Help please =] Solve system of equations using any method 2x-6y=8 and 2x-6y=3

2 answers:

6 0

Let z be any number (like x would be)

if we set z equal to 2x-6y, then we get z = 2x-6y

The system

2x-6y = 8
2x-6 = 3

turns into this new system

z = 8
z = 3

but z is a single number. It can't be both 8 and 3 at the same time. So there are no solutions

romanna [79]3 years ago

3 0

The answer is no solution. If you use elimination, by subtracting the second equation from the first, you come out with 0=5, which is a false statement. Therefore, there is no solution.

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Eight rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other ro

Brums [2.3K]

Answer:

9.11*10^-4 %

Step-by-step explanation:

To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.

For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320

In order to find the total amount of ways to place the rooks, you can just use a combinatoric:

$\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9$