If y=1/5 when x=-20, how do you find y when x=-8/5 given that y varies inversely as x?

2 Answers
Oct 2, 2017

Answer:

#y=5/2#

Explanation:

#color(blue)("The teaching bit")#

Inverse means '1 over'

So the inverse of #a# is #1/a#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the structure of the whole equation")#

Something changes the invers to the target value.

Let #k# be some constant

Then we have:

#y=1/x xxk color(white)("d")->color(white)("d")y=k/x" "...Equation(1)#

Initial condition #-> y=k/xcolor(white)("ddd")->color(white)("ddd")1/5=k/(-20)#

Multiply both sides by #color(red)(-20)larr# gets #k# on its own.

#color(green)(1/5=k/(-20)color(white)("ddd")->color(white)("ddd")1/5color(red)(xx(-20))=color(white)("d")k/(-20)color(red)(xx(-20))#

#color(white)("ddddddddddd")->color(white)("dddddd")(color(red)(-20))/color(green)(5)color(white)("ddd")=color(white)("ddd")color(green)(k)color(white)("dd")xxcolor(white)("d")(color(red)(-20))/(color(green)(-20))#

But #(-20)/(-20)=+1 and kxx1=k#

# color(white)("dddddddddddd")->color(white)("ddddd") -20/5color(white)("dd") =color(white)("d") k#

#k=-4#

So by substitution in #Equation(1)# we have:

#y=-4/x#
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#color(blue)("Determine the value requested in the question")#

Given that #x=-8/5#

#y=-4/xcolor(white)("d")->color(white)("d")y=-4-:(-8/5)#

Both signs are the same (negative) so the answer is positive (plus)

Turn the #8/5# upside down and change divide into multiply.

#y=(-4)xx(-5/8)color(white)("ddd")->color(white)("ddd")y=(-5)xx(-4/8)#

Remember that the answer id positive.

#color(white)("dddddddddddddddddddd")->color(white)("ddd")y=5xx1/2#

#color(white)("dddddddddddddddddddd")->color(white)("ddd")y=5/2#

Oct 2, 2017

Answer:

#y=2 1/2# when #x=-8/5#

Explanation:

If #color(red)x# and #color(blue)y# vary inversely then
#color(white)("XXX")color(red)x xx color(blue)y=color(magenta)c color(white)("xxxx")#for some constant #color(magenta)c#

Given that #(color(red)x,color(blue)y)=(color(red)(-20),color(blue)(1/5))# is a solution to this relation we have
#color(white)("XXX")color(red)(-20) xx color(blue)(1/5)=color(magenta)c#

#color(white)("XXX")color(brown)(-4)=color(magenta)c color(white)("xxx")#or#color(white)("xx")color(magenta)c=color(brown)(-4)#

That is
#color(white)("XXX")color(blue)x xx color(red)y=color(brown)(-4)#

So when #color(blue)x=color(blue)(-8/5)#
#color(white)("XXX")color(blue)(""(-8/5))xxcolor(red)y = color(brown)(-4)#

#color(white)("XXX")color(red)y=color(brown)(""(-4)) xx color(blue)(""(-5/8))=5/2=2 1/2#