
No solutions
Explanation:
The given system of equations is
2y = x + 9
3x - 6y = -15
To solve the system, we first need to solve the first equation for x, so
2y = x + 9
2y - 9 = x + 9 - 9
2y - 9 = x
Then, replace x = 2y - 9 on the second equation
3x - 6y = -15
3(2y - 9) - 6y = -15
3(2y) + 3(-9) - 6y = -15
6y - 27 - 6y = -15
-27 = -15
Since -27 is not equal to -15, we get that this system of equation doesn't have solutions.
