Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!
I need help with 5 equations... this is one of them.
8/x+2 + 8/2 = 5
Please help me!!!
Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!
I need help with 5 equations... this is one of them.
8/x+2 + 8/2 = 5
Please help me!!!
You haven't used parentheses, so your post is rather ambiguous.
Is this it:
$\displaystyle \frac{8}{x+2}+\frac{8}{2}=5$
If so, why not write 8/2 as 4?.
Anyway, multiply by $\displaystyle x+2$:
$\displaystyle (x+2)\frac{8}{x+2}+4(x+2)=5(x+2)$
$\displaystyle 8+4(x+2)=5(x+2)$
Now, can you finish?.
No, check yourself. If you put your answer back into the equation, do you get 5?. I don't think so.
You got a little discombobulated on your distribution.
Let's try your answer first:
$\displaystyle \underbrace{\frac{8}{\underbrace{-2+2}_{\text{oops, division by 0, a big no-no}}}+4=5}_{\text{bad, the universe will\\collapse in on itself}}$
Now, let's do it this way:
$\displaystyle 8+4(x+2)=5(x+2)$
$\displaystyle 8+4x+8=5x+10$
$\displaystyle 16+4x=5x+10$
Subtract 4x from both sides:
$\displaystyle 16=x+10$
Subtract 10 from both sides:
$\displaystyle x=6$
See?.
Actually, you're not seeing the forest for the trees. Don't let the algebra and x's psyche you out.
Look at it for a second. Follow:
$\displaystyle \frac{8}{x+2}+4=5$
What plus 4 equals 5. Well, you don't have to be a brilliant mathematician to see that. 4+1=5
Therefore, $\displaystyle \frac{8}{x+2}$ must be 1. What makes that 1?. When x=6. See?.
Relax. It'll be OK.
You're partially correct. You use the GCD, which is $\displaystyle 3x(x+6)$
See?. You have more in the denominators than just x+6. There's a 3x there, also.
$\displaystyle (x+6)(3x)\frac{2}{3x}+(x+6)(3x)\frac{2}{3}=(x+6)(3 x)\frac{8}{x+6}$
Cancel what needs cancelled and you get:
$\displaystyle 2(x+6)+2(x+6)(x)=8(3x)$
Distribute:
$\displaystyle 2x+12+2x^{2}+12x=24x$
$\displaystyle 14x+12+2x^{2}=24x$
$\displaystyle 2x^{2}-10x+12=0$
Divide by 2:
$\displaystyle x^{2}-5x+6=0$
Now, can you solve the quadratic?.
I divided by 2 because I could and wanted to. All the terms were divisible by 2
The quadratic $\displaystyle x^{2}-5x+6=0$ can be factored.
Can you factor?. You can use the quadratic formula, too, but this one is easily factorable.
$\displaystyle (x-3)(x-2)=0$
$\displaystyle \frac{x-3}{x-4}+4=\frac{3x}{x}$
$\displaystyle (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}$
I multipied them
$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)$ but I had a feeling that I was suppose to cancel out the 4's but I chose not to..
Next, $\displaystyle x^{2}-3x+4x^{2}-16x=3x^{2}-12x$
And I moved the $\displaystyle 3x^{2}-12x$
to the left side of the equation...
$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x$
$\displaystyle 5x^{2}-3x{2}-19x-12x$
I'm stuck at $\displaystyle 2x^{2}-21x$
I did something wrong, didn't I? =( I am horrible at this!
$\displaystyle (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}$
I multipied them
$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)$ but I had a feeling that I was suppose to cancel out the 4's but I chose not to..
Next, $\displaystyle x^{2}-3x+4x^{2}-16x=3x^{2}-12x$
And I moved the $\displaystyle 3x^{2}-12x$
to the left side of the equation...
$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x$$\displaystyle \leftarrow\text{good to here}$
$\displaystyle 5x^{2}-19x=3x^{2}-12x$
$\displaystyle 2x^{2}-7x=0$
Factor:
$\displaystyle x(2x-7)=0$
See the solution now?.
$\displaystyle 5x^{2}-3x{2}-19x12x$
I'm stuck at $\displaystyle 2x^{2}-21x$
I did something wrong, didn't I? =( I am horrible at this!
$\displaystyle x(2x-7) = 0$
Since the RHS is 0, at least one of x or 2x - 7 must also be 0. Thus:
$\displaystyle x = 0$ or $\displaystyle 2x - 7 = 0$
Thus $\displaystyle x = 0$ or $\displaystyle x = \frac{7}{2}$
Now be careful. Your original equation canNOT work with x = 0. (The RHS of your original equation has us dividing 0/0, which is not allowed.) So your only solution is $\displaystyle x = \frac{7}{2}$.
-Dan