第 10 类 RD Sharma 解 – 第 3 章两变量线性方程组 – 练习 3.2 |设置 3

问题 23. 以图形方式求解以下线性方程组，并在两条线和 x 轴之间的区域涂上阴影。

(i) 2x + 3y = 12 和 x – y = 1

解决方案：

Given that, 2x + 3y = 12 and x – y = 1

Now, 2x + 3y = 12

x = (12-3y)/2

When y = 2, we get x = 3

When y = 4, we get x = 0

So, the following table giving points on the line 2x + 3y = 12

x	0	3
y	4	2

Now, x – y = 1

x = y + 1

When y = 0, we get x = 1

When y = 1, we get x = 2

So, the following table giving points on the line x – y = 1

x	1	2
y	0	1

So, the graph of the equations 2x + 3y = 12 and x – y = 1:

From the graph we conclude that the two lines intersect at P (3, 2)

Hence, x = 3 and y = 2 is the solution of the given system of equations.

(ii) 3x + 2y – 4 = 0 和 2x – 3y – 7 = 0

解决方案：

Given that, 3x + 2y – 4 = 0 and 2x – 3y – 7 = 0

Now, 3x + 2y – 4 = 0

x = (4 – 2y)/3

When y = 5, we get x = – 2

When y = 8, we get x = – 4

So, the following table giving points on the line 3x + 2y – 4 = 0

x	-2	-4
y	5	8

We have,

2x – 3y – 7 = 0

x = (3y + 7)/2

When y = 1, we get x = 5

When y = -1, we get x = 2

So, the following table giving points on the line 2x – 3y – 7 = 0

x	5	2
y	1	-1

So, the graph of the equations 3x + 2y – 4 = 0 and 2x – 3y – 7 = 0:

From the graph we conclude that the two lines intersect at P(2,-1)

Hence, x = 2 and y = -1 is the solution of the given system of equations.

(iii) 3x + 2y – 11 = 0 和 2x – 3y + 10 = 0

解决方案：

Given that, 3x + 2y – 11 = 0 and 2x – 3y + 10 = 0

Now, 3x + 2y – 11 = 0

x = (11 – 2y)/3

When y = 1, we get x = – 3

When y = 4, we get x = 1

So, the following table giving points on the line 3x + 2y – 11 = 0 

x	3	1
y	1	4

Now, 

2x – 3y + 10 = 0

x = (3y-10)/2

When y = 0, we get x = – 5

When y = 2, we get x = – 2

So, the following table giving points on the line 2x – 3y + 10 = 0

x	-5	-2
y	0	2

So, the graph of the equations 3x + 2y – 11 = 0 and 2x – 3y + 10 = 0:

From the graph we conclude that the two lines intersect at P(1, 4)

Hence, x = 1 and y = 4 is the solution of the given system of equations.

问题 24. 在同一张方格纸上画出下列方程的图形：

2x + 3y = 12 和 x – y = 1

解决方案：

Given that, 2x + 3y = 12 and x – y = 1

Now, 2x + 3y = 12

x = (12 – 3y)/2

When y = 0, we get x = 6

When y = 2, we get x = 3

So, the following table giving points on the line 2x + 3y = 12:

x	6	3
y	0	2

Now,

x – y = 1

x = 1 + y

When y = 0, we get x = 1

When y = -1, we get x = 0

So, the following table giving points on the line x – y = 1:

x	1	0
y	0	-1

So, the graph of the equations 2x + 3y = 12 and x – y = 1:

From the graph we conclude that the two lines intersect at A (3, 2)

Also, we observe that the lines meet y-axis B (0, – 1) and C (0, 4)

Hence, the vertices of the required triangle are A (3, 2), B (0,-1), and C (0, 4).

问题 25. 画出 x – y + 1 = 0 和 3x + 2y – 12 = 0 的图形。确定由这些线和 x 轴组成的三角形的顶点坐标，并对三角形区域进行阴影处理。计算以这些线和 x 轴为界的区域。

解决方案：

Given that, x – y + 1 = 0 and 3x + 2y – 12 = 0

Now, x – y + 1 = 0

x = y – 1

When y = 3, we get x = 2

When y = -1, we get x = -2

So, the following table giving points on the line x – y + 1 = 0

x	2	-2
y	3	-1

We have,

3x + 2y – 12 = 0

x = (12 – 2y)/3

When y = 6, we get x = 0

When y = 3, we get x = 2

So, the following table giving points on the line 3x + 2y – 12 = 0

x	0	2
y	6	3

So, the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0:

From the graph we conclude that the two lines intersect at A(2, 3)

Also, we observe that the lines meet x-axis B(-1, 0) and C(4, 0)

So, x = 2 and y = 3 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. 

