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If $\large\frac{a}{b}=\frac{c}{d}$ how we can obtain $\displaystyle{\frac{ma+nb}{pa+qb}=\frac{mc+nd}{pc+qd}}$?

I can get $\large\frac{ma}{qb}=\frac{mc}{qd}$ and $\large\frac{nb}{pa}=\frac{nd}{pc}$ , now

If we have $\large\frac{a}{b}=\frac{c}{d}$ and $\large\frac{e}{f}=\frac{g}{h}$ is true that $\large\frac{a+e}{b+f}=\frac{c+g}{d+h}$ , I tried a particular solution and is not true at least for most cases,but for the above expression is true so how do I get there?can you help me out?

studentNk
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2 Answers2

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Hint:$$\frac{ma+nb}{pa+qb}=\frac{m\frac{a}{b}+n}{p\frac{a}{b}+q}$$

user71352
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  • ?I follow up your hint but I don't know what to do,I have so far (m(a/b)+n)/(p(a/b)+q)=(m(c/d)+n)/(p(c/d)+q) what to do next?or is this way I suppose to go? – studentNk Jul 29 '14 at 01:36
  • OK I think I got it,I substitute x=a/b=c/d and the resulting equation holds true, therefore I start with that to get my expression, thank you – studentNk Jul 29 '14 at 01:50
  • This is the way you are supposed to go. Notice that: $\frac{m\frac{c}{d}+n}{p\frac{c}{d}+q}=\frac{\frac{mc+dn}{d}}{\frac{pc+qd}{d}}= \frac{mc+nd}{pc+dq}$. – user71352 Jul 29 '14 at 01:50
  • Yes I understand that , thank you again – studentNk Jul 29 '14 at 01:54
  • My apologies. Just wanted to clarify. You're welcome. – user71352 Jul 29 '14 at 01:55
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Follows immediately by $\rm\color{#c00}{M} =$ mediant, or by denominator linearity ("Or" below)

$x = \dfrac{\color{c00}a}{\color{c00}c}= \dfrac{\color{c00}b}{\color{c00}d}\,\Rightarrow\, x = \dfrac{m\color{c00}a}{m\color{c00}c} = \dfrac{n\color{c00}b}{n\color{c00}d}\,\overset{\rm\color{#c00}{M}}\Rightarrow\, x = \dfrac{ma\!+\!nb}{mc\!+\!nd}\ \left[ = \dfrac{pa\!+\!qb}{pc\!+\!qd}\,\ \rm similarly\right]$

$\begin{align} {\bf Or}\qquad\qquad m\, &\left[cx = a\right]\\ +\ \ n\,&\left[dx = b\right]\\ \hline \Longrightarrow \ \ (mc\!+\!nd)\,x\!\!\!\!\!\!\!\! &\quad\ \ = ma\!+\!nd \end{align}$

Bill Dubuque
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  • OP = diagonal swap of above equality ie. $,\dfrac{I}{\color{#0a0}J} = \dfrac{\color{#c00}{I'}}{J'}!\iff \dfrac{I}{\color{#c00}{I'}}=\dfrac{\color{#0a0}J}{J'}\ \ $ – Bill Dubuque Sep 30 '23 at 00:49