Answer:
-13
Step-by-step explanation:
We are given:
[tex]\frac{x}{a}=4[/tex]
[tex]\frac{a}{y}=6[/tex]
[tex]a^2=9[/tex]
[tex]ab^2=-8[/tex]
Since [tex]ab^2=-8[/tex] then [tex]a[/tex] has to be negative.
Solving [tex]a^2=9[/tex] therefore gives [tex]a=-3[/tex].
(Note: [tex](-3)^2=(-3)(-3)=9[/tex].)
[tex]\frac{x}{a}=4[/tex] and [tex]a=-3[/tex] gives us:
[tex]\frac{x}{-3}=4[/tex].
Multiplying both sides by -3 gives: [tex]x=-12[/tex].
[tex]\frac{a}{y}=6[/tex] and [tex]a=-3[/tex] gives us:
[tex]\frac{-3}{y}=6[/tex].
Multiplying both sides by [tex]y[/tex] gives: [tex]-3=6y[/tex].
Divide both sides by 6 gives: [tex]\frac{-3}{6}=y[/tex].
Simplifying this gives us [tex]\frac{-1}{2}=y[/tex].
Now we are asked to find the numerical value for [tex]x+2y[/tex].
[tex]-12+2(\frac{-1}{2})[/tex]
[tex]-12+-1[/tex]
[tex]-13[/tex]
D.
