Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need help with the break down of this problem. I need to know exactly what I need to do to solve this problem. Please help!

share|cite|improve this question
Can you solve for $y$ in $-6+y=-5$? Then multiply what you got by $4$, and you're done. – Pedro Tamaroff Apr 18 '13 at 20:32

Hint: What do you have to add to $-6$ to get $-5$? What do you have to divide by $4$ to get the answer to my previous question?

share|cite|improve this answer
thank you!!!!!! – Deriah Apr 18 '13 at 21:09
Put another way: $$-6+\frac x4=-5\\-6+\frac x4=-6+1\\\frac x4=1\\\frac x4=\frac44\\x=4$$ – Cameron Buie Apr 18 '13 at 21:13
and the solution of $$\sin(x)=\sin(\varphi)$$ is $x=\varphi$ ? – miracle173 Sep 26 '13 at 21:20
@miracle173: That is a solution, certainly. The difference there is that we're no longer dealing with only one-to-one functions. – Cameron Buie Sep 27 '13 at 6:00
That is what I wanted to point to.When I read your description and your example it looks like pattern matching. But pattern matching is not sufficient to find all solution as the example $\sin(x)=\sin(\varphi)$ shows. The essential steps (subtracting $-6$ from and mutliplying $4$ to both side of the equation) are not mentioned. I think representing $-5$ as $-6 +1$ and $1$ as $\frac{4}{4}$ is absolutely dispensable and guides in the wrong direction. – miracle173 Sep 27 '13 at 9:20

I would think it through this way:

The first task is to get the x term (x times something) by itself on one side, and everything else on the other side of the equation. You can do that by adding 6 to both sides (to cancel the -6). Then you have:

$$\frac{x}{4} = 1$$

Now you want to get x all by itself, so you can multiply both sides by 4:

$$x = 4$$


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.