Parametric Equation question.?

Question

"The problem says to sketch the curve represented by te parametric equations and write the corresponding rectangular equation by eliminating the parameter." This is the problem:

x=3t-1, y=2t+1

This is how the problem is solved in the solutions manaual.:

Step 1: y=2((x+1)/3)+1
Step 2: 2x-3y+5=0

No how does step one develop into step 2??? In lost on this!

Answer 1

y=2((x+1)/3)+1
Multiply both sides by 3:
3y = 2(x + 1) + 3
3y = 2x + 2 + 3
3y = 2x + 5
Subtract 3y from each side:
2x - 3y + 5 = 0.

Answer 2

y=2((x+1)/3)+1 get rid of the () by multiplying the 2 in the numerator and put the answer over 3 (the old denominator which doesn't change) so you have:

y=(2x+2)/3 + 1 Now triple the whole equation to get rid of the 3 in the denominator.

3y=2x +2 + 3
3y=2x +5
0=2x - 3y +5 subtract 3y from both sides

Answer 3

Cartesian equations are written in the type of what you always for math, it is, in x's and y's. Parametric equations are written with t's to added improve what all of us be conscious of of purposes. The parametric equations given are: x = a cos t, y = b sin t the place -pi < t < pi So, we ought to transform this to basically x's and y's. the main suitable thank you to try this is via manipulating trigonometric identities. the only we can use right this is the Pythagorean identity: (cos x)^2 + (sin x)^2 = a million as a result we can sq. the two between the parametric equations to make it look greater like the Pythagorean identity x = a cos t -> x^2 = a^2 (cos t)^2 y = b sin t -> y^2 = b^2 (sin t)^2 next, we isolate the squared cosines and sines, respectively. (x^2) / (a^2) = (cos t)^2 (y^2) / (b^2) = (sin t)^2 next, merely plug the values back into the Pythagorean identity. (cos t)^2 + (sin t)^2 = a million, (x^2) / (a^2) = (cos t)^2 (y^2) / (b^2) = (sin t)^2, so, (x^2)/(a^2) + (y^2)/(b^2) = a million and that's the Cartersian equation of the given parametric curve (your answer.)

Answer 4

first, solve for t in the first eqation x = 3t - 1

x = 3t - 1
x + 1 = 3t
t = (x + 1)/3


then, plug t in for t in the second equation y = 2t + 1

y = 2t + 1
y = 2( (x + 1)/3 ) + 1
y = 2 (x/3 + 1/3) + 1
y = 2x/3 + 2/3 + 1
y = 2x/3 + 5/3
y = (2x + 5)/3
3y = 2x + 5
0 = 2x - 3y + 5