How do you solve #\frac { 4} { v ^ { 2} - 3v } = \frac { 1} { 5- v }#?

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Answer 1 · 2018-05-03T21:57:11

#v = -5, 4#

We start by cross-multiplying to get rid of the fractions.

#4 / (v^2 - 3v) = 1/(5-v) -> 4(5-v) = v^2 - 3v#

This can be rearranged:

#4(5-v) = v^2 - 3v -> 20 - 4v = v^2 - 3v -> v^2 + v - 20 = 0#
#-> (v + 5)(v - 4) = 0 -> v = -5, 4#.

We've found two solutions, #v = -5# and #v = 4#.

Answer 2 · 2018-05-03T22:03:29

#v = 4 or -5#

You'll want to start by cross multiplying.

#4(5-v) = 1(v^2-3v)#
#20 - 4v = v^2 - 3v#

Have like terms on the same side.
#20 = v^2 + v#

Now you can figure out in your head that #v = 4#, but if you use the quadratic formula, you'll get another possible solution.

#v^2 + v - 20 = 0#

#v = (-b +- sqrt(b^2 - 4ac))/(2a)#

Plug in your values (a = 1, b = 1, c = -20), and you get:

#v = (-1 +- sqrt(1^2 - (4 * 1 * -20)))/(2 * 1)#

Now let's simplify.

#v = (-1 +- sqrt(1 + 80))/2#

#v = (-1 +- sqrt81)/2#

#v = (-1 +- 9)/2#

#(-1 + 9)/2 = 8/2 = 4#

#(-1 - 9)/2 = 8/2 = -5#

So #v# can be either #4# or #-5#