FA17, HA21 - Forming equations from problems

An equation from a context

An equation states that two expressions represent the same amount. Define the unknown first, then use an equality such as “sum of lengths = total length” to connect the expressions.

Worked example 1: an additive relationship

Question. Two ribbons have total length $31\,\mathrm{cm}$. The longer ribbon is $5\,\mathrm{cm}$ longer than the shorter. Let $x$ be the shorter length in centimetres. Form an equation for $x$.

Plan. Write both lengths in terms of $x$, then add them to make the total.

The shorter length is $x$; the longer is $x+5$. Therefore

Answer. $2x+5=31$, where $x>0$.

Check. Solving gives $x=13$. Lengths $13$ and $18\,\mathrm{cm}$ sum to $31\,\mathrm{cm}$ and differ by $5\,\mathrm{cm}$, so the equation represents both facts.

Worked example 2: a multiplicative relationship

Question. Two ribbons have total length $32\,\mathrm{cm}$. The longer ribbon is three times as long as the shorter. Let $x$ be the shorter length in centimetres. Form an equation.

Plan. Use $3x$ for the longer ribbon, then add the lengths.

Answer. $4x=32$, where $x>0$.

Check. $x=8$ gives lengths $8$ and $24\,\mathrm{cm}$. Their total is $32\,\mathrm{cm}$, and $24=3\times8$.

Watch out

If the longer ribbon is $5\,\mathrm{cm}$ longer, writing $x+5=31$ makes the longer ribbon alone equal to the total. Include both ribbons: $x+(x+5)=31$.

Worked example 3: a permitted solution

Question. The enclosure uses exactly $54\,\mathrm m$ of fencing. Find its dimensions, with length $3\,\mathrm m$ more than twice its width and both dimensions whole numbers of metres.

Plan. Form the perimeter equation, solve for the width, then find and check the length.

Answer. Width $8\,\mathrm m$, length $19\,\mathrm m$.

Check. $8+19+8+19=54\,\mathrm m$; $19=2\times8+3$. Both dimensions are positive whole numbers, so the design is permitted.