#| This exercise has no tests.
Any solution is a right answer. |#
#lang sicp

10
; 10

(+ 5 3 4)
; 12

(- 9 1)
; 8

(/ 6 2)
; 3

(+ (* 2 4) (- 4 6))
; 6

(define a 3)
(define b (+ a 1))
(+ a b (* a b))
; 19

(= a b)
; #f

(if (and (> b a) (< b (* a b)))
  b
  a)
; 4

(cond ((= a 4) 6)
      ((= b 4) (+ 6 7 a))
      (else 25))
; 16

(+ 2 (if (> b a) b a))
;6

(* (cond ((> a b) a)
         ((< a b) b)
         (else -1))
   (+ a 1))
; 16