How do you solve #(5q^2)/6-q^2/3=72#?

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Answer 1 · 2016-10-27T12:15:10

#q=+-12#

The objective is to have only one #q#

notice that #q^2# is in both numerators so factor that out

#q^2(5/6-1/3)=72#

#q^2(5/6-2/6)=72#

#q^2((cancel(3)^1)/(cancel(6)^2))=72" "->" "q^2/2=72#

#q^2=2xx72=144#

#sqrt(q^2)=sqrt(144)#

#q=+-12#
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Footnote

#(-12)xx(-12)=+144#

#(+12)xx(+12)=+144#