How do you solve $\frac { v - 6} { v - 4} = \frac { v } { v + 1}$ ?

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Answer 1 · 2018-03-12T04:37:48

$v=-6.$

$(v-6)/(v-4)=v/(v+1)$

or, $(v-6)(v+1)=v(v-4)$

or, $v^2-5v-6=v^2-4v$

Cancelling the $v^2$ on both sides,
we have:
$-5v-6=-4v$

Simplifying further,

$-v-6=0$

or, $-v=6$
Thus, we have, $v=-6$

Plugging in the value of $v$ in the equation,
Left hand side is:
$(v-6)/(v-4)=(-6-6)/(-6-4)=(-12)/-10=6/5$

Plugging in the value of $v$ in the Right hand side:

$v/(v+1)=-6/(-6+1)=(-6)/-5=6/5$

Thus, $LHS=RHS$ in the equation.

Hope this helps!!

How do you solve \frac { v - 6} { v - 4} = \frac { v } { v + 1}\frac { v - 6} { v - 4} = \frac { v } { v + 1}?