Word Problem: "Four movie tickets in Germany plus three movie tickets in France would cost $62.27...."

sirnam505

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Feb 20, 2019
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I'm pretty frustrated as to how I would approach this problem:

Four movie tickets in Germany plus three movie tickets in France would cost $62.27. Three tickets in Germany plus four tickets in France would cost $62.19. How much would an average movie ticket cost in each of these countries?

I made x = germany and y = france but I'm not sure what other type of equation I would use to solve besides:

4x + 3y = 62.27

3x + 4y = 62.19
 

MarkFL

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Nov 24, 2012
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Hello, and welcome to FMH! :D

You're off to a fantastic start...you have two equations in two unknowns. I think what I would do is solve the first equation for \(y\), and then plug that into the second equation so that you have an equation in \(x\) only, and solve it for \(x\). Then plug that into your equation where you have \(y\) in terms of \(x\) to get the value of \(y\).

Can you proceed?
 

Jomo

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Dec 30, 2014
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I'm pretty frustrated as to how I would approach this problem:

Four movie tickets in Germany plus three movie tickets in France would cost $62.27. Three tickets in Germany plus four tickets in France would cost $62.19. How much would an average movie ticket cost in each of these countries?

I made x = germany and y = france but I'm not sure what other type of equation I would use to solve besides:

4x + 3y = 62.27

3x + 4y = 62.19
Since you do haveTWO equations and only TWO unknowns then you can try to solve this. Please use the method described by MarkFL
 

HallsofIvy

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Jan 27, 2012
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The method described by Mark is excellent but there are many ways to solve "systems of equations". For example, instead of solving the first equation for y (in terms of x) and putting that into the second equation, you could solve the second equation for y (in terms of x) and put that into the first equation. Or solve either equation for x (in terms or y) and put that into the other equation. The basic idea is to reduce from "two equations in two unknowns" to "one equation in one unknowm". What I might do is, observing that the first equation has "4x" and the second equation "3x" is multiply the first equation by 3, to get 12x+ 9y= 186.81, and multiply the second equation by 4, to get 12x+ 16y= 248.76. Since I now have "12x" in both equations, subtracting the first from the second "eliminates" x giving the single equation in y, 7y= 248.76- 186.81= 61.95 so that y= 61.95/7.
 
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