Solve, please:1) 8 x + 1 = 4 3 2) x + 3 3 x − 6 &gt; 03) 2 x + 3 = x 4 4) 1 x 2 − 4 ≤ 05) 4 x + 1 3 x = 96) x + 32 x + 6 ≤ 67) 4 x + 1 x 2 = 1 5 x 2 8) 1 + 2 x + 1 &lt; 2 x 9) x − 3 2 x + 10 + 2 x − 12 = x 2 + 3 x − 18 2 x + 10 10) 12 x 3 + 16 x 2 − 3 x − 4 8 x 3 + 12 x 2 + 10 x + 15 &lt; 0

Question

Solve, please:1)       8          x      +      1        =      4    3  2)             x      +      3              3      x      −      6        &amp;gt;  03)   2  x  +  3  =      x    4  4)       1                  x        2            −      4        ≤  05)       4    x    +      1          3      x        =  96)             x      +      32              x      +      6        ≤  67)       4    x    +      1          x      2        =      1          5              x        2            8)   1  +      2          x      +      1        &amp;lt;      2    x  9)             x      −      3              2      x      +      10        +  2  x  −  12  =                    x        2            +      3      x      −      18              2      x      +      10      10)             12              x        3            +      16              x        2            −      3      x      −      4              8              x        3            +      12              x        2            +      10      x      +      15        &amp;lt;  0

Nola Rivera · Accepted Answer

1)       8          x      +      1        =      4    3    8  ×  3  =  (  x  +  1  )  ×  4  24  =  (  x  +  1  )  ×  4  x  +  1  =  6  x  =  6  −  1  x  =  52)             x      +      3              3      x      −      6        &amp;gt;  0            x      +      3              3      (      x      −      2      )        &amp;gt;  0            3      (      x      +      3      )              3      (      x      −      2      )        &amp;gt;  0  ×  3            x      +      3              x      −      2        &amp;gt;  0The rational function is zero when   x  =  −  3 and undefined when   x  =  2. Thus, the critical points are   −  3 and   2. Now, check the sign of the function in the intervals   (  −  ∞  ,  −  3  )  ,  (  −  3  ,  2  ) and   (  2  ,  ∞  )When   x  =  −  4            x      +      3              x      −      2        =            −      4      +      3              −      4      −      2        =            −      1              −      6        =      1    6    &amp;gt;  0When   x  =  0            x      +      3              x      −      2        =            0      +      3              0      −      2        =  −      3    2    &amp;lt;  0When   x  =  3            x      +      3              x      −      2        =            3      +      3              3      −      2        =  6Thus, the solutions are:   x  &amp;lt;  −  3  ,  x  &amp;gt;  2Interval   (  −  ∞  ,  −  3  )  ∪  (  2  ,  ∞  )3)   2  x  +  3  =      x    4    2  x  =      x    4    −  3  2  x  ×  4  =      x    4    ×  4  −  3  ×  4  8  x  =  x  −  12  7  x  =  −  12  x  =  −      12    7  4)       1                  x        2            −      4        ≤  0      x    2    −  4  &amp;lt;  0 (If       1    a    ≤  0 then   a  &amp;lt;  0)      x    2    −  4  +  4  &amp;lt;  0  +  4      x    2    &amp;lt;  4  −      4    &amp;lt;  x  &amp;lt;      4   (For       k    n    &amp;lt;  a if   n is even then   −      a    n    &amp;lt;  k  &amp;lt;      a    n  )  −  2  &amp;lt;  x  &amp;lt;  2The function is not defined when   x  =  ±  2. Thus, the solution is   (  −  2  ,  2  )5)       4    x    +      1          3      x        =  9      4    x    ×  3  x  +      1          3      x        ×  3  x  =  9  ×  3  x  12  +  1  =  27  27  x  =  13  x  =      13    27  6)             x      +      32              x      +      6        ≤  6            x      +      32              x      +      6        −  6  ≤  0            −      6      (      x      +      6      )      +      x      +      32              x      +      6        ≤  0            −      5      x      −      4              x      +      6        ≤  0The function             −      5      x      −      4              x      +      6        ≤  0 is zero when   −  5  x  −  4  =  0  −  5  x  −  4  =  0  −  5  x  =  4  x  =  −      4    5  The function is undefined when   x  =  −  6And critical points are   −  6 and   −      4    5  Now, check the sign of   −            −      5      x      −      4              x      +      6       in each interval   (  −  ∞  ,  −  6  )  ,  (  −  6  −      4    5    )  ,  (  −      4    5    ,  ∞  ) Thus,   −            −      5      x      −      4              x      +      6        ≤  0 when   x  &amp;lt;  −  6 or   x  ≥  −      4    5   as the function is undefined when   x  =  −  6Thus, the solution is   (  −  ∞  ,  −  6  )  ∪  [  −      4    5    ,  ∞  )7)       4    x    +      1          x      2        =      1          5              x        2                  4    x    ×  5      x    2    +      1          x      2        ×  5      x    2    =      1          5              x        2              ×  5      x    2    20  x  +  5  =  1  20  x  =  −  4  x  =  −      1    5

