what do you have to find?
I have to solve using the elimination method.
What you need to do to start with is arrange one of the equations into the form y=... or x=... and then substitute it into the other equation. which one would you like to start with?
So far I know I multiply 3 to the top equation And -2 to the bottom equation
cool, do you know why you do this multiplication?
I have 3*2x+4y=18*3 Equals 6x+4y=54
No ..I don't :/
The point is to eliminate one of the x's or y's from the equation, then it makes it easy to solve it right?
Ok I got you .
You multiply the whole equations so that you get the same number of x or the same number of y on each line, then they can cancel each other out.
so if you multiply the top by 3, you get 12y+6x=54 and the bottom by -2 you get -6x -12y = -52
which is where you were at.. still following?
Ok I know for the 2nd equation they cancel out -2*3x+6y=26*-2 Equals 6x+6y=24y And they cancel out
Wait I'm wrong aren't i..
yeah i'm just trying to follow your math, hold on
this one is tricky because both the x's and y's cancel each other out. are you sure you have it written down right?
are you sure its elimination? substitution would be much easier
It says "solve using the elimination method"
Are you able to help me ?
it has me stumped. can you show me whole of the question?
it looks like it says substitution up the top
ok, so the way i can see this working if you start back with this: so if you multiply the top by 3, you get 12y+ 6x = 54 and the bottom by -2 you get -6x -12y = -52 you followed up to this point right?
My conclusion is that there is no solution. I'll post my working shortly.
Ok because I am so confused lol
Oh wow . Ok , so is that what I tell my instructor ?
do you know how to draw graphs?
Not really .. I'm sorry I'm just now starting to understand this math thing .
You should tell your instructor that you can't solve this using the elimination method. You've talked to some other people who have done university maths, and they can't solve it using the elimination method or the substitution method. When both of these equations are shown on a graph, there is no point where the two lines intersect - so no real values of x and y where both equations are solved. So the equations are ok on their own, but there is no solution.
Ok great , Thank you soo much !! You're awesome ! Helped me a lot
you can have a look at the graphs on dismiss.com, i've also posted the picture showing the two lines don't intersect
any other time you're trying to check your simultaneous equations, you can write them in and check them here too - its great if you're a visual learner
Yes ! I am actually a visual learner do this will definitely help me .
Do you have time for another question ? Last one