How do you solve $3 x + \frac{14}{5 x} = \frac{31}{5}$ $3x+14/(5x)=31/5$ ?

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How do you solve $3 x + \frac{14}{5 x} = \frac{31}{5}$ $3x+14/(5x)=31/5$ ?

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Answer 1 · 2018-03-17T17:15:43

$x = \frac{2}{3}, \frac{7}{5}$ $x=2/3, 7/5$

Explanation:

$\frac{15 x^{2}}{5 x} + \frac{14}{5 x} = \frac{31 x}{5 x} \to$ $(15x^2)/(5x)+14/(5x)=(31x)/(5x) rarr$ Find the least common denominator so that the fractions can be eliminated

$15 x^{2} + 14 = 31 x$ $15x^2+14=31x$

$15 x^{2} - 31 x + 14 = 0 \to$ $15x^2-31x+14=0 rarr$ Now multiply 15 and 14 together (it equals 210) and think of two numbers that add to -31 and multiply to 210

$15 x^{2} - 21 x - 10 x + 14 = 0$ $15x^2-21x-10x+14=0$

$3 x (5 x - 7) - 2 (5 x - 7) = 0$ $3x(5x-7)-2(5x-7)=0$

$(3 x - 2) (5 x - 7) = 0$ $(3x-2)(5x-7)=0$

$3 x - 2 = 0$ $3x-2=0$

$3 x = 2$ $3x=2$

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

$5 x - 7 = 0$ $5x-7=0$

$5 x = 7$ $5x=7$

$x = \frac{7}{5}$ $x=7/5$

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How do you solve $3 x + \frac{14}{5 x} = \frac{31}{5}$ $3x+14/(5x)=31/5$ ?

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How do you solve $3 x + \frac{14}{5 x} = \frac{31}{5}$ $3x+14/(5x)=31/5$ ?