第11章連立方程式と連立不等式Systems of Equations and Inequalities

11.2 Systems of Linear Equations: Three Variables11.2 連立一次方程式: 三変数

Learning Objectives 学習目標

In this section, you will:

この節を終えると、次のことができるようになる。

• Solve systems of three equations in three variables.• Identify inconsistent systems of equations containing three variables.• Express the solution of a system of dependent equations containing three variables.

・三変数三式の連立方程式を解く。・三変数を含む解の無い連立方程式を見分ける。・三変数を含む従属な連立方程式の解を表す。

A close-up shot captures a diverse assortment of US coins, including copper pennies, silver dimes, and quarters, creating a textured pile of everyday currency. — Figure 1 (credit: “Elembis,” Wikimedia Commons)

Jordi received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. Jordi invested $4,000 more in mutual funds than in municipal bonds. He earned $670 in interest the first year. How much did Jordi invest in each type of fund?

ジョルディは12,000ドルの遺産を受け取り、それを三つに分けて三通りに投じた。年利3%の短期金融資産投資信託、年利4%の地方債、年利7%の投資信託である。ジョルディは地方債より4,000ドル多く投資信託に投じた。最初の年に670ドルの利息を得た。ジョルディは各種類の基金にいくらずつ投じたか。

Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visualization.

この種の問題の立て方の正しい近づき方が分かれば、解を求めるのは型に従うだけのことになる。この節では、これと似た三式三変数の問題を解く。そのために使うのは、二変数二式の連立方程式を解くのに使ったのと似た手法である。だが三式の組の解を求めるには、もう少し整理と、少しばかりの図の思い浮かべが要る。

Solving Systems of Three Equations in Three Variables三変数三式の連立方程式を解く

In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution $(x, y, z),$ which we call an ordered triple. A system in upper triangular form looks like the following:

三変数の連立方程式、すなわち三元の組を解くために、主に使う道具はガウスの消去法という。多くの業績を残したドイツの数学者カール・フリードリヒ・ガウスにちなむ名である。演算を行う順序は決まっていないが、どんな操作ができるかについては決まった指針がある。どの手順を使ったかを追うために式に番号を付けてもよい。目標は、一度に一つずつ変数を消して上三角の形にすることである。これが三元の組にとって理想の形なのは、素直に後ろから代入して解 $(x, y, z)$ が求まるからである。この解を順序三つ組という。上三角の形の組は次のようになる。

\begin{array}{l} A x + B y + C z = D \\ E y + F z = G \\ H z = K \end{array}

The third equation can be solved for $z,$ and then we back-substitute to find $y$ and $x .$ To write the system in upper triangular form, we can perform the following operations:

三つ目の式は $z$ について解け、そして後ろから代入して $y$ と $x$ を求める。組を上三角の形に書くには、次の演算を行ってよい。

1. Interchange the order of any two equations.2. Multiply both sides of an equation by a nonzero constant.3. Add a nonzero multiple of one equation to another equation.

1. どの二つの式でも順序を入れ替える。2. 式の両辺に零でない定数を掛ける。3. 一つの式の零でない倍を、別の式に足す。

The solution set to a three-by-three system is an ordered triple ${(x, y, z)} .$ Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes.

三元の組の解の集合は順序三つ組 ${(x, y, z)}$ である。グラフで言えば、この順序三つ組は空間の三平面の交わりである点を定める。この交わりは、直方体の部屋のどれかの隅を思い浮かべれば見て取れる。隅は三つの平面で定まる。隣り合う二枚の壁と床(または天井)である。二枚の壁と床が出会うどの点も、三平面の交わりを表す。

Number of Possible Solutions 解の個数

Figure 2 and Figure 3 illustrate possible solution scenarios for three-by-three systems.

図2と図3に、三元の組のありうる解の筋書きを示す。

• Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple ${(x, y, z)} .$ Graphically, the ordered triple defines a point that is the intersection of three planes in space.• Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as $0 = 0.$ Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.• Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as $3 = 0.$ Graphically, a system with no solution is represented by three planes with no point in common.

・解がただ一つの組とは、消去のあとに順序三つ組 ${(x, y, z)}$ からなる解の集合ができるものである。グラフで言えば、この順序三つ組は空間の三平面の交わりである点を定める。・解が無限にある組とは、消去のあとに $0 = 0.$ のようなつねに正しい式ができるものである。グラフで言えば、解が無限にあるとは、空間の三平面の交わりとなる直線か、重なった平面を表す。・解が無い組とは、消去のあとに $3 = 0.$ のような矛盾する式ができるものである。グラフで言えば、解の無い組は共通の点を持たない三平面で表される。

Figure (a) displays three intersecting planes in red, green, and blue, meeting at a central point. Figure (b) shows a 2D projection of these planes with an arrow indicating a vertical dimension. — Figure 2 (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions.図2 (a)三平面が一点で交わり、解がただ一つの三元の組を表す。(b) 三平面が一直線で交わり、解が無限にある三元の組を表す。

The image displays three distinct examples of plane arrangements. Figure (a) depicts multiple planes intersecting, where an orange rectangular plane is intersected by several angled blue and teal planes. Figure (b) shows two sets of intersecting planes, with orange and blue/teal planes crossing each other in a grid-like fashion. Figure (c) illustrates three parallel planes, colored orange, dark blue, and light blue, stacked vertically. — Figure 3 All three figures represent three-by-three systems with no solution. (a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. (c) All three planes are parallel, so there is no point of intersection.図3 三つの図はどれも解の無い三元の組を表す。(a) 三平面は互いに交わるが、共通の点では交わらない。(b) 二平面が平行で三つ目の平面と交わるが、互いには交わらない。(c) 三平面すべてが平行なので、交わる点は無い。

Example 1例1

Determining Whether an Ordered Triple Is a Solution to a System順序三つ組が組の解かどうかを判定する

Determine whether the ordered triple $(3, −2, 1)$ is a solution to the system.

