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Sagot :
Okay, here we have this:
[tex]\begin{gathered} 1/4\mleft(x+2\mright)-1/3\mleft(x-1\mright)=1/2\mleft(x+1\mright) \\ \frac{x+2}{4}-\frac{x-1}{3}=\frac{1}{2}x+\frac{1}{2} \\ \frac{\left(x+2\right)\cdot\:3}{12}-\frac{\left(x-1\right)\cdot\:4}{12}=\frac{1}{2}x+\frac{1}{2} \\ \frac{\left(x+2\right)\cdot\:3-\left(x-1\right)\cdot\:4}{12}=\frac{1}{2}x+\frac{1}{2} \\ \frac{-x+10}{12}=\frac{1}{2}x+\frac{1}{2} \\ -\frac{x}{12}+\frac{10}{12}=\frac{1}{2}x+\frac{1}{2} \\ -\frac{x}{12}+\frac{5}{6}=\frac{1}{2}x+\frac{1}{2} \\ -\frac{x}{12}=-\frac{1}{3}+\frac{1}{2}x \\ -\frac{7x}{12}=-\frac{1}{3} \\ -7x=-4 \\ x=\frac{4}{7} \end{gathered}[/tex]Now, let's check the solution:
[tex]\begin{gathered} 1/4\mleft(x+2\mright)-1/3\mleft(x-1\mright)=1/2\mleft(x+1\mright) \\ 1/4\mleft(\frac{4}{7}+2\mright)-1/3\mleft(\frac{4}{7}-1\mright)=1/2\mleft(\frac{4}{7}+1\mright) \\ \frac{1}{4}(\frac{18}{7})-\frac{1}{3}(-\frac{3}{7})=\frac{1}{2}(\frac{11}{7}) \\ \frac{9}{14}+\frac{1}{7}=\frac{11}{14} \\ \frac{11}{14}=\frac{11}{14} \end{gathered}[/tex]Since the sides are equal, the result (x=4/7) is true.
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