Question

Please help me ! Reduced to lowest terms

Answer 1

Answer:

Step-by-step explanation:

7) 2ab + 4ac = 2*a*b + 2 *2 *a*c

                     = 2a(b + 2c)

$\dfrac{2ab^{2}c}{2ab+4ac}=\dfrac{2*a* b^{2}*c}{2a(b+2c)}\\\\\\\text{2a in the denominator and numerator get cancelled}\\\\ = \dfrac{b^{2}c}{b+2c}$

8) ac - a²c² = c( a - a²c)

bc - abc     = c(b - ab)

$\dfrac{ac - a^{2}c^{2}}{bc-abc^{2}}=\dfrac{c(a-a^{2}c)}{c(b-ab)}$

               $=\dfrac{a-a^{2}c}{b-abc}$

9) b²+ 4b - 5 = b² + 5b - 1b - 5

                    = b(b + 5) - 1(b + 5)

                    = (b +5)(b - 1)

b² + 8b  + 15 = b² + 5b + 3b + 15

                    = b(b + 5) + 3(b + 5)

                    = (b + 5 )(b + 3)

$\dfrac{b^{2}+4b-5}{b^{2}+8b+15}=\dfrac{(b+5)(b-1)}{(b+5)(b+3)}$

                     $=\dfrac{b-1}{b+3}$