Answer:
Step-by-step explanation:
Let c and p represent the numbers of cardigans and pullovers sold, respectively. Then the revenue equation is ...
31c +28p = 1460
Solving for p, we have ...
p = (1460 -31c)/28 = 52 -c +(4 -3c)/28
Now, define ...
a = (4 -3c)/28
and solve for c:
(4-28a)/3 = c = 1 -9a +(1 -a)/3
At this point, we can define ...
b = (1 -a)/3
Working backward, we can find c and p.
a = 1 -3b
c = 1 -9a +b = 1 -9(1 -3b) +b = -8 +28b
p = 52 -c +a = 52 -(-8 +28b) +(1 -3b) = 61 -31b
That is, for some integer b, solutions to the equation will be ...
c = 28b -8
p = 61 -31b
The only value of b that gives positive solutions for both c and p is b=1. For that value of b, we have ...
c = 20, p = 30
The boutique owner sold 20 cardigans and 30 pullovers.
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<em>Comment on the solution</em>
This is a variation of the Extended Euclidean Algorithm. Here, we have defined intermediate variables (a and b) that might not show up in a tableau solution form.
