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# operation

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The operation * is defined for non-zero integers as follows: a*b = 1/a + 1/b. If a + b = 9 and ab = 18, what is the value of a * b? Express your answer as a common fraction.

Apr 6, 2021

#1
+2

Cool short manipulation way:

You can simplify $\frac{1}{a}+\frac{1}{b}$ as $\frac{a+b}{ab}$.(If you don't believe me, expand out $\frac{a+b}{ab}$ yourself and simplify. So, now we just plug in $a*b=\frac{9}{18}=\boxed{\frac{1}{2}}$.

Long grind way:

Solving $a+b=9$, we get $a=9-b$. Substituting in, we get $(9-b)(b)=18$. Expanding out:

$(9-b)(b)=18$

$9b-b^2=18$

$-b^2+9b-18=0$

$b^2-9b+18=0$.

$(b-6)(b-3)=0$.

Now we have $b=6$ or $b=3$, and because $a+b=9$, we have $a=3$ if $b=6$, or $a=6$ if $b=3$.

Note that this does not matter because $\frac{1}{a}+\frac{1}{b}$ is symmetric. I will go with $a=3$ and $b=6$, although it does not matter.

Substituting into $\frac{1}{a}+\frac{1}{b}$ gives $\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$

Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021

#1
+2

Cool short manipulation way:

You can simplify $\frac{1}{a}+\frac{1}{b}$ as $\frac{a+b}{ab}$.(If you don't believe me, expand out $\frac{a+b}{ab}$ yourself and simplify. So, now we just plug in $a*b=\frac{9}{18}=\boxed{\frac{1}{2}}$.

Long grind way:

Solving $a+b=9$, we get $a=9-b$. Substituting in, we get $(9-b)(b)=18$. Expanding out:

$(9-b)(b)=18$

$9b-b^2=18$

$-b^2+9b-18=0$

$b^2-9b+18=0$.

$(b-6)(b-3)=0$.

Now we have $b=6$ or $b=3$, and because $a+b=9$, we have $a=3$ if $b=6$, or $a=6$ if $b=3$.

Note that this does not matter because $\frac{1}{a}+\frac{1}{b}$ is symmetric. I will go with $a=3$ and $b=6$, although it does not matter.

Substituting into $\frac{1}{a}+\frac{1}{b}$ gives $\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$

RiemannIntegralzzz Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021