Answer:
One number, let's call it x, is 6 more than twice the other number, let's say this number is y.
x= 2y+6
x+y=21
We're left with a system of equations.
If we substitute the first equation in the second we find the value of y.
(2y+6)+y=21
3y+6=21
3y=15
<u><em>y=5</em></u>
Now that we have found the value of y, we can take that value and substitute it in the second equation (since it's easier) to find the value of x.
x+ (5)=21
<u><em>x=16</em></u>
<u><em>Now to check if our answers are correct, we plug in both values into any equation and see if they equate.</em></u>
x+y=21
(16)+(5)=21
21=21
Our solution is correct!
my answer is in the image above
Do 77 times 14
770+308=1,078
Divide 1,078 by 100
=10.78
(x-2) cancels from the numerator and denominator, which is called a "hole" at x = 2.
There is a vertical asymptote at x = 4.
There is a horizontal asymptote at y = -3
