How do you solve $\frac{3}{2 x - 9} - 2 = \frac{5 x}{2 x - 9}$ $3/(2x-9)-2=(5x)/(2x-9)$ and check for extraneous solutions?

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Answer 1 · 2016-09-09T14:43:24

$x = \frac{7}{3}$ $x=7/3$

First you have to find the least common denominator $L C D$ $LCD$ that is:

$2 x - 9$ $2x-9$

Now, a denominator cannot be zero, then:

$2 x - 9 \neq 0 \Rightarrow x \neq \frac{9}{2}$ $2x-9!=0 => x!=9/2$

$:. x=9/2 !in FE$ (Field of existence of the equation), and it cannot be a solution.

Now, you simplyfy the equation using the $LCD$

$(3-2(2x-9))/cancel((2x-9))=(5x)/cancel((2x-9))$

$=> 3-4x+18=5x$

Now move all the x therms in the LHS and the other in the RHS

$-4x-5x=-21$
$-9x=-21$
$9x=21$
$x=cancel(21)^7/cancel(9)^3=7/3$

with $x in FE$

How do you solve 3/(2x-9)-2=(5x)/(2x-9)32x−9−2=5x2x−93/(2x-9)-2=(5x)/(2x-9) and check for extraneous solutions?