In the figure, sed LM || side BC, AL = x + 3, BL = x - 3, AM = x + 5 and CM = x - 2 then complete the following activity to find the value of x.

Activity :
In `Delta ABC`, sed LM || side BC,
`:.` by ________ theorem,
`(AL)/(LB) = (AM)/(square)`
`:. (x + 3)/(x - 3) = (x + 5)/(square)`
`:. (x + 3) ________ =(x + 5) (x - 3)`
`x^(2) + x - square = x^(2) + 2x - 15`
`:. x = square`

Explanation

In `Delta ABC`, seg LM|| side BC,
`:.` by basic proportionality theorem,
`(AL)/(LB) = (AM)/(MC)`
`:. (x + 3)/(x - 3) = (x + 5)/(x - 2)`
`:. (x + 3) (x - 2) = (x + 5) (x - 3)`
`:. X^(2) + x - 6 = x^(2) + 2x - 15`
`:. X = 9`