So, 

AD = y-coordinate point A (2, 3)

AD = 3 and BC = 4- ( – 1) = 4 + 1 = 5

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 5 × 3 = 15/2 = 7.5 sq. units

问题 26. 用图解法求解线性方程组：

4x – 3y + 4 = 0 和 4x + 3y – 20 = 0

找到由这些线和 x 轴界定的区域。

解决方案：

Given that, 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0

Now, 4x – 3y + 4 = 0

x = (3y – 4)/4

When y = 0, we get x = -1

When y = 4, we get x = 2

So, the following table giving points on the line 4x – 3y + 4 = 0

x	2	-1
y	4	0

Now, 

4x + 3y – 20 = 0

x = (20 – 3y)/4

When y = 0, we get x = 5

When y = 4, we get x = 2

So, the following table giving points on the line 4x + 3y – 20 = 0

x	5	2
y	0	4

So, the graph of the equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0:

From the graph we conclude that the two lines intersect at A (2, 4)

Also, we observe that the lines meet x-axis B (-1, 0) and C(5, 0)

So, x = 2 and y = 4 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. 

So, we get

AD = y-coordinate point A (2, 4)

AD = 3 and BC = 5 – (-1) = 4 + 1 = 6

Hence, the area of the shaded region =1/2 × base × altitude

1/2 × 6 × 4 = 12 sq. units

问题 27. 以图形方式求解以下线性方程组：

3x + y – 11 = 0 和 x – y – 1 = 0

阴影由这些线和 y 轴界定的区域。此外，找到由这些线和 y 轴界定的区域的面积。

解决方案：

Given that, 3x + y – 11 = 0 and x – y – 1 = 0

Now, 3x + y – 11 = 0

y = 11 – 3x

When x = 0, we get y = 11

When x = 3, we get y = 2

So, the following table giving points on the line 3x + y – 11 = 0

x	0	3
y	11	2

We have

x – y – 1 = 0

y = x – 1

When x = 0, we get y = -1

When x = 3, we get y = 2

So, the following table giving points on the line x – y – 1 = 0

x	0	3
y	-1	2

So, the graph of the equations 3x + y – 11 = 0 and x – y – 1 = 0:

From the graph we conclude that, the two lines intersect at A (3, 2)

We also observe that the lines meet y-axis B(0, 11) and C (0, – 1)

So, x = 3 and y = 2 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. Clearly we have,

AD = y-coordinate point A(2, 4)

AD = 3 and BC = 11- (- 1) = 11 + 1 = 12

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 12 × 3 = 18 sq. units

问题 28. 画出下列等式的图形：

2x – 3y + 6 = 0,

2x + 3y – 18 = 0，

y – 2 = 0

找到这样获得的三角形的顶点。另外，求三角形的面积。

解决方案：

Given that, 2x – 3y + 6 = 0, 2x + 3y – 18 = 0, and y – 2 = 0

Now, 2x – 3y + 6 = 0

x = (3y – 6)/2

When y = 0, we get x = -3

When y = 2, we get x = 0

So, the following table giving points on the line  2x – 3y + 6 = 0

x	-3	0
y	0	2

Now,

2x + 3y – 18 = 0

x = 18-3y / 2

When y = 2, we get x = 6

When y = 6, we get x = 0

So, the following table giving points on the line 2x + 3y – 18 = 0

x	6	0
y	2	6

Now, 

y – 2 = 0

y = -2

So, the graph of the equations 2x – 3y + 6 = 0, 2x + 3y – 18 = 0, and y – 2 = 0:

From the graph of three equations, we conclude that the three lines taken in pairs 

intersect each other at points A(3, 4), B(0, 2), and C(6, 2)

Hence, the vertices of the required triangle are (3, 4), (0, 2) and (6, 2)

From the graph, we have

AD = 4 – 2 = 2

BC = 6 – 0 = 6

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 6 × 2 = 6 sq. units

问题 29. 以图形方式求解以下方程组：

2x – 3y + 6 = 0 和 2x + 3y – 18 = 0

此外，找到由这些线和 y 轴界定的区域的面积。

解决方案：

Given that, 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0

Now, 2x – 3y + 6 = 0

y = (2x + 6)/3

When x = 0, we get y = 2

When x = -3, we get y = 0

So, the following table giving points on the line 2x – 3y + 6 = 0 

x	0	-3
y	2	6

Now, 

2x + 3y – 18 = 0

x = (18 – 3y)/2

When y = 2, we get x = 6

When y = 6, we get x = 0

So, the following table giving points on the line 2x + 3y – 18 = 0

x	6	0
y	2	6

So, the graph of the equations 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0:

From the graph we conclude that, the two lines intersect at A (3, 4).

