# Word Problem: "Four movie tickets in Germany plus three movie tickets in France would cost $62.27...." #### sirnam505 ##### New member 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 #### MarkFL ##### Super Moderator Staff member Hello, and welcome to FMH! 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? • Otis #### Jomo ##### Elite Member 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

• MarkFL

#### HallsofIvy

##### Elite Member
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.

• MarkFL