Finding Transformation of a Line

I'm having a bit of trouble with this one. Can anyone help me spot where I've gone wrong?

Many thanks.

**Q. **If f is the transformation (x, y) ---> (x', y'), where x' = 3x + 2y & y' = 5x + 3y, find f(K), where K = 2x + y = 0.

**Attempt: **Begin by expressing x & y in terms of x' & y': x' - y' = 3x + 2y - (5x + 3y) => x' - y' = -2x - y => y' - x' = 2x + y => y = y' - x' - 2x

Now apply this expression of y to x': x' = 3x + 2y => x' = 3x + 2(y' - x' - 2x) => x' = 3x + 2y' - 2x' -4x => x' = 2y' - 2x' -x => x = 2y' -3x'

Returning to the expression of y', factor in the expression of x to bring equation to x'/ y' format: y' = 5x + 3y => 3y = y' - 5x => y = $\displaystyle \frac{y' - 5x}{3}$ => y = $\displaystyle \frac{y' - 5(2y' - 3x')}{3}$ => $\displaystyle \frac{y' - 10y' - 15x'}{3}$ => $\displaystyle \frac{-9y' - 15x'}{3}$ => y = -3y' - 5x'.

Now apply the values of x & y to K = 2x + y = 0 => 2(2y' - 3x') + (-3y' - 5x') => 4y' - 6x' - 3y' -5x' => y' - 11x' = 0 or y - 11x = 0

**Ans.: **(From text book): x - y = 0

Re: Finding Transformation of a Line

Quote:

Originally Posted by

**GrigOrig99** y = $\displaystyle \frac{y' - 5(2y' - 3x')}{3}$ => $\displaystyle \frac{y' - 10y' - 15x'}{3}$

That should've been $\displaystyle \frac{y' - 10y' + 15x'}{3}$

Quote:

**Attempt: **Begin by expressing x & y in terms of x' & y': x' - y' = 3x + 2y - (5x + 3y) => x' - y' = -2x - y => y' - x' = 2x + y

Interestingly, you could've finished the problem here itself.

$\displaystyle y'-x'=2x+y$

So, $\displaystyle 2x+y=0\implies x'-y'=0$

Re: Finding Transformation of a Line