Given:
the sum of the reciprocals of two numbers is 1/6
Let the numbers are (x) and (y)
so,
[tex]\frac{1}{x}+\frac{1}{y}=\frac{1}{6}\rightarrow(1)[/tex]And, the product of these two numbers is -48
so,
[tex]x\cdot y=-48\rightarrow(2)[/tex]from equation (2):
[tex]\begin{gathered} y=-\frac{48}{x} \\ \\ \frac{1}{y}=-\frac{x}{48}\rightarrow(3) \end{gathered}[/tex]substitute with (1/y) from equation (3) into equation (1) then solve for (x):
[tex]\begin{gathered} \frac{1}{x}-\frac{x}{48}=\frac{1}{6} \\ \frac{48-x^2}{48x}=\frac{1}{6} \end{gathered}[/tex]Using the cross product:
[tex]\begin{gathered} 48x=6(48-x^2)\rightarrow(\div6) \\ 8x=48-x^2 \\ x^2+8x-48=0 \\ (x-4)(x+12)=0 \\ x-4=0\rightarrow x=4 \\ x+12=0\rightarrow x=-12 \end{gathered}[/tex]We will substitute the value of (x) into equation (2) to find (y)
[tex]\begin{gathered} x=4\rightarrow y=-\frac{48}{4}=-12 \\ \\ x=-12\rightarrow y=-\frac{48}{-12}=4 \end{gathered}[/tex]so, the answer will be the numbers are: -12 and 4
