A line has parametric equations of x = 3 - t, y = -2 - 4t. Determine the Cartesian equation for this line.

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The parameter for this line is t.  A standard method to go from parametric form to cartesian form is to isolate t in both the x and y equations and then set them equal to each other.

So, for the x equation, we get t = 3 - x and for the y equation we get t = (-y-2)/4

by setting them equal, we get what is called the symmetric form of the line, which is 

3-x = (-y-2) / 4

Multiplying both sides by 4, we get

12-4x = -y-2

Depending on what you want, you can get the standard form of the line by moving everything to the left side of the equation

-4x+y+14 = 0

although you will often see lines in the slope-intercept form, which means isolating y to get


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You need to solve for t the parametric equations such that:

`t = 3 - x`

You need to substitute `3 - x`  for t in equation `y = -2 - 4t`  such that:

`y = -2 - 4(3 - x)`

You need to open the brackets such that:

`y = -2 - 12 + 4x`

`y = -14 + 4x`

You need to move all terms to the left side such that:

`-4x + y + 14 = 0`

Hence, converting the parametric equations of the line in cartesian equation yields `-4x + y + 14 = 0.`

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