Answers

p=-1 , q=c , r=-2 , s=-3 , b=4 をb=p+3q+3r+sに代入
4=-1 + 3c + -6 -3
3c=14
c=14/3 だと思う

aza
taro

訂正4=-1 + 3c +( -6)+( -3)

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中央の問題
b=  p+3q+3r+s
 p+2q+r  q+2r+s
p+q   q+r    r+s

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6とaの上  6+a
aと−4の上 a+(−4)
その上は (6+a) + (a+(−4)) これが 12になる
2a + 2 = 12 
a = 5 だと思う

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