If a, b are positive real numbers such that t | vedclass

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(B) Given that the lines $ax + 9y = 5$ and $4x + by = 3$ are parallel,their slopes must be equal.For the line $ax + 9y = 5$,the slope is $m_1 = -\frac

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$12$

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(B) Given that the lines $ax + 9y = 5$ and $4x + by = 3$ are parallel,their slopes must be equal.For the line $ax + 9y = 5$,the slope is $m_1 = -\frac{a}{9}$.For the line $4x + by = 3$,the slope is $m_2 = -\frac{4}{b}$.Since the lines are parallel,$m_1 = m_2$,which implies $-\frac{a}{9} = -\frac{4}{b}$,so $ab = 36$.We need to find the minimum value of $a + b$ where $a, b > 0$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality:$\frac{a + b}{2} \geq \sqrt{ab}$Substituting $ab = 36$:$\frac{a + b}{2} \geq \sqrt{36}$$\frac{a + b}{2} \geq 6$$a + b \geq 12$.Thus,the least possible value of $a + b$ is $12$.

If $a, b$ are positive real numbers such that the lines $ax + 9y = 5$ and $4x + by = 3$ are parallel,then the least possible value of $a + b$ is

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