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連立方程式(まとめ)
2次2次連立形-現在地

== 2次-2次連立方程式の解き方 ==

1.このページでは,次のような形の連立方程式を1次-2次連立形と呼ぶことにします.
2x−y+1=0…(1)
x2+y2=1…(2)
(解き方)
1次-2次連立形の連立方程式は,1次式の方を1文字について解き,2次の方に代入すると解けます.
(解答)
1次-2次連立形の連立方程式の解は,直線と楕円,放物線,双曲線などの2次曲線との共有点を表し,一般には2組あります
(1)よりy=2x+1…(1’)
(1’)を(2)に代入する
x2+(2x+1)2=1
5x2+4x=0
x(5x+4)=0
x=0,45
x=0のとき,(1’)に戻すと,y=1
x=45のとき,(1’)に戻すと,y=35
(x,y)=(0,1),(45,35)…(答)
2次-2次連立形の連立方程式の解は,楕円,放物線,双曲線などの2次曲線と楕円,放物線,双曲線などの2次曲線との共有点を表し,4組あるのが普通ですが,それ以下の場合もあります.
2.このページでは,次のような形の連立方程式を2次-2次連立形と呼ぶことにします.
x2+y2+3xy=5…(1)
x2+y2−x+y=1…(2)

 このページでは,主に2次-2次連立形の連立方程式の解き方を解説します.
 なお,1次-2次連立形の連立方程式は,しばしば登場するもので,必ず解けなければなりませんが,2次-2次連立形の連立方程式は教科書や授業ではほとんど触れません.読者が何かの都合で解く必要を生じたときに,参考にしてください.

【要点】
2次-2次連立方程式を解くには,次のいずれかの型に当てはめます.
(A) 一方の2次方程式が因数分解できる場合
(B) 和や差により2次の項が消去できる場合
(C) 定数項が消去できる場合
(D) 対称式になっている場合
(A) 一方の2次方程式が因数分解できる場合
一方の2次方程式が因数分解できる場合は,それを利用して1次-2次連立形の連立方程式に直すことができるので,代入消去の方法で簡単に解けます.
【例A-1】
x2−xy−2y2=0…(1)
x2+y2=1…(2)
(解答)
(1)式は,(x+y)(x−2y)=0と因数分解できます.
これにより,x=−yまたはx=2yとなるので,元の連立方程式は
x=−y…(1’)
x2+y2=1…(2)
x=2y…(1”)
x2+y2=1…(2)
に分けられます.
(1’)を(2)に代入すると
2y2=1となり,y=±12=±22
(1’)に戻すと
(x,y)=(22,22),(22,22)…(答)
(1”)を(2)に代入すると
5y2=1となり,y=±15=±55
(1”)に戻すと
(x,y)=(255,55),(255,55)…(答)

【問題1】 次の連立方程式を解いてください.

x2−5xy+6y2=0…(1)
x2+y2=10…(2)
参考答案を見る参考答案を隠す


x2−y2+4y−4=0…(1)
x2+y2=10…(2)
参考答案を見る参考答案を隠す

(B) 和や差により2次の項が消去できる場合
2つの式を足したり引いたりして1次式を作ることができる場合は,1次-2次連立形に直すことができます
【例B-1】
x2+y2−2x+2y+1=0…(1)
x2+y2=1…(2)
(解答)
この問題では,次の図のように2つの円の共有点になり,解は2組になります
(1)−(2)により
−2x+2y+1=−1
−2x+2y+2=0
y=x−1…(3)
(3)を(2)に代入すると
x2+(x−1)2=1
2x2−2x=0
2x(x−1)=0
x=0, 1
(3)に戻すと
(x, y)=(0, −1), (1, 0)…(答)

【問題2】 次の連立方程式を解いてください.

