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.
+486 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 $\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$
+486 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 $\frac{1}{3}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$
