Solution 2.2:2a

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If we divide up the denominators that appear in the equation into small integer factors
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If we divide up the denominators that appear in the equation into small integer factors <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> and 2, we see that the lowest common denominator is <math>2\cdot 3\cdot 3=18</math>. Thus, we multiply both sides of the equation by <math>2\cdot 3\cdot 3</math> in order to avoid having denominators in the equation
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<math>6=2\centerdot 3</math>,
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<math>9=3\centerdot 3</math>
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and 2, we see that the lowest common denominator is
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<math>2\centerdot 3\centerdot 3=18</math>. Thus, we multiply both sides of the equation by
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<math>2\centerdot 3\centerdot 3</math>
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in order to avoid having denominators in the equation:
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{{Displayed math||<math>\begin{align}
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& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt]
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&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\
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\end{align}</math>}}
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<math>\begin{align}
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We can rewrite the left-hand side as <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, so that we get the equation
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& 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\
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& \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\
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\end{align}</math>
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{{Displayed math||<math>13x-4=9\,\textrm{.}</math>}}
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We can rewrite the left-hand side as
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We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:
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<math>3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4</math>, so that we get the equation
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<ol>
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<li>Add 4 to both sides, <math>\vphantom{x_2}13x-4+4=9+4\,,</math> which gives <math>\ 13x=13\,\textrm{.}</math></li>
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<li>Divide both sides by 13, <math>\frac{13x}{13}=\frac{13}{13}\,,</math> which gives the answer <math>\ x=1\,\textrm{.}</math></li>
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</ol>
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<math>13x-4=9</math>
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The equation has <math>x=1</math> as the solution.
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When we have obtained an answer, it is important to go back to the original equation to check that <math>x=1</math> really is the correct answer (i.e. that we haven't calculated incorrectly)
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We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get
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{{Displayed math||<math>\begin{align}
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<math>x</math>
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\text{LHS}
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by itself on one side:
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&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]
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&= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}
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1. Add 4 to both sides:
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= \text{RHS}\,\textrm{.}
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\end{align}</math>}}
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<math>13x-+4=9+4</math>
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which gives
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<math>13x=13</math>
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2. Divide both sides by 13:
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<math>\frac{13x}{13}=\frac{13}{13}</math>
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which gives the answer
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<math>x=1</math>.
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The equation has
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<math>1</math>
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as the solution.
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When we have obtained an answer, it is important to go back to the original equation to check that
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<math>x=1</math>
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really is the correct answer( i.e. that we haven't calculated incorrectly):
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LHS =
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<math>\frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}</math>
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= RHS
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Current revision

If we divide up the denominators that appear in the equation into small integer factors \displaystyle 6=2\cdot 3, \displaystyle 9=3\cdot 3 and 2, we see that the lowest common denominator is \displaystyle 2\cdot 3\cdot 3=18. Thus, we multiply both sides of the equation by \displaystyle 2\cdot 3\cdot 3 in order to avoid having denominators in the equation

\displaystyle \begin{align}

& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt] &\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\ \end{align}

We can rewrite the left-hand side as \displaystyle 3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4, so that we get the equation

\displaystyle 13x-4=9\,\textrm{.}

We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get x by itself on one side:

  1. Add 4 to both sides, \displaystyle \vphantom{x_2}13x-4+4=9+4\,, which gives \displaystyle \ 13x=13\,\textrm{.}
  2. Divide both sides by 13, \displaystyle \frac{13x}{13}=\frac{13}{13}\,, which gives the answer \displaystyle \ x=1\,\textrm{.}

The equation has \displaystyle x=1 as the solution.

When we have obtained an answer, it is important to go back to the original equation to check that \displaystyle x=1 really is the correct answer (i.e. that we haven't calculated incorrectly)

\displaystyle \begin{align}

\text{LHS} &= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt] &= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2} = \text{RHS}\,\textrm{.} \end{align}

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