Basic inequality problem

1. Aug 6, 2013

MathJakob

1. The problem statement, all variables and given/known data

$\frac{a}{4}>\frac{a}{2}+6$

3. The attempt at a solution

$\frac{2a}{2}>\frac{4a+48}{2}$

$a>2a+24$

So do I just plug random numbers in and see what I get? I realised right away that it has to be a negative number so I stuck in -30 and got

$-30>-60+24$ Well that's true.

$-24>-48+24$ This is false.

So $a<-24$ is this correct? Am I going about these the right way?

Last edited: Aug 6, 2013
2. Aug 6, 2013

F1MH

Yes you are correct, but here is a different approach to the problem:

a>2a+24

Your could simply treat the > sign as an = sign and thus, algebra plays its role;

a-2a>24
-a>24

Then divide both sides by -1 but when you multiply or divide an inequality by a negative number the sign switches from > to < or vice versa...

$\frac{-a}{-1}$ < $\frac{24}{-1}$

$\Rightarrow$ a<-24

Its the same answer but its avoids the random number input.

Hope this helped.

3. Aug 6, 2013

Staff: Mentor

No, you can continue with the simplification. For example, subtract a on both sides, and think about the 24 afterwards.

Your intermediate step looks more complicated than necessary.

4. Aug 6, 2013

MathJakob

Here is another:

$\frac{y-3}{5}<\frac{y+2}{10}$

$\frac{10y-30<5y+10}{5y+10}$

$2y-3<1$

$y<2$

is this correct?

5. Aug 6, 2013

Staff: Mentor

That does not make sense.

The result is wrong. Just try y=3, for example. Is the original inequality satisfied? Is your result satisfied?

6. Aug 6, 2013

MathJakob

$\frac{y-3}{5}-\frac{y+2}{10}$

$4y-2y-12+8$

$2y<8$

$y<4$ ?

I just checked and now it doesn't work... how can I solve these without having to try lots of numbers?

Last edited: Aug 6, 2013
7. Aug 6, 2013

F1MH

No, because its erroneous $\frac{10y-30}{5y+10}$ ≠ 2y-3

10y-5y<10+30 is what you should be doing. (Remember treat the inequality like an = sign, with the exception of dividing/multiplying with a negative number during simplification)

8. Aug 6, 2013

MathJakob

oh right I see so I can just treat them like normal fractions.

$10y-30=5y+10$

$5y=40$

$y=8$ so the correct solution is $y<8$ ? On wolfram is specifically says $y<8$ and not $y=8$ does this matter?

9. Aug 6, 2013

Staff: Mentor

y<8 is the correct solution.
There is no = in your inequality.

10. Aug 6, 2013

F1MH

Yes, it does. Remember it's an inequality not a linear equation. I said treat it LIKE an = sign in the sense that algebraic rules are still valid for inequalities. The < remains because its saying $\frac{y-3}{5}$<$\frac{y+2}{10}$ stays true for all values y<8 and it does.

Once you plug in y=8 you get 1<1 Which is false because 1 is not less than 1 its 1=1. And when you plug another number out of the inequality like say 9 you get 1.2<1.1 which again is false.

The logic is really shown in the term inequality. Your trying to find what values of y will this relationship stay true and those that produce an inequality.

Again, its not a linear equation you can't just replace the < with = its a relationship between those two sides.

11. Aug 6, 2013

haruspex

A word of warning with this approach of treating < and > as like =. Suppose you had something more complicated, like $\frac{x-3}{x+5} < 2x$. If you multiply through by x+5, as you would happily do with an equality, you do not know whether x+5 is positive or negative. So you do not know whether you should be reversing the inequality. Generally, all you can do here is treat the two possibilities separately.