Question

Algebra help! Question #5 below!

Answer 1

lol i just noticed...you're too green

Answer 2

What? You said that before ;)

Answer 3

"too" green...i mean you're all green haha

Answer 4

lol oh...I see now :D

Answer 5

Ah! that one's good.
Based on the first two equations, since \[\left\{ x_1, y_1 \right\}\] is a solution to the aforementioned two equations, we know that:
(eq1) \[3x_1+2y_1 = 1\](eq2)\[2x_1+3y_1 = 5\]
Notice that if I have some general equation
\[Ax_1+Bx_2 = C\]
Then A = 3 and B = 2 for the first one, and A = 2 ad B = 3 for the second one.
If we simply add the two equations, we get A = 5, B = 5 and C = 6. But we need C = 0, and A = a - i.e. 0 < A < 100 - and B = b - i.e. 0 < B < 100.
For C = 0, we need to add eq(1) five times for every time we subtract eq(2).
This gives us, for five eq1's and one eq2, A = 13 and B = 7 and C = 0. Yay! That's one! Now we need to find all such pairs of A's and Bs under a hundred that satisfy this

Answer 6

Using the five eq1s for one eq2, we can see that A will be the first coeff to be over a 100. Let's solve for when that will happen: 3(5n)-2n = 100, n = 7.6...

Thus we can have 7 possible solutions where we subtract eq2 for every 5 eq1s

Answer 7

But we're not done, unfortunately. You can also subtract 5 eq1s for every 1 eq2 added. Here, the B term grows fastest. We solve the same equation 3(5n)-2n = 100, n = 7.6 since the coeffs for y are the same as for b

Answer 8

Or so you would think. See, having eq2 - 5*eq1 gives A = -1. Thus the solution is not 7+7 = 14, but rather 13.

Answer 9

I'm done. 13 is the answer

Answer 10

Let's see...the answer key says that the answer s 7...

Answer 11

This is the solution they gave...not sure what it means though...

Answer 12

They did the same thing as me
When I solved for 5n(3) - n (1) = 100 I was doing the same thing as that 5k(...) - k(...) = 0 thing

Answer 13

How did they end up with 7 and you with 13 though?

Answer 14

Ah yes! scratch what I said earlier it turns that all A's and B's are negative when you have n eq2s minus 5*n eq1s

Answer 15

Not just the first one.

Answer 16

Oh. Ok. Thanks! :)

Answer 17

Yeah np.