Change the subject of formula for a- a - 1-b - 2 - 3a-b

The correct option is C 1 + 2b/2 Given equation is: (a+1) b + 2= 3a b (a+1+2b)= 3a 1+2b= 2a ∴ a=1+2b2 ...

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Division of Fractions

Change the su...

Question

Change the subject of formula for $a$ .
$\frac{a + 1}{b} + 2 = \frac{3 a}{b}$

\frac{1 - 2 b}{2}

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\frac{1 + b}{2}

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\frac{1 + 2 b}{2}

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\frac{1 - b}{2}

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1-2a-2b - 3(a+b)^ 2 factorize

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & - 2 b 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(1 + a^{2} + b^{2})}^{3}

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{n}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(1 + a^{2} + b^{2})}^{3}

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