Hence x = 3 and y = 4 is the solution of the given system of equations.

Also, from the graph, we have

AD = x-coordinate point A(3, 4) = 3

BC = 6 – 2 = 4

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 4 × 3 = 6 sq. units

问题 30. 用图解法求解下列线性方程组；

4x – 5y – 20 = 0 和 3x + 5y – 15 = 0

确定由表示上述方程的线和 y 轴形成的三角形的顶点。

解决方案：

Given that, 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Now, 4x – 5y – 20 = 0

x = (5y + 20)/4

When y = 0, we get x = 5

When y = – 4, we get x = 0

So, the following table giving points on the line 4x – 5y – 20 = 0 

x	5	0
y	0	-4

We have

3x + 5y – 15 = 0

x = (5y + 20)/5

When y = 0, we get x = 5

When y = – 4, we get x = 0

So, the following table giving points on the line 3x + 5y – 15 = 0

x	5	0
y	0	3

So, the graph of the equations 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0:

From the graph we conclude that the two lines intersect at A(5, 0). 

Hence, x – 5, y – 0 is the solution of the given system of equations.

So, the lines meet the y-axis at B(0, -4) and C(0, 3) respectively.

Hence, the vertices of the triangle are (5, 0), (0,- 4), and (0, 3)

问题 31. 画出方程 5x – y = 5 和 3x – y = 3 的图形。通过这些线和 y 轴确定三角形的顶点坐标。计算三角形的面积。

解决方案：

Given that, 5x – y = 5 and 3x – y = 3

Now, 

5x – y = 5

y = 5x – 5

So, the following table giving points on the line 5x – y = 5

x	0	1	2
y	-5	0	5

Now, 3x – y = 3

y = 3x – 3

So, the following table giving points on the line 3x – y = 3

x	0	1	2
y	-3	0	3

So, the graph of the equations 5x – y = 5 and 3x – y = 3:

It can be observed that the required triangle is ABC

Hence, the coordinates of its vertices A (1, 0) B(0, – 3) and C(0, -5)

Now, AB = 3.2, BC = 5.1

s = (a + b + c) /2 = (3.2 + 2 + 5.1)/2 = 5.2

Area of triangle ABC = √s(s – a)(s – b)(s – c)

= √5.2(5.2 – 3.2)(5.2 – 2)(5.2 – 5.1)

= √3.328

= 1.8 sq.unit

问题 32. 形成下列问题中的一对线性方程，并用图解法求其解：

(i) X 班 10 名学生参加了数学测验。如果女孩的数量比男孩的数量多4。找出参加测验的男孩和女孩的人数。

(ii) 5 支铅笔和 7 支钢笔合起来花费 50 卢比，而 7 支铅笔和 5 支钢笔合起来花费 46 卢比。

找出一支铅笔和一支钢笔的成本。

(iii) Champa 去拍卖会购买了一些裤子和裙子。当她的朋友问她每种都买了多少时，她回答说：“裙子的数量是裤子数量的两倍不到两倍”。还有，“裙子的数量是购买裤子数量的四倍不到四倍。”帮助她的朋友找出 Champa 买了多少裤子和裙子。

解决方案：

(i) Let the number of girls and boys in the class be x and y respectively.

According to the question, 

x + y = 10 and x – y = 4 are the given equations

Now, x + y = 10

x = 10 – y

So, the following table giving points on the line x + y = 10

x	4	5	6
y	6	5	4

Now, x – y = 4

x = 4 + y

So, the following table giving points on the line x – y = 4

x	5	4	3
y	1	0	-1

So, the graph of the equations x + y = 10 and x – y = 4:

From the graph we conclude that the two lines intersect each other at point (7, 3).

So, x = 7 and y = 3

Hence, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil and one pen Rs. x and Rs. y respectively.

According to the question, we get,

5x + 7y = 50

7x + 5y = 50

Now, 5x + 7y = 50

x = (50 – 7y)/5

So, the following table giving points on the line 5x + 7y = 50

x	3	10	-4
y	5	0	10

Now, 7x + 5y = 46,

x = (46 – 5y)/7

So, the following table giving points on the line 7x + 5y = 46,

x	8	3	-2
y	-2	5	12

So, the graph of the equations 5x + 7y = 50 and 7x + 5y = 50:

From the graph we conclude that the two lines intersect each other at the point (3, 5)

So, x = 3 and y = 5

Hence, the cost of the pencil and the pen are 3 and 5 respectively.

