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Solve for x:
(x-1)/2+(x+5)/2 = (x-1)/3

(x-1)/2+(x+5)/2 = ((x+5)+(x-1))/2:
(x+x-1+5)/2 = (x-1)/3

Grouping like terms, x+x-1+5 = (x+x)+(-1+5):
((x+x)+(-1+5))/2 = (x-1)/3

x+x = 2 x:
(2 x+(-1+5))/2 = (x-1)/3

5-1 = 4:
(2 x+4)/2 = (x-1)/3

Multiply both sides by 6:
(6 (2 x+4))/2 = 2 (x-1)

6/2 = (2×3)/2 = 3:
3 (2 x+4) = (6 (x-1))/3

6/3 = (3×2)/3 = 2:
3 (2 x+4) = 2 (x-1)

Expand out terms of the left hand side:
6 x+12 = 2 (x-1)

Expand out terms of the right hand side:
6 x+12 = 2 x-2

Subtract 2 x from both sides:
(6 x-2 x)+12 = (2 x-2 x)-2

6 x-2 x = 4 x:
4 x+12 = (2 x-2 x)-2

2 x-2 x = 0:
4 x+12 = -2

Subtract 12 from both sides:
4 x+(12-12) = -2-12

12-12 = 0:
4 x = -2-12

-2-12 = -14:
4 x = -14

Divide both sides of 4 x = -14 by 4:
(4 x)/4 = (-14)/4

4/4 = 1:
x = (-14)/4

The gcd of -14 and 4 is 2, so (-14)/4 = (2 (-7))/(2×2) = 2/2×(-7)/2 = (-7)/2:
Answer: |x = (-7)/2

X-(x-y-z) 

\(x-(x-y-z)\\ =x+\quad -1*(x-y-z)\\ =x+\quad -1*x\quad-1*-y\quad-1*-z\\ =x+\quad -x\quad--y\quad--z\\ =x\quad -x\qquad+y\qquad+z\\ =y+z \)

To help you comprehend why, try replacing the letters with numbers

20-(20-5-7) = 20-(8) = 12

20-(20-5-7) = 20-20+5+7 = 12

When you have a minus sign outside a bracket like this the easiest way to deal with it is

to change all the - terms to + terms and all the + terms to terms signs INSIDE the bracket AND change the negative that is outside the bracket to a positive.

20-(20-5-7) =20+(-20+5+7) =20-20+5+7 =12

\(x-(x-y-z)\\ =x-(+x-y-z)\qquad \text{I've just added the invisable + here} \\ = x+(-x+y+z)\\ =  x-x+y+z \\= y+z\)

Please Note:

I have answered your question VERY fully.

I have done that because I liked the way you asked it "help me understand"

I rarely answer so fully for 'guests' if you want priority and good answers like this one it is a good idea to become a member and get yourself known.  

We knock ourselves out trying to explain things to people who have good manners and show a guenuine desire to learn.   

It's all good! I've asked so many different people and no one could help. You're not alone! I turned in what I had and asked the teacher to let me know what I can fix, which is probably a lot. I appreciate you effort though!!!

wow that ishard for me oh nnnoooo

thxsasss

Hi Brittany, itr is REALLY good to see you again :))

Chris, I am not so sure about your answer.....

\(If\;\;y=x^2\qquad then\qquad x=\pm\sqrt y \)

If the domain of the original function was [0,2] then   \(x=+\sqrt y\)

And that would certainly have an inversese function.

BUT

given the original domain restriction of [-2,2]  the inverse becomes...

\(y=\pm\sqrt x \qquad \mbox{and this is not a function at all}\)

(It does not pass the vertical line test)

So I think, 

If f(x) = x² on the domain [-2,2] then f^-1 does not exist.

I could be wrong - I will certainly consider further discussion on the topic.  

like you post guyssss

hi all you want point sbroooo

1.24999998861

:p

6 3/10

NP BRO

thnx

SORRY THAT IS HARD FOR ME

(3/8)(3/8)(3/8)(3/8)=[3 x 3 x 3 x 3] / [8 x 8 x 8 x 8] =81 / 4,096=3/8^4=(0.375^4)

-5*-2 = 10 as a negative times a negative is always a positive.
10 * 9 = 90
Hope this makes sense!

SORRY CAN BE ABLE TO HELPP I WICH I CAN SOORRRYYY

42 OK BRO 21      

                    21   

                     42

Step 1: 0.432 = 432⁄1000 
Step 2: Simplify 432⁄1000 = 54⁄125 

THIS  SHOUNLD HELP BRO

Solve for x:
6+5 sqrt(x-5) = 10

Subtract 6 from both sides:
5 sqrt(x-5) = 4

Divide both sides by 5:
sqrt(x-5) = 4/5

Raise both sides to the power of two:
x-5 = 16/25

Add 5 to both sides:
Answer: |x = 141/25

Sqrt(-1.3) =1.140175....i.

1738