MHB Simultaneous equation word problem

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The mathematics department has a budget of 1440 euros to purchase textbooks, with volume 1 costing 70 euros and volume 2 costing 40 euros. The first equation representing their budget is 70x + 40y = 1440, where x is the number of volume 1 books and y is the number of volume 2 books. Given that the department wants twice as many copies of volume 1 as volume 2, the second equation is x = 2y. By substituting this second equation into the first, you can solve for y and subsequently find x. This approach allows for determining the exact number of each volume that can be purchased within the budget.
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"A school mathematics department has 1440 euros to buy textbooks.
Maths for All volume 1 costs 70 euros. Maths for All volume 2 costs 40 euros.
The department wants twice as many copies of volume 1 as volume 2.
How many of each volume can they buy?"

I got up to that the first equation is 70x + 40y= 1440, but I have no idea how to proceed after that. How do I get a constant in the second equation? I'm lost.
 
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You have chosen to let x represent the number of vol. 1 and y represents the number of vol. 2 purchased.

Your first equation is good, although I would divide through by 10 so the numbers are smaller. Now, we are also told:

"The department wants twice as many copies of volume 1 as volume 2."

This means x must be twice the value of y. How can you write this mathematically?
 
Would x = 2y be right?
 
That would be exactly right!:cool:

So, we now have:

$\displaystyle 7x+4y=144$

$\displaystyle x=2y$

Now, use the second equation, and substitute for x into the first equation to get an equation in y, which you can then solve. Once you have the value of y, then use the second equation to get x.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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