順序三つ組 $(3, −2, 1)$ がこの組の解かどうかを判定せよ。

\begin{array}{l} x + y + z = 2 \\ 6 x - 4 y + 5 z = 31 \\ 5 x + 2 y + 2 z = 13 \end{array}

Solution 解答

We will check each equation by substituting in the values of the ordered triple for $x, y,$ and $z .$

順序三つ組の値を $x, y$ と $z$ に代入して、各式を確かめる。

$\begin{matrix} \begin{array}{r} x + y + z = 2 \\ (3) + (−2) + (1) = 2 \\ True \end{array} & \begin{array}{r} 6 x −4 y + 5 z = 31 \\ 6 (3) −4 (−2) + 5 (1) = 31 \\ 18 + 8 + 5 = 31 \\ True \end{array} & \begin{array}{r} 5 x + 2 y + 2 z = 13 \\ 5 (3) + 2 (−2) + 2 (1) = 13 \\ 15 −4 + 2 = 13 \\ True \end{array} \end{matrix}$

式中の英語：True：成立

The ordered triple $(3, −2, 1)$ is indeed a solution to the system.

順序三つ組 $(3, −2, 1)$ は確かにこの組の解である。

How To 手順

Given a linear system of three equations, solve for three unknowns.

三式の一次の組が与えられたとき、三つの未知数について解く。

1. Pick any pair of equations and solve for one variable.2. Pick another pair of equations and solve for the same variable.3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.4. Back-substitute known variables into any one of the original equations and solve for the missing variable.

1. どれでも二つの式を選び、一つの変数について解く。2. 別の二つの式を選び、同じ変数について解く。3. これで二つの未知数の二式の組ができた。できた二元の組を解く。4. 分かった変数を元の式のどれか一つに後ろから代入し、未知の変数について解く。

Example 2例2

Solving a System of Three Equations in Three Variables by Elimination三変数三式の連立方程式を消去で解く

Find a solution to the following system:

次の組の解を求めよ。

\begin{array}{l} x −2 y + 3 z = 9 & (1) \\ - x + 3 y - z = −6 & (2) \\ 2 x −5 y + 5 z = 17 & (3) \end{array}

Solution 解答

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate $x$ by adding equations (1) and (2).

どこから始めるかには、つねにいくつかの選び方があるが、ここでいちばん見えやすい最初の一歩は、式(1)と式(2)を足して $x$ を消すことである。

\frac{\begin{array}{l} x - 2 y + 3 z = 9 & (1) \\ - x + 3 y - z = −6 & (2) \end{array}}{\begin{array}{l} y + 2 z = 3 & (4) \end{array}}

The second step is multiplying equation (1) by $−2$ and adding the result to equation (3). These two steps will eliminate the variable $x .$

二つ目の一歩は、式(1)に $−2$ を掛けてその結果を式(3)に足すことである。この二つの一歩で変数 $x$ が消える。

\begin{array}{l} \underset{____________________________________}{\begin{array}{l} −2 x + 4 y - 6 z = −18 & (1) multiplied by - 2 \\ 2 x - 5 y + 5 z = 17 & (3) \end{array}} \\ - y - z = −1 (5) \end{array}

In equations (4) and (5), we have created a new two-by-two system. We can solve for $z$ by adding the two equations.

式(4)と式(5)で、新しい二元の組ができた。この二つの式を足せば $z$ について解ける。

\frac{\begin{array}{l} \begin{array}{l} y + 2 z = 3 (4) \end{array} \\ - y - z = - 1 (5) \end{array}}{z = 2 (6)}

Choosing one equation from each new system, we obtain the upper triangular form:

新しい組からそれぞれ一つずつ式を選ぶと、上三角の形が得られる。

\begin{array}{l} x −2 y + 3 z = 9 & (1) \\ y + 2 z = 3 & (4) \\ z = 2 & (6) \end{array}

Next, we back-substitute $z = 2$ into equation (4) and solve for $y .$

次に、 $z = 2$ を式(4)に後ろから代入し、 $y$ について解く。

\begin{array}{l} y + 2 (2) = 3 \\ y + 4 = 3 \\ y = −1 \end{array}

Finally, we can back-substitute $z = 2$ and $y = −1$ into equation (1). This will yield the solution for $x .$

最後に、 $z = 2$ と $y = −1$ を式(1)に後ろから代入できる。これで $x$ の解が出る。

\begin{array}{r} x −2 (−1) + 3 (2) = 9 \\ x + 2 + 6 = 9 \\ x = 1 \end{array}

The solution is the ordered triple $(1, −1, 2) .$ See Figure 4.

解は順序三つ組 $(1, −1, 2)$ である。図4を見よ。

An illustration of three orthogonal planes, x=1 (green), y=-2 (red), and z=2 (blue), intersecting at the labeled point (1, -1, 2) in a 3D coordinate system.

Example 3例3

Solving a Real-World Problem Using a System of Three Equations in Three Variables三変数三式の連立方程式を使って現実世界の問題を解く

In the problem posed at the beginning of the section, Jordi invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. Jordi invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?

この節の初めに出した問題では、ジョルディは12,000ドルの遺産を三つの異なる基金に投じた。一部は年利3%の短期金融資産投資信託に、一部は年利4%の地方債に、残りは年利7%の投資信託に投じた。ジョルディは地方債に投じた額より4,000ドル多く投資信託に投じた。一年で得た利息の合計は670ドルだった。彼は各種類の基金にいくらずつ投じたか。

Solution 解答

To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:

この問題を解くには、与えられた情報をすべて使い、三つの式を立てる。まず、三つの投資額それぞれに変数を割り当てる。

\begin{array}{l} x = amount invested in money-market fund \\ y = amount invested in municipal bonds \\ z = amount invested in mutual funds \end{array}

The first equation indicates that the sum of the three principal amounts is $12,000.

一つ目の式は、三つの元金の和が12,000ドルであることを示す。

x + y + z = 12,000

We form the second equation according to the information that Jordi invested $4,000 more in mutual funds than he invested in municipal bonds.

