| Let p be the number of jelly beans Peter ate. |
| Let j be the number of jelly beans Jerry ate. |
| Let q be the number of jelly beans Pam ate. |
| Then, since there are 66 jelly beans altogether, p+j+q=66 |
| and, since Peter had 4 times as many as Jerry, p=4j |
| and, since Peter had 3 fewer than Pam, p=q-3. |
| We may rewrite p=q-3 as q=p+3. |
| Now, substituting (q=p+3) into the very first equation, we have |
| p+j+(p+3)=66. |
| Substituting (p=4j) into this equation, we have |
| 4j+j+(4j+3)=66. |
| 9j+3=66 |
| 9j=63 |
| j=7 |
| Thus, Jerry had 7 jelly beans. Since p=4j=28, |
| Peter had 28 jelly beans. |
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