Question

Can someone explain to me pls how you do these types of questions maybe at least a and b?? Thank you

Answer 1

a is 1/12 and b is x+3/x+1

Explanation:

this is the explanation

Answer 2

`\textsf{(a)}\quad (1)/(12)`

`\textsf{(b)}\quad (x+3)/(x+1)`

Explanation:

To divide fractions, we can multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of a fraction is obtained by swapping its numerator and denominator, resulting in a new fraction whose product with the original fraction equals one.

`\hrulefill`

Question (a)

Given expression:

`(x-4)/(8)/ (3x-12)/(2)`

Swap the numerator and denominator of the second fraction, and change the division sign for a multiplication sign:

`(x-4)/(8)* (2)/(3x-12)`

When multiplying fractions, multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator:

`((x-4)* 2)/(8* (3x-12))`

Simplify:

`(2x-8)/(24x-96)`

Factor out 2 from the numerator and 24 from the denominator:

`(2(x-4))/(24(x-4))`

Cancel the common factor (x - 4);

`(2)/(24)`

Simplify by dividing the numerator and denominator by the common factor 2:

`(2/ 2)/(24/ 2)=(1)/(12)`

Therefore:

`(x-4)/(8)/ (3x-12)/(2)=\boxed{(1)/(12)}`

`\hrulefill`

Question (b)

Given expression:

`(x+3)/(x+2)/ (x+1)/(x+2)`

Swap the numerator and denominator of the second fraction, and change the division sign for a multiplication sign:

`(x+3)/(x+2)*(x+2)/(x+1)`

When multiplying fractions, multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator:

`((x+3)(x+2))/((x+2)(x+1))`

Cancel the common factor (x + 2):

`(x+3)/(x+1)`

Therefore:

`(x+3)/(x+2)/ (x+1)/(x+2)=\boxed{(x+3)/(x+1)}`