What is the answer for this question using the elimination formula

It is convenient to add 3 times the first equation to the second. This will eliminate x.

In general, you want to find some multiple of one equation that you can combine with some multiple of the other to cause one of the variable coefficients to become zero. Here, we see that x is by itself in one equation, so it is convenient to multiply that by a suitable factor to cancel the x-term in the other equation. The x-coefficient in the second equation is -3, so when we multiply the first equation by +3 and add the result to the second equation, we get +3x-3x = 0x in the result.

3(x +5y) +(3y -3x) = 3(17) + (21)

... 18y = 72 . . . . . . collect terms

... y = 4 . . . . . . . . . divide by 18

Substitute into the first equation:

... x + 5·4 = 17

... x = -3 . . . . . . . . subtract 20

The solution is (x, y) = (-3, 4).

Here, we note the second equation has terms that all have a factor of 3 that can be removed. That is, the second equation can be rewritten as

... y - x = 7

Now, we can add this directly to the first equation to eliminate the x terms.

... (y -x) +(x +5y) = 7 + 17

... 6y = 24 . . simplify

... y = 4 . . . . same as above

We could also subtract 5 times the new second equation from the first to eliminate y.

... (x +5y) -5(y -x) = (17) -5(7)

... 6x = -18 . . . . . . simplify

x = -3 . . . . . . . . divide by the coefficient of x

Answer and Step-by-step explanation:

(A) Suppose 3n is even, this implies that the product of 3n is a multiple of two, let's say 2(3k) where k is a real number. Rearranging that, we have 3(2k), let 2k = n. Therefore, n is even.

(B) Suppose xyz is odd, then let x = 2k, y = 2p + 1 and z = 2q + 1. xyz = 2k(2p + 1)(2q + 1) and this is obviously even since it is a multiple of 2.