2a+ba+4b=3,2a+ba+4b=3,\frac{{2a + b}}{{a + 4b}} = 3{\text{,}} then find the value of a+ba+2b=?" role="presentation">a+ba+2b=?a+ba+2b=?\frac{{a + b}}{{a + 2b}} = ?" /> 2a+ba+4b=3,2a+ba+4b=3,\frac{{2a + b}}{{a + 4b}} = 3{\text{,}} then find the value of a+ba+2b=?" role="presentation">a+ba+2b=?a+ba+2b=?\frac{{a + b}}{{a + 2b}} = ?" /> 2a+ba+4b=3,2a+ba+4b=3,\frac{{2a + b}}{{a + 4b}} = 3{\text{,}} then find the value of a+ba+2b=?" role="presentation">a+ba+2b=?a+ba+2b=?\frac{{a + b}}{{a + 2b}} = ?" />

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If $\frac{2 a + b}{a + 4 b} = 3,$ $\frac{{2a + b}}{{a + 4b}} = 3{\text{,}}$ then find the value of $\frac{a + b}{a + 2 b} = ?$ $\frac{{a + b}}{{a + 2b}} = ?$

1 $\frac{5}{9}$ $\frac{5}{9}$
2 $\frac{2}{7}$ $\frac{2}{7}$
3 $\frac{10}{9}$ $\frac{{10}}{9}$
4 $\frac{10}{7}$ $\frac{{10}}{7}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{2 a + b}{a + 4 b} = 3 (Given) \\ \Rightarrow 2 a + b = 3 (a + 4 b) \\ \Rightarrow 2 a + b = 3 a + 12 b \\ \Rightarrow - a = 11 b \\ \Rightarrow a = - 11 b \\ ∴ \frac{a + b}{a + 2 b} \\ \Rightarrow \frac{- 11 b + b}{- 11 b + 2 b} \\ \Rightarrow \frac{- 10 b}{- 9 b} \\ \Rightarrow \frac{10}{9} \end{aligned}

$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr}$

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then find the value of

\frac{a + b}{a + 2 b} = ?

$\frac{{a + b}}{{a + 2b}} = ?$ ", "text": "If

\frac{2 a + b}{a + 4 b} = 3,

$\frac{{2a + b}}{{a + 4b}} = 3{\text{,}}$ then find the value of

\frac{a + b}{a + 2 b} = ?

$\frac{{a + b}}{{a + 2b}} = ?$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{2 a + b}{a + 4 b} = 3 (Given) \\ \Rightarrow 2 a + b = 3 (a + 4 b) \\ \Rightarrow 2 a + b = 3 a + 12 b \\ \Rightarrow - a = 11 b \\ \Rightarrow a = - 11 b \\ ∴ \frac{a + b}{a + 2 b} \\ \Rightarrow \frac{- 11 b + b}{- 11 b + 2 b} \\ \Rightarrow \frac{- 10 b}{- 9 b} \\ \Rightarrow \frac{10}{9} \end{aligned}

$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{2 a + b}{a + 4 b} = 3 (Given) \\ \Rightarrow 2 a + b = 3 (a + 4 b) \\ \Rightarrow 2 a + b = 3 a + 12 b \\ \Rightarrow - a = 11 b \\ \Rightarrow a = - 11 b \\ ∴ \frac{a + b}{a + 2 b} \\ \Rightarrow \frac{- 11 b + b}{- 11 b + 2 b} \\ \Rightarrow \frac{- 10 b}{- 9 b} \\ \Rightarrow \frac{10}{9} \end{aligned}

$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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