college algerba unit 2 part 1 question 5\nmargaret drove to a business appointment at 60 mph. her\naverage…

college algerba unit 2 part 1 question 5\nmargaret drove to a business appointment at 60 mph. her\naverage speed on the return trip was 50 mph. the return trip took\n\\( \\frac { 1 } { 5 } \\) hour longer because of heavy traffic. how far did she travel to\nthe appointment?\n\\( \\begin{array} { | c | c | c | c | } \\hline & { r } & { t } & { d } \\\\ \\hline \\text { morning } & { 60 } & { x } & { } \\\\ \\hline \\text { afternoon } & { 50 } & { x + \\frac { 1 } { 5 } } & { } \\\\ \\hline \\end{array} \\)

college algerba unit 2 part 1 question 5\nmargaret drove to a business appointment at 60 mph. her\naverage speed on the return trip was 50 mph. the return trip took\n\\( \\frac { 1 } { 5 } \\) hour longer because of heavy traffic. how far did she travel to\nthe appointment?\n\\( \\begin{array} { | c | c | c | c | } \\hline & { r } & { t } & { d } \\\\ \\hline \\text { morning } & { 60 } & { x } & { } \\\\ \\hline \\text { afternoon } & { 50 } & { x + \\frac { 1 } { 5 } } & { } \\\\ \\hline \\end{array} \\)

Answer

Explanation:

Step1: Use the formula (d = rt)

For the morning trip: (d = 60x). For the afternoon trip: (d=50(x + \frac{1}{5}))

Step2: Set the two distance equations equal

Since the distance to the appointment and back is the same, (60x=50(x+\frac{1}{5}))

Step3: Expand the right - hand side

(60x = 50x+10)

Step4: Solve for (x)

Subtract (50x) from both sides: (60x - 50x=50x + 10-50x), so (10x = 10), then (x = 1)

Step5: Find the distance

Substitute (x = 1) into the morning trip distance formula (d = 60x), so (d=60\times1 = 60)

Answer:

(60)