Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
2. Solving Systems Using Substitution
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Exercise 43 Page 377

Practice makes perfect
a We can use the given information to create a system of equations.
  • Let t represent the number of seconds since Michelle started running.
  • Let d represent the distance they have run.
To calculate distance traveled, we can use the formula for rate of motion. d=r * t Here d represent the distance, r represents the rate at which something travels, and t represents the time traveled. We have that Michelle runs at a speed of r=7.5 m/s. Using that information and the formula, we can find the distance she will have run after t seconds. d=7.5 t Pam starts running 1 second after Michelle. Therefore, Pam runs 1 second less than Michelle. The number of seconds she is running is ( t-1). Again, we use the rate formula to find the distance — here using that the rate is 7.8 m/s. d=7.8( t-1) By combining the equations we get a system of equations. d=7.5 t & (I) d=7.8( t-1) & (II) To determine the time at which Pam catches up to Michelle we must solve for t. To do this we will substitute the first equation into the second.
d=7.5t & (I) d=7.8(t-1) & (II)
d=7.5t 7.5t=7.8(t-1)
â–Ľ
(II): Solve for t
d=7.5t 7.5t=7.8t-7.8
d=7.5t - 0.3t=- 7.8
d=7.5t t=26
After t= 26 seconds Pam will catch up to Michelle.
b Pam overtakes Michelle after 26 seconds. We want to know if Michelle then has crossed the finishing line. In other words, is the distance covered by Michelle in 26 seconds less than 200 meters? We can represent this with an inequality.

7.5* 26? <200 ⇒ 195 <200 Since Michelle has run only 195 meters in 26 seconds, she has not finished the race yet. Therefore, Pam will overtake Michelle before the finishing line.