Class 11 RD Sharma解决方案–第15章线性不等式–练习15.2 |套装2

问题11：在R中求解以下每个方程组：4x – 1≤0，3 – 4x <0

解决方案：

Let the first equation be 4x – 1 ≤ 0

⇒ 4x ≤ 1

⇒ x ≤ 1/4

and the second equation be 3 – 4x < 0

⇒ 4x > 3

⇒ x > 3/4

According to first equation, x lies in range ( -∞, 1/4 ] and according to second equation, x lies in range ( 3/4, -∞ )

Calculating the intersection of these two intervals we get no value for x. Therefore, given set of inequations has no solution.

问题12：在R中求解以下每个方程组：x + 5> 2(x + 1)，2 – x <3(x + 2)

解决方案：

Let the first equation be x + 5 > 2(x+1)

⇒ x + 5 > 2x + 2

⇒ x < 3

and the second equation be 2 – x < 3(x+2)

⇒ 2 – x < 3x + 6

⇒ -4 < 4x

⇒ x > -1

Hence using above equations, we know x lies in range (-1,3)

问题13：在R中求解以下每个方程组：2(x – 6)<3x – 7，11 – 2x <6 – x

解决方案：

Let the first equation be 2(x – 6) < 3x – 7

⇒ 2x – 12 < 3x – 7

⇒ -x < 5

⇒ x > -5

and the second equation be 11 – 2x < 6 – x

⇒ -x < -5

⇒ x > 5

Hence using above equations, we know x lies in range (5,∞)

问题14.在R中求解以下每个方程组：5x – 7 <3(x + 3)，1 – $\frac{3x}{2}$ ≥x – 4

解决方案：

Let the first equation be 5x – 7 < 3 (x + 3)

⇒ 5x-7 < 3x + 9

⇒ 2x < 16

⇒ x < 8

and the second equation be1 –$\frac{3x}{2}$ ≥ x – 4

⇒ x + $\frac{3x}{2}$ ≤ 5

⇒ $\frac{5x}{2}$ < 5

⇒ \frac{x}{2} ≤ 1

⇒ x ≤ 2

Hence using above equations, we know x lies in range [-∞, 2 ]

问题15：求解R中的以下每个方程组： $\frac{2x - 3}{4}$ -2≥ $\frac{4x}{3}$ – 6、2(2x + 3)<6(x – 2)+ 10

解决方案：

Let the first equation be $\frac{2x - 3}{4}$ -2 ≥ $\frac{4x}{3}$ – 6

⇒ $\frac{2x - 3}{4}$ – $\frac{4x}{3}$ ≥ – 6 + 2

⇒ $\frac{3(2x-3)-16x}{12}$ ≥ -4

⇒ 6x – 9 -16x ≥ -48

⇒ 10x ≤ 39

⇒ x ≤ $\frac{39}{10}$

and the second equation be 2(2x + 3) < 6(x – 2) + 10

⇒ 4x + 6 < 6x -12 + 10

⇒ -2x < -8

⇒ x > 4

According to first equation, x lies in range ( -∞, 39/10 ] and according to second equation, x lies in range ( 4, ∞ )

Calculating the intersection of these two intervals we get no value for x. Therefore, given set of inequations has no solution.

问题16：求解R中的以下每个方程组： $\frac{7x -1}{2}$ <-3， $\frac{3x + 8}{5}$ + 11 <0

解决方案：

Let the first equation be $\frac{7x -1}{2}$ < -3

⇒ 7x – 1 < -6

⇒ 7x < -5

⇒ x < -5/7

and the second equation be $\frac{3x + 8}{5}$ + 11 < 0

⇒ 3x + 8 < -55

⇒ 3x < -63

⇒ x < -21

Hence using above equations, we know x lies in range ( -∞, -21)

问题17：在R中求解以下每个方程组： $\frac{2x + 1}{7x - 1}$ > 5 $\frac{x + 7}{x - 8}$ > 2

解决方案：

Let the first equation be $\frac{2x + 1}{7x - 1}$ > 5

⇒ 2x + 1 > 5 (7x -1)

⇒ 2x – 35x > -6

⇒  – 33x > -6

⇒ x < 2/11

Also, 7x -1 > 0 ⇒ x > 1/7

using first equation we get, x lies in range $(\frac{1}{7}, \frac{2}{11})$

and let the second equation be $\frac{x + 7}{x - 8}$ > 2

⇒ x + 7 >  2x – 16

⇒ 23 >  x

⇒ x < 23

Also, x – 8 > 0 ⇒ x > 8

Hence using above equations, we know x lies in range ( 8, -23 )

Calculating the intersection of the two intervals we get after solving equation 1 and equation 2 we get no value for x. Therefore, given set of inequations has no solution.

问题18：在R中求解以下每个方程组：0 < $\frac{-x}{2}$ <3

解决方案：

Using the equation, 0 < $\frac{-x}{2}$ < 3

⇒ 0 < -x < 6

⇒ 0 > x > -6

⇒ x > -5

Hence using above equation, we know x lies in range ( -6, 0 )

问题19：在R中求解以下每个方程组：10≤-5(x – 2)<20

解决方案：

Let the first equation be 10 ≤ -5 (x – 2) < 20

⇒ 10 ≤ -5x + 10 < 20

⇒ 0 ≤ -5x < 10

⇒ 0 ≤ x < -2

Hence using above equations, we know x lies in range ( -2, 0 ]

问题20：在R中求解以下每个方程组：-5 <2x -3 <5

解决方案：

Using the equation -5 < 2x -3 < 5

⇒ -2 < 2x < 8

⇒ -1 < x < 4

Hence using above equations, we know x lies in range ( -1, 4 )

问题21.解决R中的以下每个方程组： $\frac{4}{x+1}$ ≤3≤ $\frac{6}{x+1}$ ，x> 0

解决方案：

Using the equation $\frac{4}{x+1}$ ≤ 3 ≤ $\frac{6}{x+1}$, x>0

⇒ 4 ≤ 3 (x+1) ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

⇒ 1 ≤ 3x ≤ 3

⇒ 1/3 ≤ x ≤ 1

Hence using above equations, we know x lies in range $\left [ \frac{1}{3}, 1 \right ]$