Question

What is the solution set of the equation?
Missing Metadata



{4, 6}


{–4, 12}

{12}


{–12, 4}
https://static.k12.com/eli/bb/1627/6_92780/2_38049_6_48675/cd28c4ba4844f46ecc522c46f9a4df4d7c12cf1d/media/6ce4a2d840c69def4f524dbb4e388c847d53b7f9/mediaasset_1070997_1.gif

Answer 1

@FROST_ZERO

Answer 2

Think about what you must do to any two fractions with different denominators before you can add or subtract those fractions.

Do that to the first 2 terms. In other words, combine those two terms into one. (Note: there are other ways in which to approach this problem, but please begin with my suggestion.)

Answer 3

you make them have common denominators, but im not sure how to do that with a variable n the denominator

Answer 4

Please combine (8/x) and 2: Simplify (8/x)+2. You could think of this as (8/x)+(2/1). What is the lcd?

Answer 5

the lcd should be x, right?

Answer 6

Right.

You already have that denominator in the fraction. Multiply the 2 by x and then divide the result by x (but do not cancel / simplify).

What do you get?

Answer 7

8/x+2x/x

Answer 8

that's great. Now, you have 2 fractions with the same denominators and can therefore combine them. Plese do that now.

Answer 9

8+2x/x?

Answer 10

It's important that you use parentheses here: (8+2x)/x. But yes, that's right!

Equate this (8+2x)/x to (x+4)/(x-6).

Answer 11

\[\frac{ 8+2x }{ x }=\frac{ x+4 }{ x-6 }\]

Answer 12

You had 3 fractions before, but now you have only 2.

Do they have the same denominator?

Answer 13

no, there is a six there that is keeping it from being common

Answer 14

my bad, a (-6)

Answer 15

If we were to add 1/3 and 1/4, what would we do to obtain the LCD?

Answer 16

multiply both denominators by 2

Answer 17

and the numerators

Answer 18

would i multiply the first fraction by (-6)?

Answer 19

but that would leave you with 2 different denominators. not good.

Here's what I would do to add 1/3 and 1/4:\[\frac{ 1 }{ 3 }\frac{ 4 }{ 4 }+\frac{ 1 }{ 4 }\frac{ 3 }{ 3 }\]

Answer 20

Can you explain in words what I did and also explain the reason behind it?

Answer 21

you multiplied so that the denominator would be 12

Answer 22

that makes them both have a common denominator

Answer 23

Yes, and 12 is the LCD we were looking for.

Answer 24

\[\frac{ 8+2x }{ x }=\frac{ x+4 }{ x-6 }\]

Answer 25

Drawing on this example, consider how you would find the LCD of this equation.

Answer 26

could you multiply (-6)?

Answer 27

No. Remember, in our example, we multiplied together the 2 diff. denoms. to obtain the LCD, which turned out to be 12.

Multiply together the 2 denoms of \[\frac{ 8+2x }{ x }=\frac{ x+4 }{ x-6 }\]

Answer 28

to obtain the LCD. Method is the same as before!

The LCD is ... ?

Answer 29

so multiply (-6) to the first one and 1 to the second?

Answer 30

Multiply the current denominators as they are: x(x-6).\[\frac{ 8+2x }{ x }=\frac{ x+4 }{ x-6 }\]

Answer 31

x^2-6x

Answer 32

One of the easier ways to finish this problem would be to cross-multiply. yOU familiar with that? (No need to multiply x*(x-6). Leave it as x(x-6). )

Answer 33

yes, i have cross multiplied

Answer 34

Cross multiplication would involve multiplying (8+2x)(x-6). Do that, please.

Answer 35

2x^2-4x-48?

Answer 36

I'll take your word for it. Next, multiply the left denom. (x) by the numerator on the right: (x+4). You get ... what ?

Answer 37

x^2+4x

Answer 38

Then equate your previous product, (2x^2-4x-48) to x^2+4x.

Gather all the x-terms on the left and all the constant terms on the right.

Answer 39

3x^2-48?

Answer 40

Check your subtraction. I get x^2-8x-48.

Answer 41

And that should be an equation: x^2 - 8x - 48 = 0.

Answer 42

-4x+4x is zero though isint it?

Answer 43

You have -4x on the left already, and then must subtract 4 from both sides of the equation to move the +4x on the right to the left side of your equation. Result: -8x.

Answer 44

OH! i see, i didnt set them equal each other, i just added them

Answer 45

Assuming that x^2-8x-48 = 0 is correct, what would our next step be, towards finding x?

Answer 46

yes, i got x^2-8x-48

Answer 47

isolating x?

Answer 48

wait, reorder the equation

Answer 49

this is a quadratic equation, sean, so you'd need to apply one of the several methods for solving quadratics to obtain (isolate) x.

Answer 50

find a,b,and c. the quadratic function -b+/- the square root of b^2-4ac all over 2a, right?

Answer 51

a=1, b=-8, c=-48

Answer 52

yes. Quadratic formula.

Answer 53

OH! ok, im stuck at the square root

Answer 54

Sean, one of the most important aspects of this problem was identifying and using the LCD of the given equation. That LCD is x(x-6).
I've tried unsuccessfully (so far) to obtain the roots / solutions of x^2 + 16x -48 = 0.
I'd suggest you not spend much (if any) more time with this particular problem, but rather go back and review what we did to find the LCD and how we used it.

Answer 55

ive got to go, my father is home from work. Thank you for your help and time. I really appreciate it. i have all of the steps written down , and i think i have it now. Thank you again!

Answer 56

We have an algebraic error somewhere. I've found the equation to be x^2-8x-48, and this has the solutions 12 and -4. You could check that by substitution.

Answer 57

Tag me later on if you have further questions. Very happy to work with you. Bye for now.

Answer 58

Thank you!