Question

please can someone help asap ? thank you ! :)

Answer 1

Answer:

$\frac{x + 2}{3x - 1}$

Step-by-step explanation:

Things you should before solving this question :-

• a²- b² = (a + b)(a - b)

where a and b are any two variables.

SIMPLIFYING THE EXPRESSION

$\frac{ {x}^{2} - 9 }{3 {x}^{2} + 8x - 3} \div \frac{x - 3}{x + 2}$

using the identity discussed above

$\frac{ (x + 3)(x - 3) }{3 {x}^{2} + 8x - 3} \div \frac{x - 3}{x + 2}$

simplifying 3x² + 8x - 3

$\frac{ {x}^{2} - 9 }{3 {x}^{2} + 9x -x - 3} \div \frac{x - 3}{x + 2}$

as 8x can be written as 9x - x

$\frac{ (x - 3)(x + 3) }{3 {x}(x + 3) - 1(x + 3)} \div \frac{x - 3}{x + 2} \\ \frac{ (x - 3)(x + 3) }{(3 {x} - 1)(x + 3) } \div \frac{x - 3}{x + 2}$

flipping the fraction that follows division sign and inserting multiply in place of divide

$\frac{ (x - 3)(x + 3) }{(3 {x} - 1)(x + 3) } \times \frac{x + 2}{x - 3}$

canceling (x - 3) and (x + 3) as they're common in both denominator and numerator

$\frac{x + 2}{3x - 1}$