二つ目の式は、ジョルディが地方債に投じた額より4,000ドル多く投資信託に投じたという情報に従って作る。

z = y + 4,000

The third equation shows that the total amount of interest earned from each fund equals $670.

三つ目の式は、各基金から得た利息の合計が670ドルに等しいことを示す。

0.03 x + 0.04 y + 0.07 z = 670

Then, we write the three equations as a system.

そして、三つの式を組として書く。

\begin{array}{l} x + y + z = 12,000 \\ - y + z = 4,000 \\ 0.03 x + 0.04 y + 0.07 z = 670 \end{array}

To make the calculations simpler, we can multiply the third equation by 100. Thus,

計算を簡単にするために、三つ目の式に100を掛けてよい。よって、

\begin{array}{l} x + y + z = 12,000 & (1) \\ - y + z = 4,000 & (2) \\ 3 x + 4 y + 7 z = 67,000 & (3) \end{array}

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

第1歩。三変数を含む二つの式がそろうように、式(2)と式(3)を入れ替える。

\begin{array}{l} x + y + z = 12,000 \\ 3 x + 4 y + 7 z = 67,000 \\ - y + z = 4,000 \end{array}

Step 2. Multiply equation (1) by $−3$ and add to equation (2). Write the result as row 2.

第2歩。式(1)に $−3$ を掛けて式(2)に足す。その結果を第2行として書く。

\begin{array}{l} x + y + z = 12,000 \\ y + 4 z = 31,000 \\ - y + z = 4,000 \end{array}

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

第3歩。式(2)を式(3)に足し、その結果を式(3)として書く。

\begin{array}{l} x + y + z = 12,000 \\ y + 4 z = 31,000 \\ 5 z = 35,000 \end{array}

Step 4. Solve for $z$ in equation (3). Back-substitute that value in equation (2) and solve for $y .$ Then, back-substitute the values for $z$ and $y$ into equation (1) and solve for $x .$

第4歩。式(3)を $z$ について解く。その値を式(2)に後ろから代入し、 $y$ について解く。そして $z$ と $y$ の値を式(1)に後ろから代入し、 $x$ について解く。

\begin{array}{l} 5 z = 35,000 \\ z = 7,000 \\ y + 4 (7,000) = 31,000 \\ y = 3,000 \\ x + 3,000 + 7,000 = 12,000 \\ x = 2,000 \end{array}

Jordi invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.

ジョルディは短期金融資産投資信託に2,000ドル、地方債に3,000ドル、投資信託に7,000ドルを投じた。

Try It #1やってみよう1

Solve the system of equations in three variables.

三変数の連立方程式を解け。

\begin{array}{l} 2 x + y −2 z = −1 \\ 3 x −3 y - z = 5 \\ x −2 y + 3 z = 6 \end{array}

Identifying Inconsistent Systems of Equations Containing Three Variables三変数を含む解の無い連立方程式を見分ける

Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as $3 = 7$ or some other contradiction.

二変数の連立方程式と同じく、三変数でも解の無い連立方程式に出会うことがある。これは、三つの式すべてを満たす解が無いということである。三つの式は、三枚の平行な平面、二枚の平行な平面と交わる一枚の平面、または他の二枚と交わるが同じ場所では交わらない三枚の平面を表しうる。消去の過程で、 $3 = 7$ のような偽の式か、ほかの矛盾する式ができる。

Example 4例4

Solving an Inconsistent System of Three Equations in Three Variables三変数三式の解の無い連立方程式を解く

Solve the following system.

次の組を解け。

\begin{array}{l} x −3 y + z = 4 & (1) \\ - x + 2 y −5 z = 3 & (2) \\ 5 x −13 y + 13 z = 8 & (3) \end{array}

Solution 解答

Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).

$x$ の係数を見ると、式(1)を式(2)に足せば $x$ が消せると分かる。

\frac{\begin{array}{l} x −3 y + z = 4 (1) \\ - x + 2 y −5 z = 3 (2) \end{array}}{- y −4 z = 7 (4)}

Next, we multiply equation (1) by $−5$ and add it to equation (3).

次に、式(1)に $−5$ を掛けて式(3)に足す。

\begin{array}{l} \underset{______________________________________}{\begin{array}{l} - 5 x + 15 y - 5 z = −20 & (1) multiplied by −5 \\ 5 x - 13 y + 13 z = 8 & (3) \end{array}} \\ \begin{array}{l} 2 y + 8 z = −12 & (5) \end{array} \end{array}

Then, we multiply equation (4) by 2 and add it to equation (5).

そして、式(4)に2を掛けて式(5)に足す。

\begin{array}{l} \underset{_______________________________________}{\begin{array}{l} −2 y - 8 z = 14 (4) multiplied by 2 \\ 2 y + 8 z = - 12 (5) \end{array}} \\ 0 = 2 \end{array}

The final equation $0 = 2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

最後の式 $0 = 2$ は矛盾なので、この連立方程式には解が無く、したがって解を持たないと結論する。

Analysis 考察

In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.

この組では、各平面が他の二枚と交わるが、同じ場所では交わらない。したがってこの組に解は無い。

Try It #2やってみよう2

Solve the system of three equations in three variables.

三変数三式の連立方程式を解け。

\begin{array}{l} x + y + z = 2 \\ y −3 z = 1 \\ 2 x + y + 5 z = 0 \end{array}

Expressing the Solution of a System of Dependent Equations Containing Three Variables三変数を含む従属な連立方程式の解を表す

We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line.

二変数の連立方程式を扱ったことから、従属な連立方程式には解が無限にあると分かっている。三変数の従属な連立方程式でも同じである。解が無限にあることは、いくつかの状況から起こりうる。三平面が同じであってもよく、その場合は一つの式の解が他の二つの式の解にもなる。三つの式がすべて違っていても、一直線で交わることがあり、そのときは解が無限にある。あるいは、二つの式が同じで、三つ目と一直線で交わることもある。

Example 5例5

Finding the Solution to a Dependent System of Equations従属な連立方程式の解を求める

Find the solution to the given system of three equations in three variables.

