# Solving systems of linear equations by elimination (word problem)

• September 19th 2008, 12:51 AM
iamanoobatmath
Solving systems of linear equations by elimination (word problem)
Quote:

Fred flew from Point A to Point B at an average speed of 500km/h, and from Point B to Point C at an average speed of 600km/h. The total time in the air was 3h 24 min, and the total distance flown was 1920km. What is the distance between Point B and the other two points?
Help! I don't even know how to start this question.(Crying)
• September 19th 2008, 01:52 AM
mr fantastic
Quote:

Originally Posted by iamanoobatmath
Help! I don't even know how to start this question.(Crying)

3hr 24 min = $3 \frac{2}{5}$ hr = $\frac{17}{5}$ hours.

Let dAB = x and dBC = y.

$x + y = 1920$ .... (1)

Using $\text{time} \, = \frac{\text{distance}}{\text{speed}}$:

$\frac{17}{5} = \frac{x+y}{600}$ .... (2)

Solve equations (1) and (2) simultaneously.

Edit: Equation (2) is wrong. It should be $\frac{17}{5} = \frac{x}{500} + \frac{y}{600}$ .... (2)

Quote:

Fred flew from Point A to Point B at an average speed of 500km/h, and from Point B to Point C at an average speed of 600km/h. The total time in the air was 3h 24 min, and the total distance flown was 1920km. What is the distance between Point B and the other two points?
• September 19th 2008, 07:18 AM
iamanoobatmath
Thanks for the help. Maybe next time I shouldn't do math at 1 in the morning(Sleepy)

I managed to solve it I think
$\frac{x}{600}+\frac{7}{600}=\frac{17}{5}$
$\frac{x}{600}=\frac{-7}{600}+\frac{17}{5}$
$\frac{x}{600}=\frac{2033}{600}$
$x=2033$
$y=1920-2033$
$y=-133$
$2033,-113$
• September 19th 2008, 07:36 AM
TKHunny
Quote:

Originally Posted by iamanoobatmath
Help! I don't even know how to start this question.(Crying)

There seems to be some confusion, here. You're seriously going to settle for y = -133?! This is a real world type problem. Aircraft don't fly negative mileage. This should have alerted you that something seriously is wrong.

General Stuff

Distance = Rate * Time
Time = Distance / Rate

First Leg

Distance = d1
Rate = 500 km/h
Time = t1
d1 = (500 km/h)*t1

Second Leg

Distance = d2
Rate = 600 km/h
Time = t2
d2 = (600 km/h)*t2

Additional Information Given in the Problem Statement

"The total time in the air was 3h 24 min"

t1 + t2 = 3h 24min = 3.4h

"total distance flown was 1920km"

d1 + d2 = 1920 km

Solve - Solve - Solve

We have all this

d1 + d2 = 1920 km
d1 = (500 km/h)*t1
d2 = (600 km/h)*t2

Substituting gives:

(500 km/h)*t1 + (600 km/h)*t2 = 1920 km

We have this:

t1 + t2 = 3.4h

Can you finish? Please don't ever again think you are done when you get entirely silly results. Ever!! Work systematically and deliberately. Let the notation help you. Write CLEAR definitions.
• September 19th 2008, 08:44 AM
iamanoobatmath
$100t2=220$
$220/100=2.2$
$t2=2.2$
$t1=1.2$
$A to B=600$
$B to C=1320$
• September 19th 2008, 09:37 AM
TKHunny
Quote:

Originally Posted by iamanoobatmath
$100t2=220$
$220/100=2.2$
$t2=2.2$
$t1=1.2$
$A to B=600$
$B to C=1320$

I didn't check your arithmetic, but I do have a comment I think is needed.

[High Horse]

Without ANY discussion, you pop up with "100*t2 = 220." Where did that come from? What motivated you to write it?

This is what I am talking about when I say write clearly and let the notation help you, be deliberate and systematic. Take another look at my example, above. Did I do ANYTHING without explaining it first? This is not just for your benefit or because I think I might be teaching. It is THE way to do it - complete, thorough, deliberate, organized. Pulling stuff out of the air just is not the way to go.

[Getting Off High Horse]

Anyway, good work sticking with it. This attitude WILL get you through.
• September 19th 2008, 10:15 AM
iamanoobatmath
We have $t1+t2=3.4$
Multiply by the constant of 500 since the coefficients for t1 are different.
$(500t1+500t2)=1,700$
$(500t1+600t2=1920)-(500t1+500t2=1,700)$
500t1 cancels out, and $(600t2=1920)-(500t2=1700)=(100t2=220)$
So that means $100*t2=220$.
$220/100=2.2$
$t2=2.2$
Then you plug 2.2 into the equation to solve for t1 which in turn allows you to find the distances between the points.

• September 19th 2008, 01:26 PM
mr fantastic
Quote:

Originally Posted by TKHunny
There seems to be some confusion, here. You're seriously going to settle for y = -133?! This is a real world type problem. Aircraft don't fly negative mileage. This should have alerted you that something seriously is wrong.

General Stuff

Distance = Rate * Time
Time = Distance / Rate

First Leg

Distance = d1
Rate = 500 km/h
Time = t1
d1 = (500 km/h)*t1

Second Leg

Distance = d2
Rate = 600 km/h
Time = t2
d2 = (600 km/h)*t2

Additional Information Given in the Problem Statement

"The total time in the air was 3h 24 min"

t1 + t2 = 3h 24min = 3.4h

"total distance flown was 1920km"

d1 + d2 = 1920 km

Solve - Solve - Solve

We have all this

d1 + d2 = 1920 km
d1 = (500 km/h)*t1
d2 = (600 km/h)*t2

Substituting gives:

(500 km/h)*t1 + (600 km/h)*t2 = 1920 km

We have this:

t1 + t2 = 3.4h

Can you finish? Please don't ever again think you are done when you get entirely silly results. Ever!! Work systematically and deliberately. Let the notation help you. Write CLEAR definitions.

Thanks for the catch. I overlooked the 500 km/hr for the first leg.

To the OP: I accidently took v = 600 km/hr for the entire journey from A to C. (Nevertheless, my reply planted the seeds for what had to be done). I'll add an edit.