A few days ago
HollyW

Word Problem help in Algebra I?

Ok I don’t understand how to do any of these word problems in Algebra, please help me.

A bus traveling at an average rate of 30 mph left the city at 11:45 a.m. A car following the bus at 45 mph left the city at noon. At what time did the car catch up with the bus.

I don’t get how do solve it. or this on either.

Ellen and Kate raced on their bicycles to the library after school. They both left school at 3:00 p.m. and bicylced along the same path. Ellen rode at a speed of 12 pmh and Kate rode at 9 mph. Ellen got to the library 15 minutes before Kate.

a. How long did it take Ellen to get to the library?

b. At what time did Ellen get to the library?

Thank you for helping me understand this better, it will really help me on my test coming up Thursday!!:)

Top 2 Answers
A few days ago
LouLou

Favorite Answer

For the first one, let’s use x (in hours) to represent the total time travelled by the bus when the car catches up. So the overall distance the bus travelled would be 30*x. The car traveled the same distance in (x – 0.25) hours – – the car started travelling 15 minutes after the bus.

So:

30 * x = 45 * (x – 0.25)

Simplify both sides

30x = 45x – 11.25

Subtract 45x from both sides

-15x = – 11.25

Multiply both sides by -1

15x = 11.25

Divide by 15

x = 0.75

So the bus travelled .75 hours (or 45 minutes) when the car caught up. So the car caught up with the bus at 12:30pm.

For the second problem, we’ll have x (in hours) = Ellen’s time to get to the library. Therefore, we know the length of the trip was 12 * x. Kate travelled the same distance in (x + 0.25) hours.

So:

12 * x = 9 * (x + 0.25)

Simplify

12x = 9x + 2.25

Subtract 9x from both sides

3x = 2.25

Divide both sides by 3

x = 0.75

So it took Ellen 0.75 Hours or 45 minutes to get to the library. Which means she got there at 3:45pm, while Kate arrived at 4pm.

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4 years ago
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Hi there. Call the period “L” and the width “W”. We’re advised that the period of the subject is 2 yards greater than its width. So, if we upload 2 yards onto the width, we get the period, i.e. L=W+two. Then we are requested a query concerning the field of the subject, so permit’s discover an expression for the field. The field of a rectangle is its width increased by way of its period. i.e. AREA=L*W. But we’ve got already observed that L=W+two, so the AREA=L*W turns into AREA=(W+two)W (all we’ve got performed here’s exchange L with W+two). So, we’ve got observed an expression that offers us the field in phrases of simply W, the width of the subject. We’re advised that the field of the subject is 120yd2. Let’s write this as an equation: AREA = one hundred twenty. Now permit’s exchange AREA with the expression we’ve got simply observed: (W+two)W=one hundred twenty. Now, permit’s multiply out the brackets in (W+two)W. This offers W*W + two*W = W^two + 2W. So the prior expression turns into: W^two + 2W = one hundred twenty. Subtract one hundred twenty from either side to get: W^two + 2W -one hundred twenty = zero. Now we must factorise this to clear up this. To factorise an equation like this that entails a W^two, we wish to discover 2 numbers that multiply in combination to make -one hundred twenty and upload in combination to make two. After a little bit of inspiration, you’ll be able to confidently be capable to look that -10 and 12 are compatible the invoice. Therefore we will be able to rewrite the prior expression as (W+12)(W-10)=zero So, keep in mind while this expression is identical to 0. We’ve received 2 matters, (W+12) and (W-10), which might be 0 while increased in combination make 0. This most effective occurs while probably the most brackets is 0. This occurs while W+12=zero (so W=-12 right here) or W-10=zero (W=10 right here). So those are the widths we are watching for. Can a subject have a terrible period? Does that make feel to you? Hopefully it does not, considering that it can not! If I took you to a subject and requested you to degree the width of it, I’m definite you would not inform me it was once minus twenty yards or some thing! So the reply of “W=-12” that we observed simply does not make feel. There’s no situation with our different reply of W=10 even though. After all, a subject may also be 10 yards extensive, there is not any situation there! The equations you will have received on the backside (from “PROOF” downwards) are simply there to illustrate that your reply truthfully works. They’re the stairs that you simply would pass by way of after fixing a situation to verify that your reply suits the stipulations that the query requested for. So, we’ve got observed that W=10 after which going again to the fashioned query, the period was once two greater than the width. So L=10+two = 12 yards. Therefore the field of the subject is width x period = 10 * 12 = one hundred twenty, which we had been advised it was once! So your reply is smart! Hurrah! Does that aid? Hope so, if now not, ask approximately any detailed bits that do not make feel πŸ™‚
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