与えられた三変数三式の連立方程式の解を求めよ。

\begin{array}{r} 2 x + y −3 z = 0 & (1) \\ 4 x + 2 y −6 z = 0 & (2) \\ x - y + z = 0 & (3) \end{array}

Solution 解答

First, we can multiply equation (1) by $−2$ and add it to equation (2).

まず、式(1)に $−2$ を掛けて式(2)に足せる。

\begin{array}{l} \underset{____________________________________________}{\begin{array}{l} −4 x −2 y + 6 z = 0 equation (1) multiplied by −2 \\ ​ ​ ​ ​ 4 x + 2 y −6 z = 0 (2) \end{array}} \\ 0 = 0 \end{array}

We do not need to proceed any further. The result we get is an identity, $0 = 0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $−2,$ and adding it to equation (1). We then perform the same steps as above and find the same result, $0 = 0.$

これ以上進める必要はない。得られた結果は恒等式 $0 = 0$ であり、この組に解が無限にあることを告げている。この組を解き始めるしかたはほかにもあり、たとえば式(3)に $−2$ を掛けて式(1)に足してもよい。そのあと上と同じ手順を行うと、同じ結果 $0 = 0.$ が得られる。

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

組が従属のとき、解の一般の式が求められる。式(1)と式(3)を足すと、次を得る。

\begin{array}{l} \underset{_____________}{\begin{array}{l} \begin{array}{l} 2 x + y −3 z = 0 \end{array} \\ x - y + z = 0 \end{array}} \\ 3 x −2 z = 0 \end{array}

We then solve the resulting equation for $z .$

そして、できた式を $z$ について解く。

\begin{array}{l} 3 x −2 z = 0 \\ z = \frac{3}{2} x \end{array}

We back-substitute the expression for $z$ into one of the equations and solve for $y .$

$z$ の式を式の一つに後ろから代入し、 $y$ について解く。

\begin{array}{l} 2 x + y - 3 (\frac{3}{2} x) = 0 \\ 2 x + y - \frac{9}{2} x = 0 \\ y = \frac{9}{2} x - 2 x \\ y = \frac{5}{2} x \end{array}

So the general solution is $(x, \frac{5}{2} x, \frac{3}{2} x) .$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x .$

よって一般の解は $(x, \frac{5}{2} x, \frac{3}{2} x)$ である。この解で $x$ はどんな実数でもよい。 $y$ と $z$ の値は、 $x$ に選んだ値で決まる。

Analysis 考察

As shown in Figure 5, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

図5に示すとおり、二枚の平面が同じで、それらが三つ目の平面と一直線で交わる。解の集合は無限である。その交わりの直線上のすべての点が三つの式すべてを満たすからである。

An abstract blue and green graphic, resembling an X, paired with a system of three linear equations: x-y+z=0, -4x-2y+6z=0, and 4x+2y-6z=0.

Q&A 疑問と答え

Does the generic solution to a dependent system always have to be written in terms of $x ?$

従属な組の一般の解は、いつも $x ?$ で書かねばならないか。

No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of x and if needed $x$ and $y .$

そうではない。一般の解はどの変数で書いてもよいが、x で書き、必要なら $x$ と $y$ で書くのがふつうである。

Try It #3やってみよう3

Solve the following system.

次の組を解け。

\begin{array}{l} x + y + z = 7 \\ 3 x - 2 y - z = 4 \\ x + 6 y + 5 z = 24 \end{array}

Media 教材

Access these online resources for additional instruction and practice with systems of equations in three variables.

三変数の連立方程式についての説明と練習をさらに求めるなら、次のオンライン教材にあたるとよい。

• Ex 1: System of Three Equations with Three Unknowns Using Elimination• Ex. 2: System of Three Equations with Three Unknowns Using Elimination

・例1: 消去を使う三つの未知数の三式の組・例2: 消去を使う三つの未知数の三式の組

Verbal言葉で答える問題

1. Can a linear system of three equations have exactly two solutions? Explain why or why not

1. 三式の一次の組に解がちょうど二つあることはありうるか。なぜそうか、あるいはなぜそうでないかを説明せよ。

2. If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique.

2. 与えられた順序三つ組がこの連立方程式を解くとき、その解はただ一つか。そうならその理由を説明せよ。そうでないなら、ただ一つでない例を挙げよ。

3. If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example.

3. 与えられた順序三つ組がこの連立方程式を解かないとき、解は無いのか。そうならその理由を説明せよ。そうでないなら例を挙げよ。

4. Using the method of addition, is there only one way to solve the system?

4. 足し算の方法を使うとき、組を解くしかたは一つだけか。

5. Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not.

5. 連立一次方程式を解く方法が一つしかありえないかどうかを説明できるか。あるなら、そのような連立方程式の例を挙げよ。ないなら、なぜないかを説明せよ。

Algebraic代数の問題

For the following exercises, determine whether the ordered triple given is the solution to the system of equations.

次の各問について、与えられた順序三つ組がこの連立方程式の解かどうかを判定せよ。

6. $\begin{array}{l} 2 x −6 y + 6 z = −12 \\ x + 4 y + 5 z = −1 \\ - x + 2 y + 3 z = −1 \end{array}$ and $(0, 1, −1)$

6. $\begin{array}{l} 2 x −6 y + 6 z = −12 \\ x + 4 y + 5 z = −1 \\ - x + 2 y + 3 z = −1 \end{array}$ と $(0, 1, −1)$

7. $\begin{array}{l} 6 x - y + 3 z = 6 \\ 3 x + 5 y + 2 z = 0 \\ x + y = 0 \end{array}$ and $(3, −3, −5)$

7. $\begin{array}{l} 6 x - y + 3 z = 6 \\ 3 x + 5 y + 2 z = 0 \\ x + y = 0 \end{array}$ と $(3, −3, −5)$

8. $\begin{array}{l} 6 x −7 y + z = 2 \\ - x - y + 3 z = 4 \\ 2 x + y - z = 1 \end{array}$ and $(4, 2, −6)$

8. $\begin{array}{l} 6 x −7 y + z = 2 \\ - x - y + 3 z = 4 \\ 2 x + y - z = 1 \end{array}$ と $(4, 2, −6)$

9. $\begin{array}{l} x - y = 0 \\ x - z = 5 \\ x - y + z = −1 \end{array}$ and $(4, 4, −1)$

9. $\begin{array}{l} x - y = 0 \\ x - z = 5 \\ x - y + z = −1 \end{array}$ と $(4, 4, −1)$

10. $\begin{array}{l} - x - y + 2 z = 3 \\ 5 x + 8 y −3 z = 4 \\ - x + 3 y −5 z = −5 \end{array}$ and $(4, 1, −7)$

10. $\begin{array}{l} - x - y + 2 z = 3 \\ 5 x + 8 y −3 z = 4 \\ - x + 3 y −5 z = −5 \end{array}$ と $(4, 1, −7)$

For the following exercises, solve each system by elimination.

