Question 1199691
<pre>
Solve   2x-1 < (x+7)/(x+1)  algebraically

Contrary to what the other person states, you CAN multiply by the LCD, x + 1

Since denominator is x + 1, it follows that: {{{x <> - 1}}}
          {{{2x - 1 < (x + 7)/(x + 1)}}}
(2x - 1)(x + 1) < x + 7 ----- Multiplying by LCD, x + 1
    {{{matrix(5,1, 2x^2 + x - 1 < x + 7, 
2x^2 + x - x - 1 - 7 < 0,
2x^2 - 8 < 0,
2(x^2 - 4) < 2(0),
x^2 - 4 < 0)}}}
(x - 2)(x + 2) < 0
x - 2 < 0         OR      x + 2 < 0
x < 2             OR      x < - 2
Since {{{matrix(3,1, x <> - 1, x <> - 2, x <> 2)}}}, - 2, - 1, and 2 will be the CRITICAL points.

With the 3 CRITICAL POINTS above, we get the 4 TEST INTERVALS: 
<font color = red><font size = 2><b>  x < - 2___x = - 3    - 2 < x < - 1___x = - 1.5  -1 < x < 2__x = 0    x > 2___x = 3</font></font></b> 
 {{{matrix(4,1, 2x - 1 < (x + 7)/(x + 1), 2(- 3) - 1 < (- 3 + 7)/(- 3 + 1), - 6 - 1 < (4)/(- 2), highlight(matrix(2,1, - 7 < - 2, "(TRUE)")))}}}  {{{matrix(4,1, 2x - 1 < (x + 7)/(x + 1), 2(- 1.5) - 1 < (- 1.5 + 7)/(- 1.5 + 1), - 3 - 1 < (5.5)/(- .5), highlight(matrix(2,1, - 4 < - 11, "(FALSE)")))}}}  {{{matrix(4,1, 2x - 1 < (x + 7)/(x + 1), 2(0) - 1 < (0 + 7)/(0 + 1), - 1 < (7)/(1), highlight(matrix(2,1, - 1 < 7, "(TRUE)")))}}}  {{{matrix(4,1, 2x - 1 < (x + 7)/(x + 1), 2(3) - 1 < (3 + 7)/(3 + 1), 6 - 1 < (10)/(4), highlight(matrix(2,1, 5 < 2.5, "(FALSE)")))}}}

Solution sets: x < - 2, and - 1 < x < 2 since these are the ONLY TRUE STATEMENTS.
In INTERVAL NOTATION, we get: <font size="7"><font color = red>(*[Tex \Large - \infty, - 2])*[Tex \Large \cup](*[Tex \Large - 1, 2])</font></font></pre>