One positive number is 14 less than another positive number. if the reciprocal of the smaller number is added to five times the reciprocal of the larger number, the sum is one fourth. find the two numbers.

Let's call the two numbers and .

Given these variables, we can say: , based on the first sentence in the problem.

Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

To solve, we can simply substitute in for in the second equation and solve.

Now, notice that we have found two solutions, but the problem is only asking for one. This likely means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,

,

Since is not positive in this case, is not a solution.

Thus, is our only solution. In this case,

,

which means that the smaller number is 14 and the larger number is 28.

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes

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!