Simultanious Solutions (1 Viewer)

[chocaholiic]

hey, can't remember how to do this stuff:

y= - 1/2 x +10 and the other one is y= 1/3x

exam this wednesday

 

[table for 1]

equate the two equations, since you have the y's as the subject

ie. -1/2x + 10 = 1/3x

then continue from there, ie. mutiply out, put x on side side then solve for x

btw, is that 'a third x' or 'one over 3x' ? same for -1/2x ?

 

[chocaholiic]

its a third x

 

[table for 1]

so can you do it now? i don't want to just post up the solution, i want you to do the rest by yourself, so then you can know what part you're having trouble with

 

[Trev]

I'll post it up, lol:
y= -x/2 +10............(1)
y = x/3.......................(2)
Substitute (2) into (1) gives;
x/3 = -x/2 + 10
Multiply by 6 to get rid of denominators (well, change to denominator = 1)
2x = -3x + 60
5x = 60
x = 12;
Therefore, to find 'y'; substitute x = 12 into (2)
y = 12/3 = 4
Therefore, x = 12; y = 4.

 

[chocaholiic]

thanx dude

 

[table for 1]

I'll post it up, lol:
y= -x/2 +10............(1)
y = x/3.......................(2)
Substitute (2) into (1) gives;
x/3 = -x/2 + 10
Multiply by 6 to get rid of denominators (well, change to denominator = 1)
2x = -3x + 60
5x = 60
x = 12;
Therefore, to find 'y'; substitute x = 12 into (2)
y = 12/3 = 4
Therefore, x = 12; y = 4.

hey !! lol...!!

she's supposed to continue herself from the first step, and then see where she gets stuck, and then ask. how's she going to learn if she just copies what you write?

fine...since nobody cares, i'll shut up. =P

 

[chocaholiic]

its cool i have learnt. sorry it was my fault i didnt reply

choca

 

[PC]

I don't think anyone should need to do simultaneous equations in the General course. The only time one would do it is when drawing graphs and finding a point of intersection.

Still ... the algebraic bit is a useful skill to have up your sleeve!