How do you solve $\frac{5 q^{2}}{6} - \frac{q^{2}}{3} = 72$ $(5q^2)/6-q^2/3=72$ ?

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Answer 1 · 2016-10-27T12:15:10

$q = \pm 12$ $q=+-12$

The objective is to have only one $q$ $q$

notice that $q^{2}$ $q^2$ is in both numerators so factor that out

$q^{2} (\frac{5}{6} - \frac{1}{3}) = 72$ $q^2(5/6-1/3)=72$

$q^{2} (\frac{5}{6} - \frac{2}{6}) = 72$ $q^2(5/6-2/6)=72$

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

$q^2=2xx72=144$

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

$q=+-12$
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Footnote

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

$(+12)xx(+12)=+144$

How do you solve (5q^2)/6-q^2/3=725q26−q23=72(5q^2)/6-q^2/3=72?