# Help understanding to solve these problems

• Nov 13th 2007, 01:09 PM
slykksta
Help understanding to solve these problems
Problem 1:

1 x + 1 < 1 x + 2
6......2...12......3

Problem 2: The colored, bold numbers are exponents.

2 x2 (3 xy) (-3x2y2)

Problem 3: The colored, bold numbers are exponents. Double-parentheses denotes a much larger parentheses.

(-16 s2t3) ((3 t5))
..................8

Problem 4: The colored, bold numbers are exponents.

3 c5 d3
6 c3 d5

Problem 5: The colored, bold numbers are exponents. Double-parentheses denotes a much larger parenthesis.

(( 3 a2 b c-1 ))-2
...9 a-1 b-1 c

I appreciate any guidance in advance on how to work these out. :confused:
• Nov 13th 2007, 03:44 PM
topsquark
Quote:

Originally Posted by slykksta
1 x + 1 < 1 x + 2
6......2...12......3

$\frac{1}{6}x + \frac{1}{2} < \frac{1}{12}x + \frac{2}{3}$

First, let's get rid of those pesky fractions. The least common multiple of 6, 2, 12 and 3 is 12. So multiply both sides of the inequality by 12:
$12 \left ( \frac{1}{6}x + \frac{1}{2} \right ) < 12 \left ( \frac{1}{12}x + \frac{2}{3} \right )$

$2x + 6 > x + 8$

$2x + 6 - 6 > x + 8 - 6$

$2x > x + 2$

$2x - x > -x + x + 2$

$x > 2$

-Dan
• Nov 13th 2007, 03:56 PM
topsquark
Quote:

Originally Posted by slykksta
Problem 2: The colored, bold numbers are exponents.

2 x2 (3 xy) (-3x2y2)

Problem 3: The colored, bold numbers are exponents. Double-parentheses denotes a much larger parentheses.

(-16 s2t3) ((3 t5))
..................8

Problem 4: The colored, bold numbers are exponents.

3 c5 d3
6 c3 d5

Problem 5: The colored, bold numbers are exponents. Double-parentheses denotes a much larger parenthesis.

(( 3 a2 b c-1 ))-2
...9 a-1 b-1 c

I appreciate any guidance in advance on how to work these out. :confused:

The rest of these involve applications of the following rules: (with suitable conditions on a and b)
$a^nb^n = (ab)^n$

$a^na^m = a^{n + m}$

$(a^n)^m = a^{nm}$

$a^{-n} = \frac{1}{a^n}$

$\frac{1}{a^{-n}} = a^n$

Typically you want to get rid of the parenthesis first, then the negative exponents.

So let's look at the last problem. (If you can do that one you can do them all.)

$\frac{(3a^2bc^{-1})^{-2}}{9a^{-1}b^{-1}c}$

Parenthesis first:
$= \frac{(3)^{-2}(a^2)^{-2}(b)^{-2}(c^{-1})^{-2}}{9a^{-1}b^{-1}c}$

More parenthesis:
$= \frac{3^{-2}a^{2 \cdot -2}b^{-2}c^{-1 \cdot -2}}{9a^{-1}b^{-1}c}$

$= \frac{3^{-2}a^{-4}b^{-2}c^2}{9a^{-1}b^{-1}c}$

Now get rid of the negative powers:
$= \frac{abc^2}{3^2a^4b^2 \cdot 9c}$

$= \frac{abc^2}{81a^4b^2c}$

Now simplify:
$= \frac{c}{81a^3b}$

Each one of these lines contains simplifications given by the rules I listed above.

-Dan
• Nov 14th 2007, 11:54 AM
slykksta
Thanks!!
Thanks, that really cleared it up for me. :)