次の各問について、各組を消去で解け。

11. $\begin{array}{l} 3 x −4 y + 2 z = −15 \\ 2 x + 4 y + z = 16 \\ 2 x + 3 y + 5 z = 20 \end{array}$

12. $\begin{array}{l} 5 x −2 y + 3 z = 20 \\ 2 x −4 y −3 z = −9 \\ x + 6 y −8 z = 21 \end{array}$

13. $\begin{array}{l} 5 x + 2 y + 4 z = 9 \\ −3 x + 2 y + z = 10 \\ 4 x −3 y + 5 z = −3 \end{array}$

14. $\begin{array}{l} 4 x −3 y + 5 z = 31 \\ - x + 2 y + 4 z = 20 \\ x + 5 y −2 z = −29 \end{array}$

15. $\begin{array}{l} 5 x −2 y + 3 z = 4 \\ −4 x + 6 y −7 z = −1 \\ 3 x + 2 y - z = 4 \end{array}$

16. $\begin{array}{l} 4 x + 6 y + 9 z = 0 \\ −5 x + 2 y −6 z = 3 \\ 7 x −4 y + 3 z = −3 \end{array}$

For the following exercises, solve each system by Gaussian elimination.

次の各問について、各組をガウスの消去法で解け。

17. $\begin{array}{l} 2 x - y + 3 z = 17 \\ −5 x + 4 y −2 z = −46 \\ 2 y + 5 z = −7 \end{array}$

18. $\begin{array}{l} 5 x −6 y + 3 z = 50 \\ - x + 4 y = 10 \\ 2 x - z = 10 \end{array}$

19. $\begin{array}{l} 2 x + 3 y −6 z = 1 \\ −4 x −6 y + 12 z = −2 \\ x + 2 y + 5 z = 10 \end{array}$

20. $\begin{array}{l} 4 x + 6 y −2 z = 8 \\ 6 x + 9 y −3 z = 12 \\ −2 x −3 y + z = −4 \end{array}$

21. $\begin{array}{l} 2 x + 3 y −4 z = 5 \\ −3 x + 2 y + z = 11 \\ - x + 5 y + 3 z = 4 \end{array}$

22. $\begin{array}{l} 10 x + 2 y −14 z = 8 \\ −x −2 y −4 z = −1 \\ −12 x −6 y + 6 z = −12 \end{array}$

23. $\begin{array}{l} x + y + z = 14 \\ 2 y + 3 z = −14 \\ −16 y −24 z = −112 \end{array}$

24. $\begin{array}{l} 5 x −3 y + 4 z = −1 \\ −4 x + 2 y −3 z = 0 \\ −x + 5 y + 7 z = −11 \end{array}$

25. $\begin{array}{l} x + y + z = 0 \\ 2 x - y + 3 z = 0 \\ x - z = 0 \end{array}$

26. $\begin{array}{l} 3 x + 2 y −5 z = 6 \\ 5 x −4 y + 3 z = −12 \\ 4 x + 5 y −2 z = 15 \end{array}$

27. $\begin{array}{l} x + y + z = 0 \\ 2 x - y + 3 z = 0 \\ x - z = 1 \end{array}$

28. $\begin{array}{l} \begin{array}{l} 3 x - \frac{1}{2} y - z = - \frac{1}{2} \end{array} \\ 4 x + z = 3 \\ - x + \frac{3}{2} y = \frac{5}{2} \end{array}$

29. $\begin{array}{l} 6 x −5 y + 6 z = 38 \\ \frac{1}{5} x - \frac{1}{2} y + \frac{3}{5} z = 1 \\ −4 x - \frac{3}{2} y - z = −74 \end{array}$

30. $\begin{array}{l} \frac{1}{2} x - \frac{1}{5} y + \frac{2}{5} z = - \frac{13}{10} \\ \frac{1}{4} x - \frac{2}{5} y - \frac{1}{5} z = - \frac{7}{20} \\ - \frac{1}{2} x - \frac{3}{4} y - \frac{1}{2} z = - \frac{5}{4} \end{array}$

31. $\begin{array}{l} - \frac{1}{3} x - \frac{1}{2} y - \frac{1}{4} z = \frac{3}{4} \\ - \frac{1}{2} x - \frac{1}{4} y - \frac{1}{2} z = 2 \\ - \frac{1}{4} x - \frac{3}{4} y - \frac{1}{2} z = - \frac{1}{2} \end{array}$

32. $\begin{array}{l} \frac{1}{2} x - \frac{1}{4} y + \frac{3}{4} z = 0 \\ \frac{1}{4} x - \frac{1}{10} y + \frac{2}{5} z = −2 \\ \frac{1}{8} x + \frac{1}{5} y - \frac{1}{8} z = 2 \end{array}$

33. $\begin{array}{l} \frac{4}{5} x - \frac{7}{8} y + \frac{1}{2} z = 1 \\ - \frac{4}{5} x - \frac{3}{4} y + \frac{1}{3} z = −8 \\ - \frac{2}{5} x - \frac{7}{8} y + \frac{1}{2} z = −5 \end{array}$

