Hello,

I am in eight grade Math Honors and I would like some help on the following questions:

1. 5x + 9 = 2
2. 7 - 6d = 3
3. 1/3y - 2 = 4
4. 7 (10-3w) = 5(15 -4w)

This may look easy for you guys, but I don't have a lot of time and I don't know which place to post this.

Thanks, Help is appreciated.

1. Subtract 9 from both sides of the equation:

$\displaystyle 5x = -7$

Divide both sides by 5 to get $\displaystyle x = -\frac{7}{5}$.

2. is done the same way.

3. is a little unclear -- do you mean $\displaystyle \frac{1}{3y} - 2 = 4$ or $\displaystyle \frac{1}{3y-2} = 4$?

4. Expand both sides to get $\displaystyle 70 - 21w = 75 - 60w$. Bring all the $\displaystyle w$ terms to one side, solve for $\displaystyle w$.

Thanks. For the second one, if it is done the same way as the first one, would the answer be: D = -4/6? Also, the third one is the first one on your post. 1/3y - 2 = 4, and the two is not under the one.

Originally Posted by AbhiKap
Thanks. For the second one, if it is done the same way as the first one, would the answer be: D = -4/6? Also, the third one is the first one on your post. 1/3y - 2 = 4, and the two is not under the one.
no, that is incorrect on the second one. you should always check your work:

if d = -4/6, then:

7 - 6d = 7 - 6(-4/6) = 7 + 4 = 11, which is not 3.

i would proceed as follows (since you appear to have trouble with "signs"):

7 - 6d = 3
7 - 6d + 6d = 3 + 6d
7 + 0 = 3 + 6d
7 = 3 + 6d
7 = 6d + 3
7 - 3 = 6d + 3 - 3

...can you continue?

by the way, this *is* the wrong section, this is "pre-college algebra" not "college algebra" (as you might have guessed from being in the 8th grade, which is not college, hmm?)