Question

Given equations A and B as 2/5x+y=12 and 5/2y-x=6, respectively, which expression will eliminate variable x? A.)5/2A+2/5B, B.)5/2A-B, C.)A+2/5B, D.)5/2A-2/5B, E.)25/10A+6/10B

Answer 1

I need help,Im trying to graduate

Answer 2

Try to put it more in order cause I can't understand anything on those multiple choice lol

Answer 3

Choice a is five over two times A plus two over five times B. Choice b is five over two times A minus B. Choice c is A + two over five times B,

Answer 4

and so on

Answer 5

it needs to be written out..no one I know knows the answer and there are no examples or answers on the internet. its algebra 1 If that helps..

Answer 6

@AccessDenied

Answer 7

To eliminate x, the two equations multiplied by constants should combine to have only y, the x terms have the same coefficient but opposite signs so they add and cancel.

So I think the best way you might approach this is just by going through each example and expanding out each statement. I'll show the first one, and the other three work the same way.
(A)
5/2 A + 2/5 B
5/2 A => 5/2 ( 2/5 x + y = 12 )
=> 5/2 * 2/5 x + 5/2 y = 5/2 * 12
=> x + 5/2 y = 30
2/5 B => 2/5 ( 5/2 y - x = 6 )
=> 2/5 * 5/2 y - 2/5 x = 2/5 * 6
=> y - 2/5 x = 12/5
If we add these two, we add the left-hand and right-hand sides of the equation:
( x + 5/2 y = 30 ) + ( y - 2/5 x = 12/5 )
(x + 5/2 y) + (y - 2/5 x) = 30 + 12/5
In this case, the x values do not cancel. x and -2/5 x add together to get 3/5 x... but x is still there, so these do not eliminate x and (A) is incorrect.

Answer 8

Ah,thanks!

Answer 9

There is some lee-way if you can recognize things in a logical way, although if it doesn't make sense where I am going I won't advise it.
if you look at (B), you can quickly see that there is the positive x in the first equation and the second equation has a negative multiplied by -x or positive x, so there is no way to add those two either to eliminate x either. That speeds up the process, but it isn't as systematic and you gotta think about each case differently. :)

Answer 10

I think 5/2A-2/5B might be the answer..it has the reciprocal in it...

Answer 11

That's a good thought; 2/5 B does give you the - 2/5 x that compares to the one in A, but 5/2 A gets rid of the 2/5 on 2/5 x in the other equation... maybe check out (C) and compare it to (A).

Answer 12

So A,B, and D are out?

Answer 13

Yes, I was getting that as well.
In (C) you get that reciprocal on the other x value without losing it in the first equation, so (discounting the extraneous work, we'll only write x)
A + 2/5 B
2/5 x + 2/5(- x)

Answer 14

So, does the x cancel as well as the 2/5? :/

Answer 15

\( \dfrac{2}{5} x + \dfrac{2}{5} (-x) \)

\( \dfrac{2}{5} x - \dfrac{2}{5} x \)

Adding and subtracting the same number leaves 0, yes? Or formally, un-distribute the x:

\( \left( \dfrac{2}{5} - \dfrac{2}{5} \right) x \)

0*x

= 0

Answer 16

Ah, okay, I thought so. Thanks for your help & time!

Answer 17

You're welcome! :)