How do you simplify the expression #(y-12/(y-4))/(y-18/(y-3))#?

Question

Abigail Allen · Accepted Answer

#color(magenta)(((y+2)(y-3))/((y+3)(y-4))# #(y-12/(y-4))/(y-18/(y-3))# #:.=((y(y-4)-12)/(y-4))/((y(y-3)-18)/(y-3))# #:.=(y^2-4y-12)/(y-4) xx (y-3)/(y^2-3y-18)# #:.=((y+2)(cancel(y-6)))/((y-4)) xx ((y-3))/((y+3)(cancel(y-6))# #:.=color(magenta)(((y+2)(y-3))/((y+3)(y-4))#

Avery Alexander · Answer

I chose to multiply by a form of 1 that eliminates the complex fractions but there may be a better way, because I, later, discover a common factor that becomes 1.

Simplify: #(y-12/(y-4))/(y-18/(y-3))# Multiply the expression by 1 in the form of #((y-4)(y-3))/((y-4)(y-3))#: #(y-12/(y-4))/(y-18/(y-3))((y-4)(y-3))/((y-4)(y-3))# Multiply the two fractions: #((y(y-4)(y-3))- 12(y-3))/((y(y-4)(y-3))- 18(y-4))# Substitute #(y^2-7y+12)# for #(y-4)(y-3))#: #(y(y^2-7y+12)- 12(y-3))/(y(y^2-7y+12)- 18(y-4))# Distribute y: #(y^3-7y^2+12y- 12(y-3))/(y^3-7y^2+12y- 18(y-4))# Distribute 12 and 18: #(y^3-7y^2+12y- 12y+36)/(y^3-7y^2+12y- 18y+72)# Combine like terms: #(y^3-7y^2+36)/(y^3-7y^2- 6y+72)" [1]"# I asked WolframAlpha to factor the numerator and obtained the following answer: #(y + 2) (y - 3) (y - 6)# Then I asked WolframAlpha to factor the denominator and obtained the following answer: #(y + 3) (y - 4) (y - 6)# Please observe that #(y-6)# is common to both numerator and denominator, therefore, expression [1] becomes: #((y + 2) (y - 3))/((y + 3) (y - 4))#

Avery Alexander · Answer

# (y-12/(y-4))/(y-18/(y-3)) =((y+2)(y-3))/((y+3)(y-4)) #

We want to simplify: # (y-12/(y-4))/(y-18/(y-3)) # Firstly consider the numerator, which we can simplify by putting over a common denominator, thus: # y-12/(y-4) = (y(y-4)-12)/(y-4) # # " " = (y^2-4y-12)/(y-4) # # " " = ((y+2)(y-6))/(y-4) # Secondly consider the denominator, which we can also simplify by putting over a common denominator, thus: # y-18/(y-3) = (y(y-3)-18)/(y-3) # # " " = (y^2-3y-18)/(y-3) # # " " = ((y+3)(y-6))/(y-3) # So now we can rewrite the expression as: # (y-12/(y-4))/(y-18/(y-3)) = {((y+2)(y-6))/(y-4)}/{((y+3)(y-6))/(y-3)} # # " " = {((y+2)(y-6))/(y-4)} * {(y-3)/((y+3)(y-6))} # Then cancelling the common factor of #(y-6)# we get: # (y-12/(y-4))/(y-18/(y-3)) = {((y+2))/(y-4)} * {(y-3)/((y+3))}# # " " = ((y+2)(y-3))/((y+3)(y-4)) #

Eva Abramson · Answer

undefined To simplify the expression (y-12/(y-4))/(y-18/(y-3)), you can start by simplifying each fraction separately.

For the first fraction, y-12/(y-4), you can simplify it by factoring out a common factor of (y-4) from the numerator and denominator. This gives you (y-12)/(y-4) = (y-4-8)/(y-4) = (y-4)/(y-4) - 8/(y-4) = 1 - 8/(y-4).

For the second fraction, y-18/(y-3), you can simplify it in a similar way. Factor out a common factor of (y-3) from the numerator and denominator to get (y-18)/(y-3) = (y-3-15)/(y-3) = (y-3)/(y-3) - 15/(y-3) = 1 - 15/(y-3).

Now, you can substitute these simplified fractions back into the original expression: (1 - 8/(y-4))/(1 - 15/(y-3)).

To simplify further, you can find a common denominator for the two fractions, which is (y-4)(y-3). Multiply the numerator and denominator of the first fraction by (y-3) and the numerator and denominator of the second fraction by (y-4). This gives you ((y-3) - 8)/(y-4) and ((y-4) - 15)/(y-3).

Simplifying these fractions further, you get (y-3-8)/(y-4) = (y-11)/(y-4) and (y-4-15)/(y-3) = (y-19)/(y-3).

Finally, substitute these simplified fractions back into the expression: (y-11)/(y-4) / (y-19)/(y-3).

To divide fractions, you can multiply the first fraction by the reciprocal of the second fraction. This gives you (y-11)/(y-4) * (y-3)/(y-19).

Multiply the numerators together and the denominators together: (y-11)(y-3)/(y-4)(y-19).

This is the simplified expression: (y-11)(y-3)/(y-4)(y-19).