Линеарне једначине са једном непознатом – понављање градива

Текст задатака објашњених у видео лекцији:

Пр.1)   Решити једначине:

           а) $ - 5x + 3 =  - 12$

           б) $3\left( {x + 1} \right) =  - 9$

           в) $\frac{1}{5}x + \frac{3}{5} =  - 2$

           г) $\frac{1}{2}x - \frac{1}{3}=1$

Пр.2)   Решити једначине:

           а) $5\left( {x - 2} \right) =  - 12 + 4x$

           б) $\frac{{x - 3}}{4} + \frac{x}{2} = \frac{1}{3}$

           в) $\frac{{x + 3}}{2} - \frac{{x - 1}}{4} =  - 1$

Пр.3)   Одредити $a$, ако је број 2 решење једначине:

           а) $2a + x =  - 6$

           б) $3a + \frac{{x - 4}}{2} =  - 5$

Пр.4)   У једначини одредити $m$, тако да једначине буду еквивалентне:

            $3x + 10 = 7$ и $mx + 11 = 6$.

Пр.5)   Решити једначине:

           а) ${\left( {x - 2} \right)^2} - \left( {x + 3} \right)\left( {x - 3} \right) = {\left( {x + 4} \right)^2} - \left( {x + 5} \right)\left( {x - 5} \right)$

           б) $\frac{{{{\left( {x - 1} \right)}^2}}}{2} - \frac{{\left( {x - 3} \right)\left( {2x - 5} \right)}}{4} = 3 - \left( {x - 2} \right)$

Пр.1)   а) $ - 5x + 3 =  - 12$

\[\begin{gathered}
   - {\text{ }}5x{\text{ }} + {\text{ }}3{\text{ }} = {\text{ }} - {\text{ }}12 \hfill \\
   - {\text{ }}5x{\text{  = }} - {\text{ }}12 - 3 \hfill \\
   - {\text{ }}5x{\text{  = }} - {\text{ }}15 \hfill \\
  x = 3 \hfill \\
\end{gathered} \]

           б) $3\left( {x + 1} \right) =  - 9$

\[\begin{gathered}
  3\left( {x + 1} \right) =  - 9\left| {:3} \right. \hfill \\
  x + 1 =  - 3 \hfill \\
  x =  - 4 \hfill \\
\end{gathered} \]

           в) $\frac{1}{5}x + \frac{3}{5} =  - 2$

\[\begin{gathered}
  \frac{1}{5}x + \frac{3}{5} =  - 2\left| { \cdot 5} \right. \hfill \\
  x + 3 =  - 10 \hfill \\
  x =  - 13 \hfill \\
\end{gathered} \]

           г) $\frac{1}{2}x - \frac{1}{3}=1$

\[\begin{gathered}
  \frac{1}{2}x - \frac{1}{3} = 1\left| { \cdot 6} \right. \hfill \\
  3x - 2 = 6 \hfill \\
  3x = 8 \hfill \\
  x = \frac{8}{3} \hfill \\
  x = 2\frac{2}{3} \hfill \\
\end{gathered} \]

Пр.2)   а) $5\left( {x - 2} \right) =  - 12 + 4x$

\[\begin{gathered}
  5\left( {x - 2} \right) =  - 12 + 4x \hfill \\
  5x - 10 =  - 12 + 4x \hfill \\
  5x - 4x =  - 12 + 10 \hfill \\
  x =  - 2 \hfill \\
\end{gathered} \]

           б) $\frac{{x - 3}}{4} + \frac{x}{2} = \frac{1}{3}$

\[\begin{gathered}
  \frac{{x - 3}}{4} + \frac{x}{2} = \frac{1}{3}\left| { \cdot 12} \right. \hfill \\
  3\left( {x - 3} \right) + 6x = 4 \hfill \\
  3x - 9 + 6x = 4 \hfill \\
  9x = 13 \hfill \\
  x = \frac{{13}}{9} \hfill \\
  x = 1\frac{4}{9} \hfill \\
\end{gathered} \]

