Question

Eliminate the parameter.

x = 5t, y = t + 8

Answer 1

@worne001

Answer 2

lol, let t= x/5

Answer 3

x = 5(x/5) = x

Answer 4

@amistre64 the asker is not here to get help, so that's why I didn't start yet. However, yours is not right, hehe

Answer 5

\[t=y-8\]
\[x=5(y-8)\]
\[x=5y-40\]
\[5y=x+40\]
\[y=\frac{ x+40 }{ 5 }\]

Answer 6

mines never right .. but i do believe its correct ;)

Answer 7

surely x =x is aaaaalways right. 100% agree. hihihi....

Answer 8

but it's not what the asker wants, right?

Answer 9

they want a y=mx+b setup in the end

Answer 10

there are many ways to approach it .... erics was fine

Answer 11

yeah, I see.

Answer 12

yes @amistre64 , thank you. I dont understand the process

Answer 13

t is the parameter .. so we want to form this into a t-less construction

Answer 14

we can equate "t" from both equations

x = 5t ; t = x/5

y = t+8 ; t = y-8

since t has to equal t at all times ...

x/5 = y- 8

Answer 15

Solve either x or y for t. Now you have that variable in terms of t. Go into the other equation and replace the t with what we solved it for.

Answer 16

okay, I tried that with a practice prob and it worked when I applied it, ty guys

Answer 17

good luck :)

Answer 18

I have another problem that has a square root, x = sqrt of t, y=4t+1

Answer 19

how would I start this process?

Answer 20

just square to undo a sqrt

Answer 21

x^2 = t

Answer 22

how do you sqrt a letter?

Answer 23

a letter is just a placeholder for "some number"

Answer 24

its the number (whatever it may be) that is getting sqrted

Answer 25

so would this be the right answer? y = 4x2 - 1

Answer 26

well, +1 ... but yes

Answer 27

why + 1?

Answer 28

It was +1 to begin with wasn't it?

Answer 29

because that what they give us: y = 4t + 1

Answer 30

t = x^2 doesnt alter anything but the "t"

Answer 31

oh okay I see..

Answer 32

I appreciate the help.. im no math wiz haha

Answer 33

there is one more Im having trouble with..