# How do you solve an equation like this?

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?

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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

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

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

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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.

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

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$p = 7$? :) ${}{}$ –  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

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

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

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

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

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