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Centennial College Centennial College
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Posts: 4
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2 years ago
Question 1:

\(\frac{x+6}{2}+x=\frac{4x+3}{6}\)

LCD: 6, thus, multiply both sides by 6:

\(6\left(\frac{x+6}{2}+x\right)\ =6\left(\frac{4x+3}{6}\right)\)

\(\frac{6\left(x+6\right)}{2}+6x=\frac{6\left(4x+3\right)}{6}\)

Simplify:

\(3\left(x+6\right)+6x=4x+3\)

\(3x+6x-4x=3-18\)

\(5x=-15\)

\(x=-\frac{15}{5}=-3\)



Question 2:

\(\frac{x}{10}-\frac{x-6}{5}-\frac{x-8}{15}=0\)

LCD: 30, thus, multiply both sides by 30:

\(30\left(\frac{x}{10}\right)-30\left(\frac{x-6}{5}\right)-30\left(\frac{x-8}{15}\right)=0\)

\(3x-6\left(x-6\right)-2\left(x-8\right)=0\)

Expand:

\(3x-6x+36-2x+16=0\)

Gather like-terms:

\(3x-6x-2x=-36-16\)

\(-5x=-52\)

\(x=\frac{-52}{-5}=\frac{52}{5}\)



Question 3:

\(\frac{9n+12}{15}-\frac{4n-7}{5}-\frac{4}{10}=1\)

LCD: 30, thus, multiply both sides by 30:

\(30\left(\frac{9n+12}{15}\right)-30\left(\frac{4n-7}{5}\right)-30\left(\frac{4}{10}\right)=1\)

\(2\left(9n+12\right)-6\left(4n-7\right)-3\left(4\right)=1\)

\(18n+24-24n+42-12=0\)

\(18n-24n=12-42-24\)

\(-6n=-54\)

\(n=\frac{-54}{-6}=9\)
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