How do you do these problems ?

Question

Solve the equation.
(41)/(x) = 6 - (1)/(x)


Problem 2
Solve the equation.
(11)/(x) = 4 - (1)/(x)

Answer 1

41/x = 6 - 1/x
add 1/x on both sides

41/x + 1/x = 6 - 1/x + 1/x

-1/x + 1/x = 0
41/x + 1/x = 42/x
so we now have:
42/x = 6

multiply both sides by x
x*42/x = 6x
On the left, x/x cancels out (because x/x = 1; 1*42 = 42)

42 = 6x
divide both sides by 6
42/6 = 6x/6
7 = x

Answer 2

In both cases you would multiply the entire equation by x.

After multiplying EQN #1 by x, you get:

41 = 6x -1
42 = 6x
x =7


After multiplying EQN #2 by x, you get:

11 = 4x -1
12 = 4x
x = 3

That's it! Notice that in both equations, x cannot be equal to 0.

Answer 3

First you can get all the xs on one side so rewrite as
(41/x)+(1/x) = 6
add them
(42/x)=6
You want to get rid of the denominators. So multiply each side by x.
(42/x)* x = (6)*x
That gives you
42 = 6x
Divide by 6.
7=x

Do the same thing for problem 2.

Answer 4

The first one is really forty one over x equals 6 negative 1 over x. So the answer is 7
The second one is really 11 over x equals 4 negative one over x so then the answer is 3

Answer 5

For the first one, the answer is x=7. The second answer is x=3.

Answer 6

u shuld go 2 webmath.com they have basically everything... i think... they solve the problems 4 u and take u through the steps 2 find da answer

Answer 7

ok....so you first do is multiply the whole equation by x...like this:

(41)/(x) = 6 - (1)/(x) ..... (41x)/(x) = 6x - (1x)/(x)

then, you cancel/ simplify....like this:

41=6x-1.....which you do like a normal question:
42=6x.........x=7

the second one is the same:
(11)/(x) =4x -1 ..........(11x)/(x) =4x - (1x)/(x)

11=4x-1........12=4x...3=x

hope that helps....and it made sense ^^