I am learning the system of linear equations and my teacher game me some elimination problems. I'm really stumped because its not what I'm used to . If you can solve it please explain and show your work. Thank you

5ax+3y=5

2ax+3y=11

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I am learning the system of linear equations and my teacher game me some elimination problems. I'm really stumped because its not what I'm used to . If you can solve it please explain and show your work. Thank you

5ax+3y=5

2ax+3y=11

5ax+3y=5

2ax+3y=11

- KSwissy
**Posts:**1**Joined:**Fri Feb 08, 2013 1:02 am

KSwissy wrote:I am learning the system of linear equations and my teacher game me some elimination problems. I'm really stumped because its not what I'm used to . If you can solve it please explain and show your work. Thank you

5ax+3y=5

2ax+3y=11

Topic should be placed in Beginning Algebra subforum.

The system of equation seems to use variables x and y for the "unknown" variables. a is another variable but may be taken to act as a constant. The system is suitable to easily eliminate the terms of y to find x. As the system is in its current form, just subtract the second equation from the first equation. That would be like this:

5ax+3y=5

-2ax-3y=-11

----------------

3ax +0 =-6 (actually, I multiplied by -1 and then added it).

x=-6/(3a)

x=-2a.

Now if you want to eliminate x to find y, some multiplying of both equations is necessary.

Original system:

5ax+3y=5

2ax+3y=11

You want to multiply both equations each by its own chosen factor to have the same term of x. Simplest result is 2*5=10.

2*(5ax+3y=5) = 10ax+6y=10

5*(2ax+3y=11) = 10ax+15y=55

Now using the system in that multiplied form, subtract the first from the second:

0x+9y=44

y=44/9

The solution or answer then for the original system is x=-2a and y=44/9.

- jg.allinsymbols
**Posts:**71**Joined:**Sat Dec 29, 2012 2:42 am

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