Algebraic fractions and restrictions

1. State the restriction for $\frac{7}{x-4}$.

   2 marks
2. State all restrictions for $\frac{x+1}{x\left(x+2\right)}$.

   2 marks
3. Simplify $\frac{12a}{18a}$, retaining its restriction.

   2 marks
4. Simplify $\frac{4\left(x-5\right)}{2\left(x-5\right)}$, retaining its restriction.

   2 marks
5. Explain why $\frac{p+6}{p}$ is not equal to $7$ for every allowed $p$. Give a counterexample.

   3 marks
6. A cost model is $C=\frac{20n+40}{n}$ dollars per person. Simplify it and state a sensible domain for $n$.

   3 marks

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

1. State the restriction for $\frac{7}{x-4}$.

   2 marks

   Solve $x-4=0$.

   $x\ne 4$
2. State all restrictions for $\frac{x+1}{x\left(x+2\right)}$.

   2 marks

   Either factor makes the denominator zero.

   $x\ne 0,-2$
3. Simplify $\frac{12a}{18a}$, retaining its restriction.

   2 marks

   Cancel the nonzero $a$ and reduce $\frac{12}{18}$.

   $\frac{2}{3}$, $a\ne 0$
4. Simplify $\frac{4\left(x-5\right)}{2\left(x-5\right)}$, retaining its restriction.

   2 marks

   Cancel $x-5$ only where it is nonzero.

   $2$, $x\ne 5$
5. Explain why $\frac{p+6}{p}$ is not equal to $7$ for every allowed $p$. Give a counterexample.

   3 marks

   The numerator is a sum, not a product with a factor $p$.

   Its equivalent form is $1+\frac{6}{p}$, with $p\ne 0$.

   For $p=2$, the fraction is $4$, not $7$.
6. A cost model is $C=\frac{20n+40}{n}$ dollars per person. Simplify it and state a sensible domain for $n$.

   3 marks

   Split the numerator into $\frac{20n}{n}+\frac{40}{n}$.

   Division excludes $n=0$; a head count is positive and integral.

   $C=20+\frac{40}{n}$; $n$ is a positive whole number.