how do you find the equation of a line with the parametric equations of x=t-3, -3/5t+1?

Question

sorry thats y=3/5t+1

Answer 1

solve both equations for t:
x=t-3
t=x+3
y=-3/5t+1
y-1=-3/5t
t=-5/3y+5/3
now set the two different values for t equal to each other
x+3=-5/3y+5/3
3x+9=-5y+5
3x+5y=-4
There's your line! convert it to slope-intercept if you need to...

Hope that helps!

Answer 2

x = t-3 => x+3 = t

Substitute t = x+3 into y = 3/(5t+1),

y = 3/[5(x+3)+1]
y = 3/[5x+15+1]
y = 3/(5x+16)

Answer 3

Solve both for t.
Equation 1
x=t-3, so t=x+3
Equation 2
y=3/5t+1 so t=(y-1)5/3

Hence x+3=5/3y-5/3
or 5/3y=x+14/3, 5y=3x+14 or y=3/5x+14/5

Answer 4

I assume you mean y = -(3/5)t + 1.

Solve one of the equations for t, and then substitute that expression for t into the other equation.

In this case, t = x + 3, so
y = -(3/5)t + 1 = -(3/5)(x+3) + 1
= -(3/5)x - 4/5

Answer 5

the point(t-3,3/5t+1) lies on the required line for all values of t

first, we must eliminate t

x=t-3 >>>>> t=x+3

y=3/5t+1>>> 5y= 3t+5

>>>t= 5(y-1)/3

we now know the value of t in terms
of both x and y,so we can equate

5(y-1)/3=x+3

5y-5 =3x+9
5y-3x-14=0

therefore, the cartesian equation of the line is:

5y-3x-14=0

i hope that this helps

Answer 6

solve for t in one equation
t=x+3
substitute it into the second equation:
y=3/5t+1
y=(3/5)(x+3) +1

Answer 7

if x=t-3 n y=3/5t
then t=x+3

y=0.6(which is 3/5)(x+3)
then y= 0.6x+1.8