1. ## Working with expressions.

Please, if anyone could solve and explain to me the process of solving these two problems I would be extremely grateful. (It's not about the result, I really need to know HOW to do this, so If I can see a process of solving these problems I hope I will be able to understand it).

1) IF (b-2a)/(4a+3b)=2 (a,b!=0, 4a+3b!=0), THEN (2a^2-3ab+5b^2)/(4a^2+3b^2) EQUALS?

2) Value of the expression (25^(0,3)*5^(0,4))/12^(-1/3) belongs to the interval (which)?

2. Hi
Originally Posted by ikislav
1) IF (b-2a)/(4a+3b)=2 (a,b!=0, 4a+3b!=0), THEN (2a^2-3ab+5b^2)/(4a^2+3b^2) EQUALS?
Since $b \neq 0$ you can divide by b

Let $q = \frac{a}{b}$

$\frac{b-2a}{4a+3b}=2=\frac{1-2q}{3+4q}$
from which you can get the value for q that you can then plug into

$\frac{2a^2-3ab+5b^2}{4a^2+3b^2}=\frac{2q^2-3q+5}{4q^2+3}$

3. Since I have started writing,

For 1), the advice above is equivalent to expressing b through a from the first equation and substituting it into the second expression.

For 2), does 0,3 mean 3/10 or an interval from 1 to 3? In the English-speaking world, period is most commonly used as a decimal mark. If it means 3/10, then the value of this expression belongs to many intervals. For example, $1\le 25^{0.3}\cdot5^{0.4}/12^{-1/3}=25^{0.3}\cdot5^{0.4}\cdot12^{1/3}<25\cdot5\cdot12$. To get a tighter upper bound, $25^{0.3}\cdot5^{0.4}\cdot12^{1/3}<25^{1/3}\cdot5^{1/2}\cdot12^{1/3}=\sqrt[3]{25\cdot12}\sqrt{5}<7\cdot2.5=15$.

4. Originally Posted by running-gag
Hi

Since $b \neq 0$ you can divide by b

Let $q = \frac{a}{b}$

$\frac{b-2a}{4a+3b}=2=\frac{1-2q}{3+4q}$
from which you can get the value for q that you can then plug into

$\frac{2a^2-3ab+5b^2}{4a^2+3b^2}=\frac{2q^2-3q+5}{4q^2+3}$
Ok, I understand the principle... thank you very much.
But I dont understand the following: If I divide the second expression by b, how do I get from -3ab to -3q? Wouldn't it be -3a?

5. You should divide the second expression by b^2.

6. Originally Posted by emakarov
You should divide the second expression by b^2.
Yes, I understand now. Thank you.