INVERSE VARIATION
Problem:
» If a varies inversely as b, and a=12 when b=4, find a when b=-16
Answer:
- [tex] \large \tt \color{hotpink}{a = - 3}[/tex]
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Given:
- [tex] \sf \: a_{1} =12 \: \: \: \: \: \: \: \: \: b_{1} = 4\\ \: \: \: \: \: \: \sf \: \: a_{2} = \: \: \: ? \: \: \: \: \: \: \ \: \: b_{2} = - 16\\ [/tex]
Step-by-step explanation:
First, write the equation of variation and find the constant (k) by substituting the values of a₁ and b₁.
[tex] \: \: \sf \: a_{1} = \frac{k}{ b_{1}} \\ \sf \: 12 = \frac{k}{4} \\ \: \: \: \: \: \: \: \: \: \: \sf \:k= 12 \: (4) \\ \: \: \: \: \: \: \: \: \sf k = \underline{ \boxed{ \purple{ \sf \: 48}}}[/tex]
Then, solve for the value of a when [tex] \sf{ b_{2} = - 16}[/tex]
- [tex] \sf{ a_{2} = \frac{48}{b_{2} } } \\ \: \: \: \sf{ a_{2} = \frac{48}{ - 16 } } \\ \: \: \: \: \sf{ a_{2} = \underline{ \boxed{ \purple{ \sf \: - 3}} }}[/tex]
Thus, the value of a is -3.
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