LINR 1 | Lesson 2 | Try This! (Graphs of Linear Situations Solutions)

The Situation and Equation:  Solutions

Paul is draining his swimming pool to replace the liner.  Paul’s pool contains \(15,000\) gallons of water.  He is draining it with a pump that removes \(800\) gallons per hour.  We can use \(x\) to represent the number of hours during which the water is pumped and \(y\) to represent the number of gallons of water remaining in the pool.

1.   Slope: -800 gallons per hour

2.   \(y\)-intercept: ( 0, 15,000 )

3.    \(y=-800+15,000\)


Table

2.


Graph

3.


Interpreting the Situation

4.  To find the number of hours:

Set the \(y\)-value in the equation, \(y=-800x+15,000\), equal to zero and solve for \(x\).

\[0=-800x+15,000\]\[800x=15,000\]\[\frac{800}{800}x=\frac{15,000}{800}\]\[x=18.75\]

It will take 18 hours and 45 minutes to drain the pool.


5.  The negative \(y\)-values in the table would represent a condition where there would be  negative gallons or that it drained below empty.  Therefore, negative values do not make sense in this context.


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