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.
 
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