34. $\begin{array}{l} - \frac{1}{3} x - \frac{1}{8} y + \frac{1}{6} z = - \frac{4}{3} \\ - \frac{2}{3} x - \frac{7}{8} y + \frac{1}{3} z = - \frac{23}{3} \\ - \frac{1}{3} x - \frac{5}{8} y + \frac{5}{6} z = 0 \end{array}$

35. $\begin{array}{l} - \frac{1}{4} x - \frac{5}{4} y + \frac{5}{2} z = −5 \\ - \frac{1}{2} x - \frac{5}{3} y + \frac{5}{4} z = \frac{55}{12} \\ - \frac{1}{3} x - \frac{1}{3} y + \frac{1}{3} z = \frac{5}{3} \end{array}$

36. $\begin{array}{l} \frac{1}{40} x + \frac{1}{60} y + \frac{1}{80} z = \frac{1}{100} \\ - \frac{1}{2} x - \frac{1}{3} y - \frac{1}{4} z = - \frac{1}{5} \\ \frac{3}{8} x + \frac{3}{12} y + \frac{3}{16} z = \frac{3}{20} \end{array}$

37. $\begin{array}{l} 0.1 x −0.2 y + 0.3 z = 2 \\ 0.5 x −0.1 y + 0.4 z = 8 \\ 0.7 x −0.2 y + 0.3 z = 8 \end{array}$

38. $\begin{array}{l} 0.2 x + 0.1 y −0.3 z = 0.2 \\ 0.8 x + 0.4 y −1.2 z = 0.1 \\ 1.6 x + 0.8 y −2.4 z = 0.2 \end{array}$

39. $\begin{array}{l} 1.1 x + 0.7 y −3.1 z = −1.79 \\ 2.1 x + 0.5 y −1.6 z = −0.13 \\ 0.5 x + 0.4 y −0.5 z = −0.07 \end{array}$

40. $\begin{array}{l} 0.5 x −0.5 y + 0.5 z = 10 \\ 0.2 x −0.2 y + 0.2 z = 4 \\ 0.1 x −0.1 y + 0.1 z = 2 \end{array}$

41. $\begin{array}{l} 0.1 x + 0.2 y + 0.3 z = 0.37 \\ 0.1 x −0.2 y −0.3 z = −0.27 \\ 0.5 x −0.1 y −0.3 z = −0.03 \end{array}$

42. $\begin{array}{l} 0.5 x −0.5 y −0.3 z = 0.13 \\ 0.4 x −0.1 y −0.3 z = 0.11 \\ 0.2 x −0.8 y −0.9 z = −0.32 \end{array}$

43. $\begin{array}{l} 0.5 x + 0.2 y −0.3 z = 1 \\ 0.4 x −0.6 y + 0.7 z = 0.8 \\ 0.3 x −0.1 y −0.9 z = 0.6 \end{array}$

44. $\begin{array}{l} 0.3 x + 0.3 y + 0.5 z = 0.6 \\ 0.4 x + 0.4 y + 0.4 z = 1.8 \\ 0.4 x + 0.2 y + 0.1 z = 1.6 \end{array}$

45. $\begin{array}{l} 0.8 x + 0.8 y + 0.8 z = 2.4 \\ 0.3 x −0.5 y + 0.2 z = 0 \\ 0.1 x + 0.2 y + 0.3 z = 0.6 \end{array}$

Extensions発展問題

For the following exercises, solve the system for $x, y,$ and $z .$

次の各問について、この組を $x, y$ と $z$ について解け。

46. $\begin{array}{l} x + y + z = 3 \\ \frac{x −1}{2} + \frac{y −3}{2} + \frac{z + 1}{2} = 0 \\ \frac{x −2}{3} + \frac{y + 4}{3} + \frac{z −3}{3} = \frac{2}{3} \end{array}$

47. $\begin{array}{l} 5 x −3 y - \frac{z + 1}{2} = \frac{1}{2} \\ 6 x + \frac{y −9}{2} + 2 z = −3 \\ \frac{x + 8}{2} −4 y + z = 4 \end{array}$

48. $\begin{array}{l} \frac{x + 4}{7} - \frac{y −1}{6} + \frac{z + 2}{3} = 1 \\ \frac{x −2}{4} + \frac{y + 1}{8} - \frac{z + 8}{12} = 0 \\ \frac{x + 6}{3} - \frac{y + 2}{3} + \frac{z + 4}{2} = 3 \end{array}$

49. $\begin{array}{l} \frac{x −3}{6} + \frac{y + 2}{2} - \frac{z −3}{3} = 2 \\ \frac{x + 2}{4} + \frac{y −5}{2} + \frac{z + 4}{2} = 1 \\ \frac{x + 6}{2} - \frac{y −3}{2} + z + 1 = 9 \end{array}$

50. $\begin{array}{l} \frac{x −1}{3} + \frac{y + 3}{4} + \frac{z + 2}{6} = 1 \\ 4 x + 3 y −2 z = 11 \\ 0.02 x + 0.015 y −0.01 z = 0.065 \end{array}$

Real-World Applications現実世界への応用

51. Three even numbers sum up to 108. The smaller is half the larger and the middle number is $\frac{3}{4}$ the larger. What are the three numbers?

51. 三つの偶数の和が108である。小さいほうは大きいほうの半分で、真ん中の数は大きいほうの $\frac{3}{4}$ である。この三つの数はいくらか。

52. Three numbers sum up to 147. The smallest number is half the middle number, which is half the largest number. What are the three numbers?

52. 三つの数の和が147である。いちばん小さい数は真ん中の数の半分で、真ん中の数はいちばん大きい数の半分である。この三つの数はいくらか。

53. At a family reunion, there were only blood relatives, consisting of children, parents, and grandparents, in attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance?

53. ある親族の集まりには、子・親・祖父母からなる血縁者だけが出席していた。合計400人だった。親は祖父母の2倍おり、子は親より50人多かった。子・親・祖父母はそれぞれ何人出席していたか。

54. An animal shelter has a total of 350 animals comprised of cats, dogs, and rabbits. If the number of rabbits is 5 less than one-half the number of cats, and there are 20 more cats than dogs, how many of each animal are at the shelter?

