Bunuel wrote:
If l is not equal to −3 or −4, then \(\frac{l}{l + 4} + \frac{3}{l + 3} =\)
A. 1
B. \(\frac{7}{l + 4}\)
C. \(\frac{l+3}{2l + 7}\)
D. \(\frac{3l}{(l + 4)(l + 2)}\)
E. \(\frac{l^2 + 6l + 12}{(l + 4)(l + 3)}\)
This one becomes relatively simple if you
plug in 1 for the
variable l.
If you plug in 1, the answer to the addition question is \(\frac{19}{20}\)
You can quickly eliminate answer A and B, since they clearly do not equal \(\frac{19}{20}\)
Answer E is the most logical choice to try next, because you know that you must have 19 in the numerator (at glance you can see that the numerators C and D
will not amount to 19).
BAM!
\(\frac{1^2 + 6(1) + 12}{(1+ 4)(1 + 3)}\)=\(\frac{1+6+12}{(5)(4)}\) =\(\frac{19}{20}\)
Answer: E________
Kudo if you like it...
Critique is welcome. I am trying to learn too!!
Warm Regards,
Rumi