Solution 2.2:2b

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First, we multiply both sides in the equation by
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First, we multiply both sides in the equation by <math>4\cdot 7=28</math>, so that we get rid of the denominators in the equation,
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<math>4\centerdot 7=28</math>, so that we get rid of the denominators in the equation,
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{{Displayed math||<math>\begin{align}
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& 4\cdot{}\rlap{/}7\cdot\frac{8x+3}{\rlap{/}7} - \rlap{/}4\cdot 7\cdot\frac{5x-7}{\rlap{/}4} = 4\cdot 7\cdot 2\\[5pt]
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&\qquad\Leftrightarrow\quad 4\cdot (8x+3) - 7\cdot (5x-7) = 56\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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We can simplify the left-hand side to <math>4\cdot (8x+3) - 7\cdot (5x-7) = 32x+12-35x+49 = -3x+61\,</math>. Hence, the equation is
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& 4\centerdot 7\centerdot \frac{8x+3}{7}-4\centerdot 7\centerdot \frac{5x-7}{4}=4\centerdot 7\centerdot 2 \\
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& \Leftrightarrow 4\centerdot \left( 8x+3 \right)-7\centerdot \left( 5x-7 \right)=56 \\
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\end{align}</math>
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{{Displayed math||<math>-3x+61=56\,\textrm{.}</math>}}
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We can simplify the left-hand side to ,
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We solve this equation by subtracting 61 from both sides and then dividing by -3,
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{{Displayed math||<math>\begin{align}
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-3x+61-61&=56-61\,,\\[5pt]
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-3x&=-5\,,\\[5pt]
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\frac{-3x}{-3}&=\frac{-5}{-3}\,,\\[5pt]
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x&=\frac{5}{3}\,\textrm{.}
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\end{align}</math>}}
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<math>4\centerdot \left( 8x+3 \right)-7\centerdot \left( 5x-7 \right)=32x+12-35x+49=-3x+61</math>
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The answer is <math>x={5}/{3}\,</math>.
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As the final part of the solution, check the answer by substituting <math>x={5}/{3}</math> into the original equation
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Hence, the equation is
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{{Displayed math||<math>\begin{align}
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\text{LHS}
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&= \frac{8\cdot\frac{5}{3}+3}{7}-\frac{5\cdot\frac{5}{3}-7}{4}
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<math>-3x+61=56</math>
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= \frac{\bigl(8\cdot\frac{5}{3}+3\bigr)\cdot 3}{7\cdot 3} - \frac{\bigl( 5\cdot \frac{5}{3}-7\bigr)\cdot 3}{4\cdot 3}\\[5pt]
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&= \frac{8\cdot 5+3\cdot 3}{7\cdot 3}-\frac{5\cdot 5-7\cdot 3}{4\cdot 3}
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= \frac{40+9}{21}-\frac{25-21}{12}\\[5pt]
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We solve this equation by subtracting
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&= \frac{49}{21}-\frac{4}{12}
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<math>61</math>
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= \frac{7\cdot 7}{3\cdot 7} - \frac{2\cdot 2}{2\cdot 2\cdot 3}
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from both sides and then dividing by
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= \frac{7}{3}-\frac{1}{3}
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<math>-3</math>,
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= \frac{7-1}{3}
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= \frac{6}{3} = 2 = \text{RHS.}
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\end{align}</math>}}
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<math>\begin{align}
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& -3x+61-61=56-61 \\
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& -3x=-5 \\
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& \frac{-3x}{-3}=\frac{-5}{-3} \\
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& x=\frac{5}{3} \\
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\end{align}</math>
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The answer is
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<math>x={5}/{3}\;</math>.
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As the final part of the solution, check the answer by substituting
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<math>x={5}/{3}\;</math>
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into the original equation
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<math>\begin{align}
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& \text{LHS}\quad =\quad \frac{8\centerdot \frac{5}{3}+3}{7}-\frac{5\centerdot \frac{5}{3}-7}{4}=\frac{\left( 8\centerdot \frac{5}{3}+3 \right)\centerdot 3}{7\centerdot 3}-\frac{\left( 5\centerdot \frac{5}{3}-7 \right)\centerdot 3}{4\centerdot 3} \\
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& \\
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& =\frac{8\centerdot 5+3\centerdot 3}{7\centerdot 3}-\frac{5\centerdot 5-7\centerdot 3}{4\centerdot 3}=\frac{40+9}{21}-\frac{25-21}{12} \\
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& \\
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& =\frac{49}{21}-\frac{4}{12}=\frac{7\centerdot 7}{3\centerdot 7}-\frac{2\centerdot 2}{2\centerdot 2\centerdot 2}=\frac{7}{3}-\frac{1}{3}=\frac{7-1}{3} \\
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& \\
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& =\frac{6}{3}=2\quad =\quad \text{RHS} \\
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\end{align}</math>
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Current revision

First, we multiply both sides in the equation by \displaystyle 4\cdot 7=28, so that we get rid of the denominators in the equation,

\displaystyle \begin{align}

& 4\cdot{}\rlap{/}7\cdot\frac{8x+3}{\rlap{/}7} - \rlap{/}4\cdot 7\cdot\frac{5x-7}{\rlap{/}4} = 4\cdot 7\cdot 2\\[5pt] &\qquad\Leftrightarrow\quad 4\cdot (8x+3) - 7\cdot (5x-7) = 56\,\textrm{.} \end{align}

We can simplify the left-hand side to \displaystyle 4\cdot (8x+3) - 7\cdot (5x-7) = 32x+12-35x+49 = -3x+61\,. Hence, the equation is

\displaystyle -3x+61=56\,\textrm{.}

We solve this equation by subtracting 61 from both sides and then dividing by -3,

\displaystyle \begin{align}

-3x+61-61&=56-61\,,\\[5pt] -3x&=-5\,,\\[5pt] \frac{-3x}{-3}&=\frac{-5}{-3}\,,\\[5pt] x&=\frac{5}{3}\,\textrm{.} \end{align}

The answer is \displaystyle x={5}/{3}\,.

As the final part of the solution, check the answer by substituting \displaystyle x={5}/{3} into the original equation

\displaystyle \begin{align}

\text{LHS} &= \frac{8\cdot\frac{5}{3}+3}{7}-\frac{5\cdot\frac{5}{3}-7}{4} = \frac{\bigl(8\cdot\frac{5}{3}+3\bigr)\cdot 3}{7\cdot 3} - \frac{\bigl( 5\cdot \frac{5}{3}-7\bigr)\cdot 3}{4\cdot 3}\\[5pt] &= \frac{8\cdot 5+3\cdot 3}{7\cdot 3}-\frac{5\cdot 5-7\cdot 3}{4\cdot 3} = \frac{40+9}{21}-\frac{25-21}{12}\\[5pt] &= \frac{49}{21}-\frac{4}{12} = \frac{7\cdot 7}{3\cdot 7} - \frac{2\cdot 2}{2\cdot 2\cdot 3} = \frac{7}{3}-\frac{1}{3} = \frac{7-1}{3} = \frac{6}{3} = 2 = \text{RHS.} \end{align}

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