Equations with algebraic fractions and rational coefficients

1. Solve $\frac{x}{3}+\frac{1}{2}=\frac{7}{6}$.

   2 marks
2. Solve $\frac{5}{x-2}=1$.

   2 marks
3. Solve $\frac{2}{x+1}=\frac{3}{x+4}$.

   2 marks
4. Solve $\frac{x-1}{x+2}=\frac{-3}{x+2}$.

   2 marks
5. A cost of 72 dollars is shared equally. Adding two people reduces each share from $\frac{72}{n}$ to $\frac{72}{n+2}$. If the original share is 12 dollars, find $n$ and the new share.

   3 marks
6. Solve $\frac{2x+4}{x+2}=2$ and state the full solution set.

   3 marks

Equivalent correct methods earn the corresponding marks. Include units where asked.

1. Solve $\frac{x}{3}+\frac{1}{2}=\frac{7}{6}$.

   2 marks

   Multiply every term by $6$: $2x+3=7$.

   $x=2$
2. Solve $\frac{5}{x-2}=1$.

   2 marks

   $x\ne 2$; multiply by $x-2$ to get $5=x-2$.

   Check $\frac{5}{7-2}=1$.

   $x=7$
3. Solve $\frac{2}{x+1}=\frac{3}{x+4}$.

   2 marks

   $x\ne -1,-4$; $2\left(x+4\right)=3\left(x+1\right)$.

   This gives $2x+8=3x+3$, hence $x=5$.

   $x=5$
4. Solve $\frac{x-1}{x+2}=\frac{-3}{x+2}$.

   2 marks

   $x\ne -2$. Clearing gives $x-1=-3$, hence $x=-2$.

   The candidate is excluded.

   No solution
5. A cost of 72 dollars is shared equally. Adding two people reduces each share from $\frac{72}{n}$ to $\frac{72}{n+2}$. If the original share is 12 dollars, find $n$ and the new share.

   3 marks

   $\frac{72}{n}=12$ gives $n=6$, a positive whole number.

   The new group has $8$ people, so the new share is $\frac{72}{8}=9$.

   $n=6$; new share $9$.
6. Solve $\frac{2x+4}{x+2}=2$ and state the full solution set.

   3 marks

   The denominator requires $x\ne -2$.

   The numerator is $2\left(x+2\right)$, so both sides are $2$ for every allowed value.

   All real $x$ except $-2$.