Question

two families are taking a vacation together. they each leave at 9 a.m. on saturday morning, driving their cars along the same route to a cabin on a lake. - the jacksons live 40 miles closer to the cabin than the milligans. - the jacksons drive at an average speed of 50 miles per hour, and the milligans drive at an average speed of 60 miles per hour. an equation can be written to represent this situation for each family relating the time, x, they drive and the distance, y, they are from the cabin. decide if each statement is true or false. 1. the equation for the jacksons is ( y = 40x + 50 ) and the equation for the chases is ( y = x + 60 ). true false 2. the cars will reach the same point in 4 hours. true false 3. if the cabin is 400 miles away, it will take the jacksons 7 hours and 12 minutes to get there. true false 4. the two cars will reach the same point at noon. true false

Explanation:

1. The equation for the Jacksons is \( y = 40x + 50 \) and the equation for the Chases is \( y = x + 60 \).

Step1: Analyze Jackson's equation

The Jacksons live 40 miles closer (so initial distance from cabin is less, but wait, actually, let's re - evaluate. Wait, the problem says "The Jacksons live 40 miles closer to the cabin than the Milligans" (assuming Chases/Milligans? Maybe a typo, but let's go with speed: Jacksons' speed is 50 mph. Let \( x \) be time, \( y \) be distance from cabin. So distance from cabin for Jacksons: if they start \( d \) miles from cabin, \( y=d - 50x \) (since they are moving towards cabin). But the given equation is \( y = 40x+50 \), which doesn't match the speed (50 mph) and the "40 miles closer" context. For Milligans (Chases?) speed is 60 mph. The given equation for Chases is \( y=x + 60 \), which is incorrect as speed should be 60, so equation should be related to speed (distance from cabin \( y = D-60x \), where \( D \) is initial distance). So the given equations are wrong.

Step2: Conclusion

The statement is False.

Step1: Set up distance equations

Let the Milligans' (Chases') initial distance from cabin be \( D \), then Jacksons' initial distance is \( D - 40 \). Jacksons' speed \( v_J=50 \) mph, Milligans' speed \( v_M = 60 \) mph. Distance from cabin for Jacksons: \( y_J=(D - 40)-50x \). Distance from cabin for Milligans: \( y_M=D - 60x \). Set \( y_J=y_M \): \( (D - 40)-50x=D - 60x \).

Step2: Solve for x

Simplify the equation: \( D-40 - 50x=D - 60x \). Subtract \( D \) from both sides: \( - 40-50x=-60x \). Add \( 60x \) to both sides: \( - 40 + 10x = 0 \). Add 40 to both sides: \( 10x=40 \), so \( x = 4 \) hours.

Step1: Calculate time for Jacksons

Jacksons' speed is 50 mph. Wait, no: Wait, the Jacksons live 40 miles closer. If cabin is 400 miles from Milligans, then Jacksons' distance is \( 400 - 40=360 \) miles. Time \( t=\frac{\text{distance}}{\text{speed}}=\frac{360}{50}=7.2 \) hours. \( 0.2 \) hours \(=0.2\times60 = 12 \) minutes. So time is 7 hours and 12 minutes.

Step2: Conclusion

The statement is True.

Answer:

False

2. The cars will reach the same point in 4 hours.