Q. 359 - Intermediate Algebra

Question

In the following exercises,translate to a system of equations and solve.

Bob left home,rising his bike at a rate of 10 miles per hour to lake.cheryl ,his wife,left 45 minutes(3/4)hours Later,driving her car at a rate of 25 miles per hour.How long will it take cheryl  to catch up to Bob?

Step-by-Step Solution

Answer

$\frac{1}{2}$an hour to take for cheryl to catch up to Bob

Step 1.Given information

Let x represents the length of time of the Bob.

Let y represents the length of time of the Cheryl.

The following chart will help us organize the data.

Name	Rate	Time	Distance D=r.t
Bob	10	x	10x
Cheryl	25	y	25y

To,make the system of equations,you must recognize that sarah and her sister will drive the same distance.

So,$10x=25y$

Since Cheryl,his wife left her house after 45 minutes,her time will be $\frac{3}{4}$hour less than Bob time.

So,

$$\begin{gathered} y=x-\frac{3}{4} \\ x=y+\frac{3}{4} \end{gathered}$$

Step 2.Find the time to take for Cheryl to catch up to Bob.

Substitute $x=y+\frac{3}{4}$in the equation $10x=25y$ and solve for y.

$$\begin{gathered} 10x=25y \\ 10(y+\frac{3}{4})=25y \\ 10y+\frac{15}{2}=25y \\ 10y-25y=-\frac{15}{2} \\ -15y=-\frac{15}{2} \\ y=\frac{1}{2} \end{gathered}$$

Substitute $y=\frac{1}{2}$ into $x=y+\frac{3}{4}$ and solve for x.

$$\begin{gathered} x=y+\frac{3}{4} \\ x=\frac{1}{2}+\frac{3}{4} \\ x=\frac{2+3}{4} \\ x=\frac{5}{4} \end{gathered}$$

Step 3.Check

Substitute $x=\frac{5}{4},y=\frac{1}{2}$

Bob:- 10x=10($\frac{5}{4}$)

                =$\frac{25}{2}$

Cheryl:-25y=25($\frac{1}{2}$)

                    =$\frac{25}{2}$

yes,they will have the same distance travelled when they meet

Therefore,$\frac{1}{2}$ an hour to take for cheryl  to catch up to Bob.