Answer
$$(14\cdot3-6)\frac{8+2}{5\cdot4-10}-12\frac{7+3}{20-8}+15=41$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(14\cdot3-6)\frac{8+2}{5\cdot4-10}-12\frac{7+3}{20-8}+15& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(42-6)\frac{8+2}{5\cdot4-10}-12\frac{7+3}{20-8}+15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}36 \cdot \frac{8+2}{5\cdot4-10}-12\frac{7+3}{20-8}+15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}36\cdot\frac{10}{20-10}-12\cdot\frac{10}{20-8}+15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}36\cdot\frac{10}{10}-12\cdot\frac{10}{12}+15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}26+15 \xlongequal{ } \\[1 em] & \xlongequal{ }41\end{aligned} $$ | |
| ① | $$ 14 \cdot 3 = 42 $$ |
| ② | Combine like terms: $$ \color{blue}{42} \color{blue}{-6} = \color{blue}{36} $$ |
| ③ | Simplify numerator $$ \color{blue}{8} + \color{blue}{2} = \color{blue}{10} $$Simplify numerator $$ \color{blue}{7} + \color{blue}{3} = \color{blue}{10} $$ |
| ④ | Simplify denominator $$ \color{blue}{20} \color{blue}{-10} = \color{blue}{10} $$Simplify denominator $$ \color{blue}{20} \color{blue}{-8} = \color{blue}{12} $$ |
| ⑤ | Combine like terms |
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