COMBINED VARIATION
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» (a) varies directly as (b) and inversely as (c).if a=6 when b=3 and c=4,what is the value of b when a=10 and c=4?
» Solve:
[tex] \: : \implies \sf a = \frac{kb}{c} \\ [/tex]
[tex] \tt \red{given} \begin{cases} \sf \: a = 6 \\ \sf \: b = 3 \\ \sf \: c = 4\end{cases}[/tex]
» Substitute the value to find the constant (k).
[tex]\implies \sf 6= \frac{k(3)}{4} \\ [/tex]
[tex]\implies \sf 6 \times \frac{4}{3} = \frac{k(3)}{4} \times \frac{4}{3} \\ [/tex]
[tex]\implies \sf \frac{24}{3} = \frac{k( \cancel{12})}{ \cancel{12}} \\ [/tex]
[tex]\implies \sf k = \frac{24}{3} \\ [/tex]
[tex]\implies \sf k = 8[/tex]
» Now that we have the value of the constant (k = 8). Find (b) when a=10 and c=4:
[tex]\tt \red{given} \begin{cases} \sf \: a = 10 \\ \sf \: k = 8 \\ \sf \: c = 4\end{cases}[/tex]
[tex]\implies \sf 10 = \frac{(8)b}{4} \\ [/tex]
[tex]\implies \sf 10 = 2b[/tex]
[tex]\implies \sf b = \frac{10}{2} \\ [/tex]
[tex]\implies \sf b = 5[/tex]
Final Answer:
[tex] \tt \large » \: \purple{b=5}[/tex]
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