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If l is not equal to –3 or –4, then l/l + 4 + 3/l + 3 =
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07 Jun 2017, 02:29
07 Jun 2017, 02:29
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Question Stats:
89% (01:09) correct
10% (01:15) wrong based on 96 sessions
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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) \(\frac{3l}{(l+4)(l+3)}\)
E) \(\frac{l^2+6l+12}{(l+4)(l+3)}\)
Re: If l is not equal to –3 or –4, then l/l + 4 + 3/l + 3 =
[#permalink]
27 Sep 2017, 07:36
27 Sep 2017, 07:36
This is pure algebra
\(\frac{l}{l+4}+\frac{3}{l+3}=\frac{l^2+3l+3l+12}{(l+4)(l+3)}=\frac{l^2+6l+12}{(l+4)(l+3)}\)
Answer E!
\(\frac{l}{l+4}+\frac{3}{l+3}=\frac{l^2+3l+3l+12}{(l+4)(l+3)}=\frac{l^2+6l+12}{(l+4)(l+3)}\)
Answer E!
Re: If l is not equal to 3 or 4, then l/l + 4 + 3/l + 3 =
[#permalink]
11 Nov 2025, 00:28
11 Nov 2025, 00:28
1
\(\frac{l}{l+4} + \frac{3}{l+3}\)
Adding the two fractions we get,
\(\frac{l(l+3) + 3(l+4)}{(l+4)(l+3)}\)
\(\frac{l^2 +3l +3l+ 12}{l^2 + 3l + 4l + 12}\)
\(\frac{l^2 + 6l + 12}{l^2 + 7l + 12}\)
The answer is Choice E.
Adding the two fractions we get,
\(\frac{l(l+3) + 3(l+4)}{(l+4)(l+3)}\)
\(\frac{l^2 +3l +3l+ 12}{l^2 + 3l + 4l + 12}\)
\(\frac{l^2 + 6l + 12}{l^2 + 7l + 12}\)
The answer is Choice E.