54. ある動物保護所には、猫・犬・兎からなる合計350匹の動物がいる。兎の数が猫の数の半分より5少なく、猫が犬より20多いとすると、この保護所には各動物が何匹ずついるか。

55. Your roommate, Shani, offered to buy groceries for you and your other roommate. The total bill was $82. She forgot to save the individual receipts but remembered that your groceries were $0.05 cheaper than half of her groceries, and that your other roommate’s groceries were $2.10 more than your groceries. How much was each of your share of the groceries?

55. 同居人のシャニが、あなたともう一人の同居人のために食料品を買うと申し出た。請求額の合計は82ドルだった。彼女はそれぞれの控えを取っておくのを忘れたが、あなたの食料品が彼女の食料品の半分より0.05ドル安く、もう一人の同居人の食料品があなたの食料品より2.10ドル高かったことは覚えていた。三人それぞれの食料品の分はいくらだったか。

56. Your roommate, John, offered to buy household supplies for you and your other roommate. You live near the border of three states, each of which has a different sales tax. The total amount of money spent was $100.75. Your supplies were bought with 5% tax, John’s with 8% tax, and your third roommate’s with 9% sales tax. The total amount of money spent without taxes is $93.50. If your supplies before tax were $1 more than half of what your third roommate’s supplies were before tax, how much did each of you spend? Give your answer both with and without taxes.

56. 同居人のジョンが、あなたともう一人の同居人のために家庭用品を買うと申し出た。あなたたちは三つの州の境の近くに住んでおり、各州の売上税は違う。使った金額の合計は100.75ドルだった。あなたの品は5%、ジョンの品は8%、三人目の同居人の品は9%の売上税で買われた。税を除いた金額の合計は93.50ドルである。税抜きのあなたの品が、税抜きの三人目の同居人の品の半分より1ドル高かったとすると、それぞれいくら使ったか。答えは税込みと税抜きの両方で示せ。

57. Three coworkers work for the same employer. Their jobs are warehouse manager, office manager, and truck driver. The sum of the annual salaries of the warehouse manager and office manager is $82,000. The office manager makes $4,000 more than the truck driver annually. The annual salaries of the warehouse manager and the truck driver total $78,000. What is the annual salary of each of the co-workers?

57. 三人の同僚が同じ雇い主のもとで働いている。仕事は倉庫の管理者、事務所の管理者、貨物車の運転手である。倉庫の管理者と事務所の管理者の年俸の和は82,000ドルである。事務所の管理者は貨物車の運転手より年に4,000ドル多く稼ぐ。倉庫の管理者と貨物車の運転手の年俸の合計は78,000ドルである。各同僚の年俸はいくらか。

58. At a carnival, $2,914.25 in receipts were taken at the end of the day. The cost of a child’s ticket was $20.50, an adult ticket was $29.75, and a senior citizen ticket was $15.25. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold?

58. ある祭りで、一日の終わりに2,914.25ドルの売り上げがあった。子ども券の値段は20.50ドル、大人券は29.75ドル、高齢者券は15.25ドルだった。入場した高齢者は大人の2倍で、子どもは高齢者より20人多かった。子ども券・大人券・高齢者券は何枚ずつ売れたか。

59. A local band sells out for their concert. They sell all 1,175 tickets for a total purse of $28,112.50. The tickets were priced at $20 for student tickets, $22.50 for children, and $29 for adult tickets. If the band sold twice as many adult as children tickets, how many of each type was sold?

59. ある地元の楽団の演奏会が満員になった。1,175枚の券をすべて売り、売り上げの合計は28,112.50ドルだった。券の値段は学生券20ドル、子ども券22.50ドル、大人券29ドルだった。この楽団が子ども券の2倍の大人券を売ったとすると、各種類は何枚ずつ売れたか。

60. In a bag, a child has 325 coins worth $19.50. There were three types of coins: pennies, nickels, and dimes. If the bag contained the same number of nickels as dimes, how many of each type of coin was in the bag?

60. ある子どもが袋に19.50ドル分の硬貨を325枚持っている。硬貨は1セント・5セント・10セントの三種類だった。袋の中の5セント硬貨と10セント硬貨の枚数が同じだったとすると、各種類の硬貨は何枚ずつ入っていたか。

61. Last year, at Haven’s Pond Car Dealership, for a particular model of BMW, Jeep, and Toyota, one could purchase all three cars for a total of $140,000. This year, due to inflation, the same cars would cost $151,830. The cost of the BMW increased by 8%, the Jeep by 5%, and the Toyota by 12%. If the price of last year’s Jeep was $7,000 less than the price of last year’s BMW, what was the price of each of the three cars last year?

61. 去年、ヘイブンズ・ポンド自動車販売店では、BMW・ジープ・トヨタのある型について、三台すべてを合計140,000ドルで買えた。今年は物価の上昇のため、同じ車が151,830ドルになる。BMWの値段は8%、ジープは5%、トヨタは12%上がった。去年のジープの値段が去年のBMWの値段より7,000ドル安かったとすると、去年の三台それぞれの値段はいくらだったか。

62. When his youngest child moved out, Deandre sold his home and made three investments using gains from the sale. He invested $80,500 into three accounts, one that paid 4% simple interest, one that paid $3 \frac{1}{8} %$ simple interest, and one that paid $2 \frac{1}{2} %$ simple interest. He earned $2,670 interest at the end of one year. If the amount of the money invested in the second account was four times the amount invested in the third account, how much was invested in each account?

62. 末子が家を出たとき、ディアンドレは家を売り、売却益を使って三つの投資をした。彼は80,500ドルを三つの口座に投じた。一つは単利4%、一つは単利 $3 \frac{1}{8} %$ 、一つは単利 $2 \frac{1}{2} %$ である。一年の終わりに2,670ドルの利息を得た。二つ目の口座に投じた額が三つ目の口座に投じた額の4倍だったとすると、各口座にいくらずつ投じたか。

63. You inherit one million dollars. You invest it all in three accounts for one year. The first account pays 3% compounded annually, the second account pays 4% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $34,000 in interest. If you invest four times the money into the account that pays 3% compared to 2%, how much did you invest in each account?

