【数学I】例題3.3.13：2次の連立方程式（One More）★★

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% 例題I3.3.13：2次の連立方程式（One More）★★
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次の連立方程式を解け．

(1) $\{\begin{array}{l}4x-y=5\\x^2-4x-y=-2\end{array}$

(2) $\{\begin{array}{l}x^2-4xy+3y^2=0\\x^2+y^2+2x-2y=6\end{array}$

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% 例題I3.3.13の解答（One More）★★
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(1) $\{\begin{array}{l}4x-y=5\cdots(\mathrm{i})\\x^2-4x-y=-2\cdots(\mathrm{ii})\end{array}$

(i) より，$y=4x-5$

これを(ii)に代入すると，$x^2-4x-(4x-5)=-2$

整理すると，$x^2-8x+7=0$

したがって，$(x-1)(x-7)=0$

ゆえに，$x=1,7$

$y=4x-5$に代入すると，$x=1$のとき，$y=4\cdot 1-5=-1$

$x=7$のとき，$y=4\cdot 7-5=23$

よって，$(x,y)=(1,-1),(7,23)$

(2) $\{\begin{array}{l}x^2-4xy+3y^2=0\cdots(\mathrm{i})\\x^2+y^2+2x-2y=6\cdots(\mathrm{ii})\end{array}$

(i) より，$(x-y)(x-3y)=0$

したがって，$x=y$または$x=3y$

（ア）$x=y\cdots(\mathrm{iii})$のとき，(iii)を(ii)に代入して整理すると，$2y^2=6$

したがって，$y=\pm\sqrt{3}$

ゆえに，(iii)より，$y=\sqrt{3}$のとき，$x=\sqrt{3}$，$y=-\sqrt{3}$のとき，$x=-\sqrt{3}$

（イ）$x=3y\cdots(\mathrm{iv})$のとき，(iv)を(ii)に代入して整理すると，$10y^2+4y-6=0$

したがって，$(5y-3)(y+1)=0$となり，$y=\frac{3}{5},-1$

ゆえに，(iv)より，$y=\frac{3}{5}$のとき，$x=\frac{9}{5}$，$y=-1$のとき，$x=-3$

よって，（ア），（イ）より，求める解は，

\[
(x,y)=(\sqrt{3},\sqrt{3}),(-\sqrt{3},-\sqrt{3}),(\frac{9}{5},\frac{3}{5}),(-3,-1)
\]

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% 問題I3.3.13：2次の連立方程式（One More）★★
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次の連立方程式を解け．

(1) $\{\begin{array}{l}3x-y=4\\x^2-3x-y=-1\end{array}$

(2) $\{\begin{array}{l}x^2-3xy+2y^2+3y-9=0\\x^2-y^2+x+y=0\end{array}$

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% 問題I3.3.13の解答（One More）★★
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(1) $\{\begin{array}{l}3x-y=4\cdots(\mathrm{i})\\x^2-3x-y=-1\cdots(\mathrm{ii})\end{array}$

(i) より，$y=3x-4$

これを(ii)に代入すると，$x^2-3x-(3x-4)=-1$

整理すると，$x^2-6x+5=0$

したがって，$(x-1)(x-5)=0$

ゆえに，$x=1,5$

$y=3x-4$に代入すると，$x=1$のとき，$y=3\cdot 1-4=-1$

$x=5$のとき，$y=3\cdot 5-4=11$

よって，$(x,y)=(1,-1),(5,11)$

(2) $\{\begin{array}{l}x^2-3xy+2y^2+3y-9=0\cdots(\mathrm{i})\\x^2-y^2+x+y=0\cdots(\mathrm{ii})\end{array}$

(ii) より，$(x+y)(x-y)+(x+y)=0$

したがって，$(x+y)(x-y+1)=0$

ゆえに，$y=-x$または$y=x+1$

（ア）$y=-x\cdots(\mathrm{iii})$のとき，(iii)を(i)に代入して整理すると，

\[
2x^2-x-3=0
\]

したがって，$(2x-3)(x+1)=0$となり，$x=\frac{3}{2},-1$

ゆえに，(iii)より，$x=\frac{3}{2}$のとき，$y=-\frac{3}{2}$，$x=-1$のとき，$y=1$

（イ）$y=x+1\cdots(\mathrm{iv})$のとき，(iv)を(i)に代入して整理すると，$4x=4$

したがって，$x=1$

ゆえに，(iv)より，$x=1$のとき，$y=2$

よって，（ア），（イ）より，求める解は，

\[
(x,y)=(\frac{3}{2},-\frac{3}{2}),(-1,1),(1,2)
\]

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% 例題I3.3.13の側注：2次の連立方程式（One More）★★
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% 側注1
$y$の係数が$-1$であるので，$y$の方が文字を消去しやすい．

% 側注2
なお，$(1,-1),(7,23)$は，与えられた連立方程式(i)，(ii)を整理した$y=4x-5$（直線），$y=x^2-4x+2$（放物線）の交点を表している．

% 側注3
なお，$y=x,y=\frac{x}{3}$として場合分けをしても同じ結果が得られる．

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% 問題I3.3.13の側注：2次の連立方程式（One More）★★
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% 側注1
$(1,-1),(5,11)$は，与えられた連立方程式(i)，(ii)を整理した$y=3x-4$（直線），$y=x^2-3x+1$（放物線）の交点を表している．

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% 例題I3.3.13の考え方・補足など：2次の連立方程式（One More）★★
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% 考え方
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(1)は，1次式を利用して文字の消去を考える．(2)は，$x^2-4xy+3y^2$が因数分解できることに注目する．