Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have an equation:

$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$

How do you solve these types of equations? For example, if we have:

$\dfrac{1}{x} = \dfrac{3}{2} $, we use:

$1\times 2 = 3\times x$

$x = 1.5$

What is a similair approach to my equation?

share|improve this question
    
Yes, exactly. First forget about the middle term. Next, forget about the $\frac35$. And then you're done. –  Berci Nov 1 '12 at 19:29

4 Answers 4

up vote 4 down vote accepted

Essentially, you can solve this using the process you describe, but twice, to generate two equations in two variables, and then solving for each variable as a "system of equations".

We have:

$$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$

Equation 1: $\quad 5(p-1) = 3q \iff 5p - 3q = 5$

Equation 2: $\quad 5(q-1) = 3(2p +1) \iff -6p +5q=8$

So your system of two equations in two unknowns becomes:

$$5p - 3q = 5\tag{1}$$ $$-6p +5q=8\tag{2}$$

Can you take it from there?

You can express (1) as a function of p (isolate p), and then substitute the expression obtained for p, into p in (2), and then solve for q, then p,

or

You can use "row operations": multiply (1) by 5 (both sides), and (2) by 3 (both sides):

$$25p-15q=25\tag{1}$$ $$-18p+15q = 24\tag{2}$$

Now add the equations (q disappears), solve for p, then "plug" p into one of the original equations and solve for q:

$$7p = 49n\implies p = 7$$

Now...From (1), originally, above

$$5p-3q=5 \implies 5(7) - 3q = 5\implies 35 - 3q = 5$$ $$\implies -3q=-30 \implies q = 10$$

share|improve this answer
    
Thank you sir, I am familair with row operations so I won't have trouble with these types of equations again. –  JohnPhteven Nov 1 '12 at 19:49
    
(+1) Step by step getting the answer. –  Babak S. Aug 6 '13 at 10:15

You have two equation with two variables: $$1) \quad \frac{p-1}{q} = \frac{3}{5}$$ and $$2) \frac{q-1}{2p+1}=\frac{3}{5} $$

$1)$ Implies that $5(p-1)= 3q$, which gives $q= \frac{5(p-1)}{3}$. Then you can insert $q= \frac{5(p-1)}{3}$ in $(2)$. Then find what is $p$, and then you can find what $q$ is.

share|improve this answer

$$ \frac{p-1}{q} = \frac{q-1}{2p+1} = \frac {3}{5}$$ so $$\frac{p-1}{q}=\frac {3}{5}$$ and $$\frac {3}{5}=\frac{q-1}{2p+1}.$$ \begin{eqnarray} 5p-5&=3q&\\ 6p+3&=&5q-5 \end{eqnarray} $$5p-3q=5$$ $$6p-5q=-8$$ $$30p-18q=30$$ $$30p-25q=-40$$ $$-7q=-70$$ $$q=10$$ $$p=7.$$

share|improve this answer
1  
$p = 7$? :) ${}{}$ –  Alex Nov 1 '12 at 19:37
    
is there a problem ? –  Iuli Nov 1 '12 at 19:47

From $$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$ follow the system of linear equations with two unknowns

$$5(p-1)=3q$$ $$5(q-1)=3(2p+1)$$ or $$5p-3q=5$$ $$6p-5q=-8$$

wich can be solved using for example Cramer rule

$\Delta=-25+18=-7$

$\Delta_p=-25-24=-49$

$\Delta_q=-40-30=-70$

$$p=\frac{\Delta_p}{\Delta}=\frac{-49}{-7}=7$$,$$ q=\frac{\Delta_q}{\Delta}=\frac{-70}{-7}=10$$

share|improve this answer

Your Answer

 
discard

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.