63. あなたは百万ドルを相続する。それをすべて三つの口座に一年間投じる。一つ目の口座は年複利3%、二つ目の口座は年複利4%、三つ目の口座は年複利2%である。一年後、34,000ドルの利息を得る。2%の口座に比べて3%の口座に4倍の金を投じたとすると、各口座にいくらずつ投じたか。

64. An entrepreneur sells a portion of their business for one hundred thousand dollars and invests it all in three accounts for one year. The first account pays 4% compounded annually, the second account pays 3% compounded annually, and the third account pays 2% compounded annually. After one year, the entrepreneur earns $3,650 in interest. If they invested five times the money in the account that pays 4% compared to 3%, how much did they invest in each account?

64. ある起業家が自分の事業の一部を十万ドルで売り、それをすべて三つの口座に一年間投じる。一つ目の口座は年複利4%、二つ目の口座は年複利3%、三つ目の口座は年複利2%である。一年後、この起業家は3,650ドルの利息を得る。3%の口座に比べて4%の口座に5倍の金を投じたとすると、各口座にいくらずつ投じたか。

65. The top three countries in oil consumption in a certain year are as follows: the United States, Japan, and China. In millions of barrels per day, the three top countries consumed 39.8% of the world’s consumed oil. The United States consumed 0.7% more than four times China’s consumption. The United States consumed 5% more than triple Japan’s consumption. What percent of the world oil consumption did the United States, Japan, and China consume?1

65. ある年の石油消費量の上位三か国は次のとおりである。合衆国、日本、中国。一日あたりの百万バレルで見ると、この上位三か国は世界で消費された石油の39.8%を消費した。合衆国は中国の消費量の4倍より0.7%多く消費した。合衆国は日本の消費量の3倍より5%多く消費した。合衆国、日本、中国は世界の石油消費量の何パーセントずつを消費したか。1

“Oil reserves, production and consumption in 2001,” accessed April 6, 2014,

http://scaruffi.com/politics/oil.html

66. The top three countries in oil production in the same year are Saudi Arabia, the United States, and Russia. In millions of barrels per day, the top three countries produced 31.4% of the world’s produced oil. Saudi Arabia and the United States combined for 22.1% of the world’s production, and Saudi Arabia produced 2% more oil than Russia. What percent of the world oil production did Saudi Arabia, the United States, and Russia produce?2

66. 同じ年の石油生産量の上位三か国はサウジアラビア、合衆国、ロシアである。一日あたりの百万バレルで見ると、この上位三か国は世界で生産された石油の31.4%を生産した。サウジアラビアと合衆国を合わせると世界の生産量の22.1%で、サウジアラビアはロシアより2%多く石油を生産した。サウジアラビア、合衆国、ロシアは世界の石油生産量の何パーセントずつを生産したか。2

“Oil reserves, production and consumption in 2001,” accessed April 6, 2014,

http://scaruffi.com/politics/oil.html

67. The top three sources of oil imports for the United States in the same year were Saudi Arabia, Mexico, and Canada. The three top countries accounted for 47% of oil imports. The United States imported 1.8% more from Saudi Arabia than they did from Mexico, and 1.7% more from Saudi Arabia than they did from Canada. What percent of the United States oil imports were from these three countries?3

67. 同じ年の合衆国の石油輸入元の上位三か国はサウジアラビア、メキシコ、カナダだった。この上位三か国が石油輸入の47%を占めた。合衆国はメキシコからより1.8%多くサウジアラビアから輸入し、カナダからより1.7%多くサウジアラビアから輸入した。合衆国の石油輸入のうち、この三か国からのものは何パーセントずつだったか。3

“Oil reserves, production and consumption in 2001,” accessed April 6, 2014,

http://scaruffi.com/politics/oil.html

68. The top three oil producers in the United States in a certain year are the Gulf of Mexico, Texas, and Alaska. The three regions were responsible for 64% of the United States oil production. The Gulf of Mexico and Texas combined for 47% of oil production. Texas produced 3% more than Alaska. What percent of United States oil production came from these regions?4

68. ある年の合衆国の石油生産地の上位三つはメキシコ湾、テキサス、アラスカである。この三つの地域が合衆国の石油生産の64%を担った。メキシコ湾とテキサスを合わせると石油生産の47%だった。テキサスはアラスカより3%多く生産した。合衆国の石油生産のうち、これらの地域からのものは何パーセントずつか。4

“USA: The coming global oil crisis,” accessed April 6, 2014,

http://www.oilcrisis.com/us/

69. At one time, in the United States, 398 species of animals were on the endangered species list. The top groups were mammals, birds, and fish, which comprised 55% of the endangered species. Birds accounted for 0.7% more than fish, and fish accounted for 1.5% more than mammals. What percent of the endangered species came from mammals, birds, and fish?

69. かつて合衆国では、398種の動物が絶滅危惧種の一覧に載っていた。上位の群は哺乳類・鳥類・魚類で、絶滅危惧種の55%を占めた。鳥類は魚類より0.7%多く、魚類は哺乳類より1.5%多かった。絶滅危惧種のうち哺乳類・鳥類・魚類はそれぞれ何パーセントだったか。

70. Meat consumption in the United States can be broken into three categories: red meat, poultry, and fish. If fish makes up 4% less than one-quarter of poultry consumption, and red meat consumption is 18.2% higher than poultry consumption, what are the percentages of meat consumption?5

70. 合衆国の食肉の消費は三つに分けられる。赤身肉、家禽肉、魚である。魚が家禽肉の消費量の4分の1より4%少なく、赤身肉の消費量が家禽肉の消費量より18.2%多いとすると、食肉消費の割合はそれぞれ何パーセントか。5

“The United States Meat Industry at a Glance,” accessed April 6, 2014,

http://www.meatami.com/ht/d/sp/i/47465/pid/47465