Question

(a) complete the story below so that it can be represented by the equation 5000 - 50x = 4800 - 30x. pool a started with liters of water, and liters per minute select pool b started with liters of water, and liters per minute select the amount of water after x minutes is select (b) solve for x. for this equation (and story): x =

Explanation:

Part (a)

To form the story, we analyze the equation \(5000 - 50x=4800 - 30x\). The left - hand side represents Pool A: the initial amount of water is \(5000\) liters, and it is losing water at a rate of \(50\) liters per minute (so we subtract \(50x\) where \(x\) is the number of minutes). The right - hand side represents Pool B: the initial amount of water is \(4800\) liters, and it is losing water at a rate of \(30\) liters per minute (so we subtract \(30x\)). The phrase "The amount of water after \(x\) minutes is" should be "equal" (because the equation is setting the two amounts equal to each other). So:

• Pool A started with \(5000\) liters of water, and \(50\) liters per minute (is being drained).
• Pool B started with \(4800\) liters of water, and \(30\) liters per minute (is being drained).
• The amount of water after \(x\) minutes is equal.

Part (b)

Step 1: Isolate the variable terms

Start with the equation \(5000 - 50x=4800 - 30x\). Add \(50x\) to both sides of the equation to get all the \(x\) terms on one side.
\(5000-50x + 50x=4800-30x + 50x\)
Simplifying, we have \(5000=4800 + 20x\).

Step 2: Isolate the constant term

Subtract \(4800\) from both sides of the equation.
\(5000 - 4800=4800+20x - 4800\)
Simplifying, \(200 = 20x\).

Step 3: Solve for \(x\)

Divide both sides of the equation \(200 = 20x\) by \(20\).
\(\frac{200}{20}=\frac{20x}{20}\)
Which gives \(x = 10\).

Answer:

Part (a)

• Pool A started with \(5000\) liters of water, and \(50\) liters per minute (drained).
• Pool B started with \(4800\) liters of water, and \(30\) liters per minute (drained).
• The amount of water after \(x\) minutes is equal.

Part (b)

\(x=\boxed{10}\)