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For x ≠ –2 and x ≠ –4,
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15 Apr 2018, 03:44
15 Apr 2018, 03:44
Expert Reply
00:00
Question Stats:
92% (00:39) correct
7% (03:44) wrong based on 26 sessions
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For x ≠ –2 and x ≠ –4, \(\frac{x}{x+4}+\frac{-3}{x+2}=\)
A. \(\frac{x^2-x-12}{(x+4)(x+2)}\)
B. \(\frac{-3}{(x+4)(x+2)}\)
C. \(\frac{x-3}{2x+6}\)
D. \(\frac{1}{x+4}\)
E. \(-2\)
A. \(\frac{x^2-x-12}{(x+4)(x+2)}\)
B. \(\frac{-3}{(x+4)(x+2)}\)
C. \(\frac{x-3}{2x+6}\)
D. \(\frac{1}{x+4}\)
E. \(-2\)
Drill 2
Question: 4
Page: 510-511
Question: 4
Page: 510-511
Re: For x ≠ –2 and x ≠ –4,
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15 Oct 2018, 21:40
15 Oct 2018, 21:40
Expert Reply
sandy wrote:
For x ≠ –2 and x ≠ –4, \(\frac{x}{x+4}+\frac{-3}{x+2}=\)
A. \(\frac{x^2-x-12}{(x+4)(x+2)}\)
B. \(\frac{-3}{(x+4)(x+2)}\)
C. \(\frac{x-3}{2x+6}\)
D. \(\frac{1}{x+4}\)
E. \(-2\)
A. \(\frac{x^2-x-12}{(x+4)(x+2)}\)
B. \(\frac{-3}{(x+4)(x+2)}\)
C. \(\frac{x-3}{2x+6}\)
D. \(\frac{1}{x+4}\)
E. \(-2\)
Drill 2
Question: 4
Page: 510-511
Question: 4
Page: 510-511
we can do such questions in two ways..
1) Algebraic
\(\frac{x}{x+4}+\frac{-3}{x+2}=\frac{(x+2)x-3(x+4)}{(x+2)(x+4)}=\frac{x^2+2x-3x-12}{(x+4)(x+2)}=x^2-x-12/(x+2)(x+4)\)
A
2) Substitution
take x as 0
\(\frac{x}{x+4}+\frac{-3}{x+2}=\frac{0}{0+4}+\frac{-3}{0+2}=0+\frac{-3}{2}=\frac{-3}{2}\)
let us see the choices:-
A. \(\frac{x^2-x-12}{(x+4)(x+2)}.......\frac{0^2-0-12}{4*2}=\frac{-3}{2}\)..yes
B. \(\frac{-3}{(x+4)(x+2)}.......\frac{-3}{4*2}=\frac{-3}{8}\)....no
C. \(\frac{x-3}{2x+6}........\frac{0-3}{2*0+6}=\frac{-1}{2}\)..no
D. \(\frac{1}{x+4}.........\frac{1}{4}\)....no
E. \(-2\)...no
A
