5x+6y=510x+12y=10 what is the solution of the system?

Question

Please help.

Answer 1

5x+6y=10 and 510x+12y=10 implies 255x+6y=5

from the first equation we find that x=(10-6y)/5. in 2nd equation we can replace x by this value. thus we have 255(10-6y)/5 +6y=5

51(10-6y) +6y=5

510 - 306y+6y = 510-300y = 5

y=505/300 =101/60 thus x=(10-6(101/60))/5 = -1/50

done!!!!! hope it is correct and i have helped u.... :)

Answer 2

well since you know

5x +6y = 510x + 12y

and 10 = 510x + 12y

you know 5x + 6y = 10  [eqn 1]

also 510x + 12y = 10

and 5x + 6y =10

so 5x + 6y = 510x + 12y

505x + 6y = 0 [eqn 2]

since you now have 2 eqns you can either solve by substitution or elimination (I would use elimination here) so subtract eqn 1 from eqn 2 so that one of the variables will be either 0x or 0y and can be eliminated:

505x + 6y = 0

-(5x + 6y = 10)

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

500x +0y = -10

so 500x = -10

x = -1/50

sub x into eqn 1 to get y:

-5/50 + 6y = 10

6y = 10 + 1/10

y = (101/10)/6

y = 101/60

Answer 3

Am I about to do your homework for you?

If 5x + 6y = 510x + 12y then 505x + 6y = 0 by subracting the left hand side from the right.

If 505x + 6y = 0 then y = -(505/6)x.

Lets plug that new y into one of the first equations such as 5x + 6y = 10 and get 5x + 6 * -(505/6)x = 10. Simplifying gives us 5x - 505x =  10, so -500x = 10 and our x is now -(1/50), or .02.

Lets solve for y now that we know x; y = -(505/6)x so plug our .02 in and get y = -(505/6)(.02), so y = -505/300.

x = -1/50 or .02

y = -505/300 or 1.683

Fundamentaly 5x + 6y = 10 has an infinite number of solutions, 510x + 12y = 10 also has infinitely many. However if both x and y have to be the same for both problems then only one solution exists after combining the two equations.

Answer 4

I would suggest using the Addition Method...

http://www.mathclues.com/MAdditionMethod...

Hope this helps you!  Please email anytime for further questions.

Answer 5

5x+6y=510x+12y=10



6y=505x+12y=10

505x+6y=10

505/10x=6y/10