simple gr 9 algebra question?

Let's call this unknown number x, for convenience. Then we can write the statement above in the form of an algebraic expression:

(1/3)x + 8 = (1/2)x - 5

The key in solving this type of equation (where we're solving for the variable "x") is to get the x-terms on one side, and everything else on the other side, therefore making it much easier to solve for x, as you will see.
So let's subtract (1/3)x from both sides and add 5 to both sides, as shown:

(1/3)x - (1/3)x + 8 + 5 = (1/2)x - (1/3)x - 5 + 5

Obviously the (1/3)x's on the left will cancel out and the 5's on the right will cancel out. This leaves:

13 = (1/2)x - (1/3)x

Now we need to combine like terms (the x-terms). What needs to be done in order to do that is add the coefficients of the x-terms together, thus obtaining the coefficient of the resulting x-term. Simply put, (1/2)x - (1/3)x is the same as [(1/2) - (1/3)]x.
Now what needs to be done is the evaluation of (1/2) - (1/3).

(1/2) - (1/3) = (3/6) - (2/6) = 1/6

First I found the LCD (Least Common Denominator), and then gave each fraction the same denominator. Then I was able to subtract the numerator's over the denominator.

So now, we can simplify the earlier expression from:

13 = (1/2)x - (1/3)x

to:

13 = (1/6)x

Now 1/6 must be divided from both sides. This will give:

13/(1/6) = x

When dividing by a fraction, you must multiply the numerator by the reciprocal of the fraction (the denominator) in order to obtain the quotient (logically). Thus:

x = 13*6 = 78

Thus x=78.

Just to make sure, let me substitute 78 in for x in the original equation: (1/3)x + 8 = (1/2)x - 5

(1/3)78 + 8 = (1/2)78 - 5

(78/3) + 8 = (78/2) - 5

26 + 8 = 39 - 5

34 = 34

The equation works, therefore x=78.

There you go, I'm glad I could help. :)

let the no be x and proceed as follows:
x/3 + 8=x/2 - 5
(x+24)/3 = (x- 10)/2
by cross multiplication, you get:
2x+48=3x-30

x=78 =>this gives you the answer

Let x be the unknown number

In equation form you question is

x/3 + 8 = x/2 - 5

Isolate x to one side by adding 5 to both sides and subtracting x/3 from both sides

13 = x/6

multiply both sides by 6

x = 78

Check:

78/3 + 8 = 34
78/2 - 5 = 34
multiply both sids

Let x be the number.

x/3 + 8 = x/2 - 5 (multiply both sides by 6)
2x + 48 = 3x - 30 (add 30 to both sides)
2x + 78 = 3x (subtract 2x from both sides)
78 = x

ANSWER: 78

x/3 + 8 = x/2 - 5
Multiply both sides by 6,
2x + 48 = 3x -30
Add 30-2x to both sides
78 = x
Answer: x = 78

a million: (2m^3n^2)^4 / (-4mn)^2 = 16m^12n^8 / 16m^2n^2 = m^10n^6 (except m = 0 or n = 0) 2: (3-D^4m^3 * 8d^2m^5) / (2nd^2m^2 * 6d^3m^2) = 24d^6m^8 / 12d^5m^4 = 2dm^4 (except d = 0 or m = 0) 3: (2g^2h^3 * (-3g^2h^2)^2) / (3gh * 6g^2h^2) = (2g^2h^3 * 9g^4h^4) / 18g^3h^3 = 18g^6h^7 / 18g^3h^3 = g^3h^4 (except g = 0 or h = 0)

x/3 + 8 = x/2 - 5
2x + 48 = 3x - 30
78 = x

x = the number

1/3(x) + 8 = 1/2(x) - 5
x/3 + 8 = x/2 - 5
x/3 - x/2 = -8 - 5
6(x/3 - x/2) = 6(-13)
2x - 3x = -78
-x = -78
x = -78/-1
x = 78

∴ the number is 78.

The guy above me is correct. Please make sure you know how to do this as it rears its ugly head many times on the SAT.