Need help with this simultaneous equation question!! (1 Viewer)

kevinsta

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y=x^3
y=x^2

Use substitution
I know the answers are x=1 y=1 and x=0 y=0
but i dont know how people got x=1 y=1 with working out. Obviously it could be general knowledge but how did people get it. Please help
 

D94

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y=x^3
y=x^2

So:

x^3 = x^2
x^3 - x^2 = 0
x^2 (x-1) = 0 (factorise)

So x^2 = 0 or x-1 = 0

Solving the latter gives x = 1, and making the substitution into either y equation gives y = 1.
 

kevinsta

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y=x^3
y=x^2

So:

x^3 = x^2
x^3 - x^2 = 0
x^2 (x-1) = 0 (factorise)

So x^2 = 0 or x-1 = 0

Solving the latter gives x = 1, and making the substitution into either y equation gives y = 1.
thanks i get it now :)
How about
5s-3t-13=0
3s-7t-13=0

elimination method
 

MrBeefJerky

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To eliminate s, multiply the top equation by 3 and the bottom by 5 and then subtract both equations. Note that you want to get the lowest common multiple.
15s-9t-39=0 -
15s-35t-65=0

26t + 26 = 0
t= -1
You can either sub t=-1 into any of the equations or use elimination again to eliminate t and find s.
To find s, multiply top by 7 and bottom by 3 and subtract.
 

D94

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The idea is to form the same coefficient for either s or t in both equations. So if we want to eliminate s, I guess the easiest way is to find the common multiple, which would be 15, so:

5s - 3t - 13 = 0 (each term x3)
3s - 7t - 13 = 0 (each term x5)

15s - 9t - 39 = 0
15s - 35t - 65 = 0

So subtracting the second equation from the first gives:

26t + 26 = 0

(It may be easier to label your equations, maybe with (1) and (2) for the first and second equations, then write: (1) - (2) to show you are subtracting the two equations)

(Also note, the 15s and 15s cancel each other out, which is idea of the elimination method. This leaves just one unknown variable)

Clearly t = -1, then you can sub that into either of the original equations which gives s = 2.

edit: too slow
 

kevinsta

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To eliminate s, multiply the top equation by 3 and the bottom by 5 and then subtract both equations. Note that you want to get the lowest common multiple.
15s-9t-39=0 -
15s-35t-65=0

26t + 26 = 0
t= -1
You can either sub t=-1 into any of the equations or use elimination again to eliminate t and find s.
To find s, multiply top by 7 and bottom by 3 and subtract.
Oh i realized my mistake now thanks ;)
 

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