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TheAkuma
Jun18-09, 05:07 AM
Hey everyone. Okay, first of all I'm not sure if these equations are simultaneous or linear so please correct me if I'm wrong. I need to know a simple procedure to tackle on these equations.Thanks.

a) y=x-6 and y=2x-14
b) y=0.5x-7 and y=3x-22
c) 9y-4x+6=0 and 4x-2y-7=0

CompuChip
Jun18-09, 05:18 AM
Well, the first two are very easy because you have just "y" on the left hand side. Since obviously, y cannot have two different values at the same time, you must have y = y and therefore in the first case also x - 6 = 2x - 14 and in the second case 0.5x - 7 = 3x - 22. From this you can solve x in both cases, and then you just plug that value back in to also find y.

The third case is a little more tricky. I suggest rewriting one of the equations to isolate either x or y, then plug that into the other equation so you get a single equation in a single variable. For example, you can rewrite the second one to 4x = .... and then replace the 4x in the first equation by that expression (watch out for the minus sign in front!)

Borek
Jun18-09, 05:26 AM
c) 9y-4x+6=0 and 4x-2y-7=0

Alternatively - note, that on the left hand side of both equation you have 4x and -4x. If you add these equations side by side, x will cancel out.

Quite often this can be done just by multiplying one of the equations by a constant. For example

x + 2y = 7
2x + 3y = 6

multiply first equation by -2:

-2x - 4y = -14

add equations:

-2x - 4y = -14
2x + 3y = 6
----------------
-y = -8

You may as well multiply second equation by 2 and then subtract it from the first equation, it will give the same effect.

TheAkuma
Jun18-09, 05:31 AM
Well, the first two are very easy because you have just "y" on the left hand side. Since obviously, y cannot have two different values at the same time, you must have y = y and therefore in the first case also x - 6 = 2x - 14 and in the second case 0.5x - 7 = 3x - 22. From this you can solve x in both cases, and then you just plug that value back in to also find y.

The third case is a little more tricky. I suggest rewriting one of the equations to isolate either x or y, then plug that into the other equation so you get a single equation in a single variable. For example, you can rewrite the second one to 4x = .... and then replace the 4x in the first equation by that expression (watch out for the minus sign in front!)

Thank you so very much. That third one really made me scratching my head. Are you able to give me the order of operation to rewrite the 4x= please?

TheAkuma
Jun18-09, 05:33 AM
Alternatively - note, that on the left hand side of both equation you have 4x and -4x. If you add these equations side by side, x will cancel out.

Quite often this can be done just by multiplying one of the equations by a constant. For example

x + 2y = 7
2x + 3y = 6

multiply first equation by -2:

-2x - 4y = -14

add equations:

-2x - 4y = -14
2x + 3y = 6
----------------
-y = -8

You may as well multiply second equation by 2 and then subtract it from the first equation, it will give the same effect.

THanks. So you're saying all i have to do is just add the equations for c) together?

Borek
Jun18-09, 05:48 AM
Try it.

TheAkuma
Jun18-09, 07:04 AM
Try it.

I don't know what i keep doing wrong. I keep getting 7y=1

TheAkuma
Jun18-09, 07:51 AM
Try it.

Oi dude, my Maths B exam is tomorrow and i need to know this question please

TheAkuma
Jun18-09, 08:30 AM
Oi dude, my Maths B exam is tomorrow and i need to know this question please

its alright, ii finally got it. Thanks

Borek
Jun18-09, 08:49 AM
I keep getting 7y=1

Good :smile:

I was AFK, but I see you managed to get it right on your own.

HallsofIvy
Jun18-09, 10:05 AM
For (c), the simplest way is to notice that the two equations, 9y-4x+6=0 and 4x-2y-7=0 Have "4x" and "-4x". If you add the two equations, they cancel leaving (9-2)y+ (6-7)= 0 or 7y- 1= 0. Then 7y= 1 so y= 1/7. Replace y in either of the original equations to get a single equation to solve for x.

By the way, the equations are both "simultaneous" (because there is more than one equation) and "linear" (because you have no powers of x and y except x^1= 1 and y^1= y or more complicated functions of x and y.