INVERSE VARIATION
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» Solve:
[tex] \: : \implies \sf \large (x - 4) = \Large \frac{k}{(y + 3)} [/tex]
[tex]\large \tt \red{given} \begin{cases} \sf \: x = 8 \\ \sf \: y = 2 \end{cases}[/tex]
» Substitute to identify the constant (k).
[tex]\implies \sf \large (8 - 4) = \Large \frac{k}{(2 + 3)} [/tex]
[tex]\implies \sf \large 4 = \Large \frac{ \: k \: }{5} [/tex]
[tex]\implies \sf \large 4 \times 5= \Large \frac{k}{ \: \cancel5 \: } \large \times \cancel5[/tex]
[tex]\implies \sf \large20=k[/tex]
[tex]\implies \sf \large k = 20[/tex]
» Now identify the (x) if (y) is -1.
[tex]\large \tt \red{given} \begin{cases} \sf \: k = 20 \\ \sf \: y = - 1 \end{cases}[/tex]
[tex]\implies \sf \large (x - 4) = \Large \frac{20}{( - 1 + 3)} [/tex]
[tex]\implies \sf \large (x - 4) = \Large \frac{20}{2} [/tex]
[tex]\implies \sf \large (x - 4) = 10[/tex]
[tex]\implies \sf \large x = 10 + 4[/tex]
[tex]\implies \sf \large x = 14 \\ \\ [/tex]
Final Answer:
[tex] \tt \huge » \: \purple{14}[/tex]
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