# Answer to Question #20647 in Algebra for Javier

Question #20647

Ed has 4 more pretzels than Cal. They have 10 pretzels in all. How many pretzels doe Cal have?

Expert's answer

Let Ed has E pretzels and Cal has C ones. Then we've got a system of linear equations:

E + C = 10, (1)

E - 4 = C. (2)

Let's solve it:

(1) ==> E + (E - 4) = 10,

or

2E = 14 ==> E = 7.

Therefore, Ed has 7 pretzels and Cal has C = 10 - 7 = 3 pretzels.

E + C = 10, (1)

E - 4 = C. (2)

Let's solve it:

(1) ==> E + (E - 4) = 10,

or

2E = 14 ==> E = 7.

Therefore, Ed has 7 pretzels and Cal has C = 10 - 7 = 3 pretzels.

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