2x2+2x−y=15…(1)
x2−x+y=−1…(2)
参考答案を見る参考答案を隠す


xy+x=6…(1)
2xy−y=5…(2)
参考答案を見る参考答案を隠す

(C) 定数項が消去できる場合
2つの式を足したり引いたりして定数項を消去できる場合,次のような同次形の方程式になり,x=kyの形に解けます
ax2+bxy+cy2=0
a(xy)2+b(xy)+c=0
2次方程式の解の公式などを用いて
xy=b±b24ac2a
x=b±b24ac2ay
【例C-1】
2x2−3xy+y2=3…(1)
x2+2xy−3y2=5…(2)
(解答)
(1)×5−(2)×3により定数項を消去する
10x2−15xy+5y2=15
− )3x2+6xy−9y2=15
7x2−21xy+14y2=0
x2−3xy+2y2=0
(x−y)(x−2y)=0
x=y, x=2y
ア) x=yのとき,(1)に代入すると
2y2−3y2+y2=3
0=3
これは成り立たないから,この場合からは解は出ない
イ) x=2yのとき,(1)に代入すると
8y2−6y2+y2=3
3y2=3
y2=1
y=±1
(x, y)=(2, 1),(−2, −1)…(答)

【問題3】 次の連立方程式を解いてください.

x2−2xy−y2=2…(1)
2x2+xy+y2=2…(2)
参考答案を見る参考答案を隠す


x2−xy+y2=1…(1)
xy+y2=2…(2)
参考答案を見る参考答案を隠す

(D) 対称式になっている場合
対称式になっている場合は,基本対称式p=x+y, q=xyで表し,p, qを求めると
x, yx2−px+q=0の2つの解になる(解と係数の関係)ことから,x, yを求めることができます.
【例D-1】
x2+y2+x+y=2…(1)
x2+xy+y2=1…(2)
(解答)
p=x+y, q=xyとおくと
(1)→p2−2q+p=2…(1’)
(2)→p2−q=1…(2’)
(1’)−(2’)により2次の項を消去すると
−q+p=1
p=q+1…(3)
(3)を(2’)に代入すると
(q+1)2−q=1
q2+q=0
q(q+1)=0
q=0, −1
ア) q=0のとき,(3)に代入すると
p=1
このとき,x, yx2−x=0の2つの解だから
(x, y)=(0, 1),(1, 0)…(答)
イ) q=−1のとき,(3)に代入すると
p=0
このとき,x, yx2−1=0の2つの解だから
(x, y)=(1, −1),(−1, 1)…(答)

【問題4】 次の連立方程式を解いてください.

xy+x+y=5…(1)
2xy−x−y=1…(2)
参考答案を見る参考答案を隠す


2x2+xy+2y2=5…(1)
x2+xy+y2=10…(2)
参考答案を見る参考答案を隠す

(筆者の本音)
 以上の解説は,筆算で解くことを想定した,伝統的な解き方を述べたものです.
 筆算を前提とする限り,(A)「整数係数の問題で,かつ,厳密解を美的に求めること」が重視されているように見えますが,そういう世界ばかりではありません.
 さまざまな観測や実験の結果を分析する場合には,問題も結果も近似値になり,解はコンピュータで求めることになります.
 例えば,次のような双曲線と楕円
x2+0.3xy−3.0y2+1.1x−2.7y+12.3=0
5.1x2+0.1xy+1.2y2+1.2x+3.1y−8.5=0

の交点を有効数字16桁で求めると
x=0.1048608622954735
y=1.638491189427312
x=−0.5017452715317802
y=1.58055152394775
x=1.183747412008281
y=−2.661024498886414
x=−1.372735116479724
y=−2.638490926456542
 さらに,赤で示した定数が時々刻々と変化するときに,解を自動的に更新して交点を追跡できるようにしたいといった場合には,厳密解とか美的という要素は重要ではなく,(B)「すぐに解けて図示できること」が重視されていることになります.
 筆者の予想では,今日の生活では(B)の方が重要な場面が多いと考えられます.

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