Algebra - Applications of Linear Equations

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Section 2.3 : Applications of Linear Equations

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3. Two planes start out 2800 km apart and move towards each other meeting after 3.5 hours. One plane flies at 75 km/hour slower than the other plane. What was the speed of each plane?

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Let’s start with a diagram of what is going on in this situation.

Show Step 2

We can next set up a word equation for this situation.

\[\left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Plane A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Plane B}}\end{array} \right) = 2800\]

We know that Distance = Rate X Time so this gives to following word equation.

\[\left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{Plane A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{Plane A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{Plane B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{Plane B}}\end{array} \right) = 2800\] Show Step 3

Let’s let \(r\) be the speed of the faster plane. Therefore, the speed of the slower plane is \(r - 75\). We also know that each plane travels for 3.5 hours. Plugging all this information into the word equation above gives the following equation.

\[\begin{align*}\left( r \right)\left( {3.5} \right) + \left( {r - 75} \right)\left( {3.5} \right) & = 2800\\ 3.5r + 3.5\left( {r - 75} \right) & = 2800\end{align*}\] Show Step 4

Now we can solve this equation for the speed of the faster plane.

\[\begin{align*}3.5r + 3.5\left( {r - 75} \right) & = 2800\\ 7r - 262.5 & = 2800\\ 7r & = 3062.5\\ r & = 437.5\end{align*}\]

So, the faster plane is traveling at 437.5 km/hour while the slower plane is traveling at 362.5 km/hour (75 km/hour slower than faster plane).