1. ## Basic algebra

1) 1=12+n

2) 3=-5-10x

3) 4=5/3n

4) 8^-1/y +3^-1=-3

5) 10c=-3-3c

6) 6=-10-5m

7) 11=y/2^-1 -7

8) 10/11=3-8y+8

9) 5=60n^-1 -1

10) 4/8=10n/6

2. Originally Posted by jerryramos
1) 1=12+n
Subtract 12 from both sides to get: $\displaystyle n=1-12=11$

Try the other ones, and then tell me what ones you need help on...

3. Originally Posted by jerryramos
1) 1=12+n

2) 3=-5-10x

3) 4=5/3n

4) 8^-1/y +3^-1=-3

5) 10c=-3-3c

6) 6=-10-5m

7) 11=y/2^-1 -7

8) 10/11=3-8y+8

9) 5=60n^-1 -1

10) 4/8=10n/6
2) 3=-5-10x

5 + 3 = 5 + -5-10x

8 = -10x

$\displaystyle \frac{8}{-10} = \frac{-10x}{-10}$

$\displaystyle x = -\frac{8}{10} = -\frac{4}{5}$

-Dan

4. Originally Posted by jerryramos
3) 4=5/3n
$\displaystyle 4 = \frac{5}{3n}$

$\displaystyle 3n \cdot 4 = 3n \cdot \frac{5}{3n}$

$\displaystyle 12n = 5$

$\displaystyle \frac{12n}{12} = \frac{5}{12}$

$\displaystyle n = \frac{5}{12}$

-Dan

5. Originally Posted by jerryramos
4) 8^-1/y +3^-1=-3
Note that [tex]a^{-1} = \frac{1}{a}[tex] ($\displaystyle a \neq 0$.)

So
$\displaystyle \frac{8^{-1}}{y} +3^{-1}=-3$

$\displaystyle 8^{-1} \cdot \frac{1}{y} +3^{-1}=-3$

$\displaystyle \frac{1}{8} \cdot \frac{1}{y} + \frac{1}{3} = -3$

Let's get rid of these pesky fractions, shall we? The least common multiple of 3 and 8 is 24, so multiply both sides of the equation by 24:

$\displaystyle 24 \left ( \frac{1}{8} \cdot \frac{1}{y} + \frac{1}{3} \right ) = 24 \cdot (-3)$

$\displaystyle 3 \cdot \frac{1}{y} + 8 = -72$

$\displaystyle 3 \cdot \frac{1}{y} + 8 - 8= -72 - 8$

$\displaystyle 3 \cdot \frac{1}{y} = -80$

$\displaystyle 3 \cdot \frac{1}{y} \cdot y = -80 \cdot y$

$\displaystyle 3 = -80y$

$\displaystyle \frac{3}{-80} = \frac{-80y}{80}$

$\displaystyle y = -\frac{3}{80}$

-Dan