(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :

y = 2x – 2    …..(i)

y = 4x – 2   ……(ii)

The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. 

So, they are given in the following table.

x	2	0
y = 2x – 2	2	-2

Hence, the graphic representation is as follows:

The two lines intersect at the point (1, 0). 

So, x = 1, y = 0 is the required solution of the pair of linear equations, 

i.e., the number of parts she purchased is 1 and she did not buy any skirt.

问题 33. 以图形方式求解以下方程组：

对线条和 y 轴之间的区域进行着色

(i) 3x – 4y = 7 和 5x + 2y = 3

(ii) 4x – y = 4 和 3x + 2y = 14

解决方案 ：

(i) Given that, 3x – 4y = 7 and 5x + 2y = 3

Now, 3x – 4y = 7,

y = (3x – 7)/4

When x = 1, we get y = – 1

When x = – 3, we get y = – 4

So, the following table giving points on the line 3x – 4y = 7

x	1	-3
y	-1	-4

Now, 5x + 2y = 3,

y = (3 – 5x)/2

When x = 1, we get y = – 1

When x = 3, we get y = – 6

So, the following table giving points on the line 5x + 2y = 3

x	1	3
y	-1	-6

So, the graph of the equations 3x – 4y = 7 and 5x + 2y = 3:

From the graph we conclude that the two lines intersect at A(1,-1)

Hence, x = 1 and y = -1 is the solution of the given system of equations.

(ii) Given that, 4x – y = 4 and 3x + 2y = 14

Now, 4x – y = 4

y = 4x – 4

When x = 0, we get y = – 4

When x = -1, we get y = – 8

So, the following table giving points on the line 4x – y = 4 

x	0	-1
y	-4	-8

Now, 3x + 2y = 14,

y = (14 – 3x)/2

When x = 0, we get y = 7

When x = 4, we get y = 1

So, the following table giving points on the line 3x + 2y = 14

x	0	4
y	7	1

So, the graph of the equations 4x – y = 4 and 3x + 2y = 14:

From the graph we conclude that the two lines intersect at A (2, 4)

Hence, x = 2 and y = 4 is the solution of the given system of equations.

问题 34. 用图形表示以下方程组，并写出直线与 y 轴相交的点的坐标

x + 3y = 6 和 2x – 3y = 12

解决方案：

Given that, x + 3y = 6 and 2x – 3y = 12

Now, x + 3y = 6,

y = (6 – x)/3

When x = 0, we get y = 2

When x = 3, we get y = 1

So, the following table giving points on the line x + 3y = 6 

x	0	3
y	2	1

Now, 2x – 3y = 12,

y = (2x – 12)/3

When x = 0, we get y = – 4

When x = 6, we get y = 0

So, the following table giving points on the line 2x – 3y = 12

x	0	6
y	-4	0

So, the graph of the equations x + 3y = 6 and 2x – 3y = 12

From the graph we conclude that the two lines meet the y-axis at B(0, 2) and C(0, -4) respectively.

Hence, the required coordinates are (0, 2) and (0, -4)

问题 35. 给定线性方程 2x + 3y – 8 = 0，在两个变量中写出另一个，这样形成的对的几何表示是 (i) 相交线 (ii) 平行线 (iii) 重合线。

解决方案：

(i) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be intersecting. 

We must have, a1/a2 ≠ b1/b2

So the other linear equation can be 5x + 6y – 16 = 0

a1/a2 = 2/5

b1/b1 = 3/6 = 1/2

c1/c2 = -8/-16 = 1/2

(ii) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be parallel

We must have, a1/a2 = b1/b2 ≠ c1/c2

So, the other linear equation can be 6x + 9y + 24 = 0

a1/a2 = 2/6 = 1/3

b1/b2 = 3/9 = 1/3

c1/c2 = -8/-24 = 1/3 

(iii) For the two lines a1x + b1x + c1 = 0  and a2x + b2x + c2 = 0 to be coincident 

We must have, a1/a2 = b1/b2 = c1/c2

So, the other linear equation can be 6x + 9y + 24 = 0

a1/a2 = 2/8 = 1/4

b1/b2 = 3/12 = 1/4

c1/c2 = -8/-32 = 1/4