(-10/7) + 1/6 , once you find the lcd, do you add or subtract?

Since both fractions have the same denominator, you can simply add up (or substract) the numerators and keep the denominator unchanged :

x2+3x+2x2−16+3x+6x2−16=(x2+3x+2)+(3x+6)x2−16

=x2+6x+8x2−16.

Then, you can factorize it :

x2+6x+8x2−16=(x+2)(x+4)(x+4)(x−4)=x+2x−4.

In the case of a substraction :

x2+3x+2x2−16−3x+6x2−16=(x2+3x+2)−(3x+6)x2−16

=x2−4x2−16=(x+2)(x−2)(x+4)(x−4).

add an image

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!