Question

simultaneously solve y=-0.05t(t-20) and y=-0.1t(t-28)

Answer 1

even if you remotely know how to do this. Any help is always welcome.

Answer 2

@jb1515g could you please help me!!

Answer 3

Solve for t in one of the equations, then plug that expression into the other equation to solve for y (which will get you a number). Then plug that number in to solve for t's value.

Answer 4

y=-0.05t(t-20) and
y=-0.1t(t-28)

so the equations can be solved the same way u do normal substitution
y = -0.05t(t-20)
= t - 0.05t^2

y = -0.1t(t-28)
= 2.8t - 0.1t^2

t - 0.05t^2 = 2.8t - 0.1t^2

divide both sides by t

1 - 0.05t = 2.8 - 0.1t

therefore t = ...?

Answer 5

@hmaug4 follow Jack's explanation its pretty sound!

Answer 6

I have a question about the divide everything by t, what does this do?
@Miracrown and @Jack1

Answer 7

as everything is a multiple of t, the solution t = 0 is valid (therefore y = 0)

Answer 8

@hmaug4 it proves the y can equal 0 at t = 0, other than that it merely simplifies the equation, in the same way that 400x + 200x^2 = 800x
you can divide both sides of the equation by 100 or by x or by both and you'll still arrive at the same answer (x = 2)
4x + 2x^2 = 8x (divide by 100)
400 + 200x = 800 (divide by x)
4 + 2x = 8 (divide by both)

Answer 9

now if the above question was phrased as y = 400x +200x^2 and y = 800x
then the solution x = 0 would still be accurate (as y = 0 if x = 0 is TRUE for both statements)

Answer 10

so 2 possible answers, x = 2 and x = 0

Answer 11

okay thankyou @Jack1 I really appreciate this. Thanks to u I will now be able to pass my maths assignment. Thanks for the explanation!!!!!!!!!!!!!!!

Answer 12

all good dude, yw hey