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== 方程式の作り方 ==

※ 問題文に適するものを1つ選びなさい。※消費税などによる影響は考えません。
問1 和菓子6 個をまとめて50 円の箱に詰めてもらったら,合計770 円であった。和菓子1 個の値段を x (円)とすると,x の満たす方程式は
x+50=770
6x+50=770
6(x+50)=770
となる。

問2 角砂糖 3 個を 88 (g) の水に溶かしたら 100 (g) の砂糖水になった。角砂糖 1 個の重さを x (g) とすると,x の満たす方程式は
x+88=100
3+x+88=100
3x+88=100
3(x+88)=100
となる。

問3 トマトの苗 5 株と 400 円のプランター 1 個を買うとちょうど 1000 円だった。トマトの苗 1株の値段を x (円)とすると,x の満たす方程式は
x+400=1000
5+x+400=1000
5x+400=1000
5(x+400)=1000
となる。

問4 1個 80 円のコッペパンと1個 120 円のあんぱんを合計 15 個買って 1600 円にしたい。あんぱんを x (個)買うものとすると,x の満たす方程式は
80x+120x=1600
80x+120(15-x)=1600
80(15-x)+120x=1600
となる。

問5 1個 100 円のパンと1個 150 円のプリンを合計 7 個買って,ちょうど 1000 円にしたい。パンを x (個)買うものとすると,x の満たす方程式は
100x+150x=1000
100x+150(7-x)=1000
100(7-x)+150x=1000
(100+150)x=1000
となる。


問6 1本 100 円のお茶と1本 120 円の缶コーヒーを合計 20 本買うと,2300 円になった。缶コーヒーを x (本)買ったものとすると,x の満たす方程式は
100x+120x=2300
100x+120(20-x)=2300
100(20-x)+120x=2300
(100+120)x=2300
となる。

問7 和菓子が洋菓子よりも 3 個多くあって,合計 39 個ある。和菓子の数を x (個)とすると,x の満たす方程式は
x+3=39
x-3=39
(x+3)+x=39
x+(x−3)=39
となる。

問8 ある長方形は周囲の長さが 60 (cm)で,縦の長さが横の長さの 2 倍ある。横の長さをx (cm)とすると,x の満たす方程式は
x=2(60-x)
2x=60
x+2x=60
2(x+2x)=60
となる。

問9 ある数よりも 7 だけ大きな数は,もとの数の 2 倍よりも 5 小さいという。もとの数を x とすると,x の満たす方程式は
x−7=2x+5
x+7=2x−5
7x=2x−5
x+7=5−2x
となる。

(速さ)×(時間)=(距離)
■時間を未知数 x とするとき■
問10 A地点からB地点まで移動するのに,分速 100 (m)で歩くと予定よりも 10 分早く到着し,分速 80(m)で歩くと予定よりも 5 分遅くなる。予定の時間の長さを x (分)とすると,x の満たす方程式は
100x−10=80x+5
100x+10=80x−5
100(x−10)=80(x+5)
100(x+10)=80(x−5)
となる。


■距離を未知数 x とするとき■
問11 A地点からB地点まで移動するのに,分速 250 (m)の自転車で行くと,分速 100 (m)で歩く場合よりも 15 分早く到着する。A地点からB地点までの距離を x (m)とすると,x の満たす方程式は
100x=250x−15

100x=250x+15

.x100nnn = .x250nnn_−15

.x100nnn = .x250nnn_ +15
となる。

問12 分速 100 (m)で歩く人がスタートしてから 6 分後に分速 250 (m)の自転車で追いかけるとき,自転車が追いつくまでの時間を x (分)とすると,x の満たす方程式は
100(6+x)=250x
100x+6=250x
100x=250(x−6)
100x=250x−6
となる。

問13 お菓子を子どもたちに分けるとき,一人 5 個ずつ分けるにはお菓子は 3 個足りない。また,一人 4 個ずつ分けるとお菓子は 2 個余る。子どもの人数を x (人)とすると,x の満たす方程式は
.x5n_−3=.x4n_ +2

.x5n_+3=.x4n_−2

5x+3=4x−2

5x−3=4x+2
となる。

問14 コンパの予算で1種類のジュースをなるべくたくさん買いたい。9 本買うには 80 円足りないが, 8 本にすれば 40 円余る。ジュース1本の値段を x (円)とすると,x の満たす方程式は
9x−80=8x+40
9x+80=8x−40
.x+809nnnn = .x−408nnnn
.x−809nnnn = .x+408nnnn
となる。

問15 体育館の長いすに生徒が座るとき,1脚当り 3 人ずつ座ると 16 人分足りない。また,1脚当り 4 人ずつ座ると 64 人分空席ができる。長いすの数を x (脚)とすると,x の満たす方程式は
3x−16=4x+64
3x+16=4x−64
.x3n +16 = .x4n_−64
.x3n_−16 = .x4n +64
となる。


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