>
数学
>
已知不等式组
5a−1>3(a+1)
1
2
a−1<7−
3
2
a
的整数解a满足方程组
ax−2y=−7
2x+3y=4
,求x
2
+y
2
的值.
人气:169 ℃ 时间:2019-08-20 09:08:53
解答
解不等式①得:a>2
解不等式②得:a<4
所以不等式组的解集是:2<a<4
所以a的整数值为3.
把a=3代入方程组,得
3x−2y=−7
2x+3y=4
解得
x=−1
y=2
所以x
2
+y
2
=(-1)
2
+2
2
=5.
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