Solve the simultaneous equation- 2x+8y=10 and 3x+2y=5

Firstly, lets name each equation A and B- this will help us later. If I told you what x was, you would be able to use either equation to find y- easy! so lets first find one unknown (x or y) then use that to find the other unknown! Now there are two easy ways to do this and your exam board/ test paper may want you to do this in a way specified. Lets start with my personal preference because of ease! Lets start with A, what we want to do is get our x on its own so we can then sub it into our B equation therefore eliminating 'x' leaving us with just the y terms. So lets put our terms that arent x terms to the right so now we have 2x=10-8y, now divide everything by 2 and we have x=5-4y This is our new version of A lets call it A'. Now we can sub A' into B for the 'x'. B is now 3(5-4y) + 2y = 5. If we simplify this by first expanding the brackets 15-12y+2y=5 and then collect our y terms on one side and numbers on the other we have 10y=10 so of course y=1. Now we can sub this y value into A or B (preferably both to check) if we do this to A we have 2x+8(1)=10 giving x=1. Our solution is now x=1 y=1 ! The second way to do this is to compare A and B and see if we can add or subtract a multiple of A or B to the other in order to eliminate x or y. If we multiply B by 4 we get 12x+8y=20 now we see in equation A we have '8y' in there. If we subtract our A from 4B we get 10x+0=10 so obviously x=1 again. If we once again sub this value into both A and B we will get y=1 like we did the first time. We can use 4B because we are multiplying both the LHS of the equation and the RHS so the equation is kept in tact! I hope this helped enough to reuse and apply this to different equations similar to this one, prepare for fractions and negatives aswell!