How do you write #2-(8a+5x)/(3a)+(2a+7x)/(9a)# as a single fraction?

1 Answer
Dec 27, 2016

#(-4a - 8x)/(9a)#

or

#(-4(a + 2x))/(9a)#

Explanation:

To add two fractions we need to have the fractions over a common denominator, in this case #color(blue)(9a)#.

Therefore we need to multiple the constant and first fraction by the appropriate form of #1#:

#(color(red)(9a)/color(blue)(9a) * 2) - (color(red)(3)/color(blue)(3) * (8a + 5x)/(3a)) + (2a + 7x)/(9a)#

#color(red)(18a)/color(blue)(9a) - (color(red)(3) (8a + 5x))/color(blue)(9a) + (2a + 7x)/color(blue)(9a)#

We can now expand the terms within parenthesis for the second fraction:

#(18a)/color(blue)(9a) - (24a + 15x)/color(blue)(9a) + (2a + 7x)/color(blue)(9a)#

We then can combine the numerators over a single common denominator:

#(18a - 24a - 15x + 2a + 7x)/color(blue)(9a)#

Next we can group like terms:

#(18a - 24a + 2a - 15x + 7x)/(9a)#

Now we can combine like terms:

#((18 - 24 + 2)a + (-15 + 7)x)/(9a)#

#(-4a - 8x)/(9a)#

or

#(-4(a + 2x))/(9a)#