           в) $\frac{{x + 3}}{2} - \frac{{x - 1}}{4} =  - 1$

\[\begin{gathered}
  \frac{{x + 3}}{2} - \frac{{x - 1}}{4} =  - 1\left| { \cdot 4} \right. \hfill \\
  2\left( {x + 3} \right) - \left( {x - 1} \right) =  - 4 \hfill \\
  2x + 6 - x + 1 =  - 4 \hfill \\
  x =  - 11 \hfill \\
\end{gathered} \]

Пр.3)  а) $2a + x =  - 6$

\[\begin{gathered}
  2a + x =  - 6 \hfill \\
  2a + 2 =  - 6 \hfill \\
  2a =  - 8 \hfill \\
  a =  - 4 \hfill \\
\end{gathered} \]

           б) $3a + \frac{{x - 4}}{2} =  - 5$

\[\begin{gathered}
  3a + \frac{{x - 4}}{2} =  - 5 \hfill \\
  3a + \frac{{2 - 4}}{2} =  - 5 \hfill \\
  3a + \frac{{ - 2}}{2} =  - 5 \hfill \\
  3a - 1 =  - 5 \hfill \\
  3a =  - 4 \hfill \\
  a =  - \frac{4}{3} \hfill \\
\end{gathered} \]

Пр.4)   У једначини одредити $m$, тако да једначине буду еквивалентне:

            $3x + 10 = 7$ и $mx + 11 = 6$.

\[\begin{gathered}
  3x + 10 = 7 \hfill \\
  3x = 7 - 10 \hfill \\
  3x =  - 3 \hfill \\
  x =  - 1 \hfill \\
\end{gathered} \]

\[\begin{gathered}
  mx + 11 = 6 \hfill \\
  m \cdot \left( { - 1} \right) + 11 = 6 \hfill \\
   - m = 6 - 11 \hfill \\
  m = 5 \hfill \\
\end{gathered} \]

Пр.5)   а) ${\left( {x - 2} \right)^2} - \left( {x + 3} \right)\left( {x - 3} \right) = {\left( {x + 4} \right)^2} - \left( {x + 5} \right)\left( {x - 5} \right)$

\[\begin{gathered}
  {\left( {x - 2} \right)^2} - \left( {x + 3} \right)\left( {x - 3} \right) = {\left( {x + 4} \right)^2} - \left( {x + 5} \right)\left( {x - 5} \right) \hfill \\
  {x^2} - 4x + 4 - \left( {{x^2} - {3^2}} \right) = {x^2} + 8x + 16 - \left( {{x^2} - {5^2}} \right) \hfill \\
  {x^2} - 4x + 4 - {x^2} + 9 = {x^2} + 8x + 16 - {x^2} + 25 \hfill \\
   - 4x + 13 = 8x + 41 \hfill \\
   - 4x - 8x = 41 - 13 \hfill \\
   - 12x = 28 \hfill \\
  x = \frac{{28}}{{ - 12}} \hfill \\
  x = \frac{7}{{ - 3}} \hfill \\
  x =  - 2\frac{1}{3} \hfill \\
\end{gathered} \]

           б) $\frac{{{{\left( {x - 1} \right)}^2}}}{2} - \frac{{\left( {x - 3} \right)\left( {2x - 5} \right)}}{4} = 3 - \left( {x - 2} \right)$

\[\begin{gathered}
  \frac{{{x^2} - 2x + 1}}{2} - \frac{{2{x^2} - 5x - 6x + 15}}{4} = 3 - x + 2 \hfill \\
  \frac{{{x^2} - 2x + 1}}{2} - \frac{{2{x^2} - 11x + 15}}{4} = 5 - x\left| { \cdot 4} \right. \hfill \\
  2\left( {{x^2} - 2x + 1} \right) - \left( {2{x^2} - 11x + 15} \right) = 20 - 4x \hfill \\
  2{x^2} - 4x + 2 - 2{x^2} + 11x - 15 = 20 - 4x \hfill \\
  11x = 33 \hfill \\
  x = 3 \hfill \\
\end{gathered} \]