How do you solve $\frac{x + 3}{3} = \frac{6}{x}$ $(x+3)/3 = 6/x$ ?

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Answer 1 · 2018-05-01T04:29:24

$x = 3 or x = - 6$ $x= 3 or x= -6$

$\frac{x + 3}{3} = \frac{6}{x}$ ${x+3}/3 = 6/x$

$x (x + 3) = 6 (3)$ $x(x+3) = 6(3)$

$x^{2} + 3 x = 18$ $x^2 + 3x = 18$

$x^{2} + 3 x - 18 = 0$ $x^2 +3 x - 18 = 0$

$(x - 3) (x + 6) = 0$ $(x- 3)(x+6) = 0$

$x = 3 or x = - 6$ $x= 3 or x= -6$

Check:

$x = 3$ $x=3$

$\frac{3 + 3}{3} = 2$ ${3+3}/3 = 2$

$6/3 = 2 quad sqrt$

${-6 + 3}/3= -1$

$6/-6 = -1 quad sqrt$

How do you solve (x+3)/3 = 6/xx+33=6x(x+3)/3 = 6/x?