Davon Irwin · Accepted Answer

(8) Given   1  +      2          x      +      1        &amp;lt;      2    x              x      +      3              x      +      1        &amp;lt;      2    x              x      +      3              x      +      1        −      2    x    &amp;lt;  0            (      x      +      3      )      x              x      (      x      +      1      )        −            2      (      x      +      1      )              x      (      x      +      1      )        &amp;lt;  0            x      (      x      +      3      )      −      2      (      x      +      1      )              x      (      x      +      1      )        &amp;lt;  0            (      x      −      1      )      (      x      +      2      )              x      (      x      +      1      )        &amp;lt;  0The function             (      x      −      1      )      (      x      +      2      )              x      (      x      +      1      )       is undefined when   x  =  0 or   x  =  −  1Thus, we have the critical points:   x  =  −  2  ,  −  1  ,  0  ,  1Check the sign of the function in the intervals   (  −  ∞  ,  −  2  )  ,  (  −  2  ,  −  1  )  ,  (  −  1  ,  0  )  ,  (  0  ,  1  )  ,  (  1  ,  ∞  )Hence,             (      x      −      1      )      (      x      +      2      )              x      (      x      +      1      )        &amp;lt;  0 when   −  2  &amp;lt;  x  &amp;lt;  −  1 or   0  &amp;lt;  x  &amp;lt;  1The solution is   (  −  2  ,  −  1  )  ∪  (  0  ,  1  )(9) Given             x      −      3              2      x      +      10        +  2  x  −  12  =                    x        2            +      3      x      −      18              2      x      +      10                  x      −      3              2      x      +      10        (  2  x  +  10  )  +  2  x  (  2  x  +  10  )  −  12  (  2  x  +  10  )  =                    x        2            +      3      x      −      18              2      x      +      10        (  2  x  +  10  )  x  −  3  +  2  x  (  2  x  +  10  )  −  12  (  2  x  +  10  )  =      x    2    +  3  x  −  18  4      x    2    −  3  x  −  123  =      x    2    +  3  x  −  18  4      x    2    −  3  x  −  105  =      x    2    +  3  x  4      x    2    −  6  x  −  105  =      x    2    3      x    2    −  6  x  −  105  =  0  x  =            −      (      −      6      )      ±              (        −        6                  )          2                −        4        ×        3        (        −        105        )                    2      ×      3        =            6      ±      36        6        x    1    =      42    6  ,       x    2    =  −      30    6        x    1    =  7  ,      x    2    =  −  5The function is not defined when   2  x  +  10  =  0  2  x  =  −  10  x  =  −      10    2    x  =  −  5Thus,   x  =  −  5 is an extraneous solution. This means, the solution of the equation is   x  =  7(10) Given             12              x        3            +      16              x        2            −      3      x      −      4              8              x        3            +      12              x        2            +      10      x      +      15        &amp;lt;  0            (      3      x      +      4      )      (      2      x      +      1      )      (      2      x      −      1      )              (      2      x      +      3      )      (      4              x        2            +      5      )        &amp;lt;  0The function             (      3      x      +      4      )      (      2      x      +      1      )      (      2      x      −      1      )              (      2      x      +      3      )      (      4              x        2            +      5      )        &amp;lt;  0 is zero when   (  3  x  +  4  )  (  2  x  +  1  )  (  2  x  −  1  )  =  0   3  x  +  4  =  0 or   2  x  +  1  =  0 or   2  x  −  1  =  0  x  =  −      4    3    ,  x  =  −      1    2    ,  x  =      1    2  and undefined when   (  2  x  +  3  )  (  4      x    2    +  5  )  =  0  2  x  +  3  =  0 or   4      x    2    +  5  =  0 (has no real solution)  x  =  −      3    2  Hence, we have the critical points:   x  =  −      3    2    ,  −      4    3    ,  −      1    2    ,      1    2  The intervals are   (  −  ∞  ,  −      3    2    )  ,  (  −      3    2    ,  −      4    3    )  ,  (  −      4    3    ,  −      1    2    )  ,  (  −      1    2    ,      1    2    )Now, check the intervals where             (      3      x      +      4      )      (      2      x      +      1      )      (      2      x      −      1      )              (      2      x      +      3      )      (      4              x        2            +      5      )        &amp;lt;  0The solution is   −      3    2    &amp;lt;  x  &amp;lt;  −      4    3   or   −      1    2    &amp;lt;  x  &amp;lt;      1    2  Interval notation:   (  −      3    2    ,  −      4    3    )  ∪  (  −      1    2    ,      1    2    )

Solve, please: 1) 8 x + 1 </mrow> </mfrac> = 4 3

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