Help. Math worded problem?

Let:

P = pens

W = wallets

X = watches

P = W

90P + 120P + 140X = 2000

210P + 140X = 2000

Two unknown variables but only one equation => not enough information given.

Let x = # of pens, y = # of wallets, and z = # of watches

Then the total he spent is

2000 = 90x + 120y + 140z

Since he bought the same number of pens and watches, x = z, so

2000 = 90x + 120y + 140x = 230x + 120y

Now, we need to know how many watches he bought (i.e., we need to solve for x).

In this case, we would find that x = 4, y = 9 (try plotting the line

y = -23/12 x + 50/3

and look for points the graph passes through with integer coordinates, where x and y are both positive integers; also, neither x nor y can be 0, because 230 doesn't divide 2000, nor does 120).

So he bought 4 watches, and since he bought x + y + z items in total, and x = z, he bought

x + x + y = 4 + 4 + 9 = 17 items in total.

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Addendum:

The answer above me didn't use the fact that #(pens) = #(watches). And while in general, 2 unknowns in a single equation can't be solved uniquely, restricting ourselves to integer solutions can sometimes lead to a unique solution (as it does in this case).

And of course, we _must_ restrict ourselves to only integer non-negative solutions in this case. After all, it doesn't make any real-world sense to walk into a store and ask for 85/12 wallets (which is what would happen, for instance, if Mr. Smith bought 5 watches and 5 pens). In the same manner, it wouldn't make sense to try to buy -8 watches, which would lead to a fractional number of wallets.

So while, in general, of course 3 unknowns in 2 equations, or 2 unknowns in 1 equation can't be solved uniquely, restriction to integer solutions yields a unique (realistic) solution.

let p, w, and c be number of pens, wallets, and watches.

c = p

90p + 120w + 140c = 2000

90p + 120w + 140p = 2000

230p + 120w = 2000

23p + 12w = 200

p cannot be greater than 9, and it has to be even

so p is restricted to 2, 4, 6, 8

trying p = 2, then 23(2) + 12w = 200

12w = 200 - 46 = 154

w = 12.833333.. not possible

so p = 4, then 23(4) + 12w = 200

12w = 200 - 92 = 108

w = 9

this is a possible solution pens and watches = 4, wallets = 9

p = 6, then 23(6) + 12w = 200

12w = 200 - 138 = 62

w = 5.16666.. not possible

p = 8, then 23(8) + 12w = 200

12w = 200 - 184 = 16

12w = 16 not possible

Only answer that works is pens = 4 watches = 4 wallets = 9

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p could also possibly be zero, but if you check, zero does not work.

He bought 2 pens, 7 watches and 7 wallets.

90 x 4 = 360

120 x 9 = 1080

140 x 4 = 560

360 + 1080 + 560 = 2000

In total he bought 17 things.

Oh, thank you person underneath my answer. I have realised that I misread the question - also thank you for finding a mathematical way of finding this answer as requested. You are clearly of higher year level than me. :)

call them x, y, z, for number of pens, wallets, and watches

x = z

90x + 120y + 140z = 2000

not enough info, you have 2 equations and 3 unknowns.