Question 760205
[2(x-3y)+5] [3(x-3y)-2] = 0

Step 1: let u = x - 3y

so the equation will be [2(u) + 5][3(u) - 2] = 0

Step 2: Expand

6(u^2) + 11(u) - 10 = 0 

Step 3: substitute the value of "u" which is "x - 3y" (u = x - 3y) in the expanded equation .

6(x-3y)^2 + 11(x-3y) - 10 = 0

Step 4: Simplify

6(x^2 - 6y + 9y^2) + 11x - 33y - 10 = 0

6x^2 - 36y + 54y^2 + 11x - 33y - 10 = 0

6x^2 + 54y^2 - 69y + 11x - 10 = 0  ANSWER