Solution

Consider first two inequalities,

$\frac{4}{x+1} \leq 3 \\ $

$\Rightarrow 4 \leq 3(x+1) \\ $

$\Rightarrow 4 \leq 3 x+3 \\ $

$\Rightarrow 4-3 \leq 3 x \\ $

$\Rightarrow 1 \leq 3 x \\ $

$\therefore x \geq \frac{1}{3} $

$ \Rightarrow \quad 4-3 \leq 3 x \quad \text { subtracting } 3 \text { on both sides } $

and consider last two inequalities,

$ 3 \leq \frac{6}{x+1} $

$ \begin{matrix} \Rightarrow & 3(x+1) \leq 6 \\ \Rightarrow & 3 x+3 \leq 6 \\ \Rightarrow & 3 x \leq 6-3 \\ \Rightarrow & 3 x \leq 3 \\ \therefore & x \leq 1 \end{matrix} $

[subtracting 3 to both sides]

[dividing by 3 ]

…(ii)

From Eqs. (i) and (ii),

$ \begin{aligned} & x \in \frac{1}{3}, 1 \\ & \frac{1}{3} \leq x \leq 1 \end{aligned} $

Solution

Let

$ \begin{aligned} &|x-2|=y \\ & \frac{y-1}{y-2} \leq 0 \\ & y-1=0 \text { and } y-2=0 \\ & y=1 \text { and } y=2 \\ & \underset{-\infty}{ } \quad \end{aligned} $

$ \begin{matrix} \Rightarrow & y-1=0 \text { and } y-2=0 \\ \Rightarrow & y=1 \text { and } y=2 \end{matrix} $

$ \begin{matrix} \Rightarrow & 1 \leq y<2 \\ \Rightarrow & 1 \leq|x-2|<2 \\ \Rightarrow & 1 \leq|x-2| \text { and }|x-2|<2 \\ \Rightarrow & x-2 \leq-1 \\ \Rightarrow & x-2 \geq 1 \\ \text { and } & -2<x-2<2 \\ \Rightarrow & x \leq 1 \text { or } x \geq 3 \text { and } 0<x<4 \\ \Rightarrow & x \in(0,1] \cup[3,4) \end{matrix} $

Solution

Given,

$ \frac{1}{|x|-3} \leq \frac{1}{2} $

$\begin{aligned} & \Rightarrow \quad|x|-3 \geq 2 \quad \because \frac{1}{a}<\frac{1}{b} \Rightarrow a>b \\ & \Rightarrow \quad |x| \geq 5 \quad \text { [adding } 3 \text { to both sides] } \\ & \Rightarrow \quad x \leq-5 \text { or } x \geq 5 \quad[\because|x| \geq a \Rightarrow|x| \leq-a \Rightarrow|x| \geq a] \ & \end{aligned}$

$\begin{array}{ll} \Rightarrow & x \in(-\infty,-5] \cup[5, \infty)…(i) \\ \text { But } & |x|-3 \neq 0 \\ \text { Either } & |x|-3<0 \text { or }|x|-3>0 \\ \Rightarrow & |x|<3 \text { or }|x|>3 \\ \Rightarrow & -3<x<3 \text { or } x<-3 \text { or } x>3 …(ii) \\ & \quad[\because|x|<a \Rightarrow-a<x<\text { and }|x|>a \Rightarrow x<-a \text { (i) } x>a] \end{array}$

On combining results of Eqs. (i) and (ii), we get

$ x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty) $

Solution

On combining Eqs. (i) and (ii), we get

$ x \in(-4,-2] \cup[2,6] $

Solution

We have,

$ \begin{aligned} -5 & \leq \frac{2-3 x}{4} \\ -20 & \leq 2-3 x \\ 3 x & \leq 2+20 \\ 3 x & \leq 22 \\ x & \leq \frac{22}{3} \\ \frac{2-3 x}{4} & \leq 9 \\ 2-3 x & \leq 36 \\ -3 x & \leq 36-2 \\ -3 x & \leq 34 \\ 3 x & \geq-34 \\ x & \geq-\frac{34}{3} \\ -\frac{34}{3} & \leq x \leq \frac{22}{3} \\ x & \in \frac{-34}{3}, \frac{22}{3} \end{aligned} $

$ \text { and } $

[multiplying by 4 on both sides]

Solution

We have,

$ \begin{aligned} 4 x+3 & \geq 2 x+17 \\ 4 x-2 x & \geq 17-3 \Rightarrow 2 x \geq 14 \\ x & \geq \frac{14}{2} \\ x & \geq 7 \\ 3 x-5 & <-2 \\ 3 x & <-2+5 \Rightarrow 3 x<3 \\ x & <1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \\ & \Rightarrow \\ & \Rightarrow \\ & \text { Also, we have } \\ & \Rightarrow \end{aligned} $

$\Rightarrow$

On combining Eqs. (i) and (ii), we see that solution is not possible because nothing is common between these two solutions. (i.e., $x<1, x \geq 7$ ).

Solution

Cost function, C(x) = 26000 + 30x

and revenue function, R(x) = 43x

For profit, R(x) > C(x)

$\Rightarrow$ 26000 + 30x < 43x

$\Rightarrow$ 30x - 43x < -26000

$\Rightarrow$ -13x < -26000

$\Rightarrow$ 13x > 26000

$\Rightarrow$ $x > \frac{26000}{13}$

$\therefore x > 2000$

$\therefore$ Hence, more than 2000 cassettes must be produced to get profit.

Solution

Given,

and

first $pH$ value $=8.48$

second $pH$ value $=8.35$

Let third $pH$ value be $x$.

Since, it is given that average $pH$ value lies between 8.2 and 8.5 .

$ \begin{matrix} \therefore & 8.2<\frac{8.48+8.35+x}{3}<8.5 \\ \Rightarrow & 8.2<\frac{16.83+x}{3}<8.5 \\ \Rightarrow & 3 \times 8.2<16.83+x<8.5 \times 3 \\ \Rightarrow & 24.6<16.83+x<25.5 \\ \Rightarrow & 24.6-16.83<x<25.5-16.83 \\ \Rightarrow & 7.77<x<8.67 \end{matrix} $

Thus, third pH value lies between 7.77 and 8.67 .

Solution

Let $x L$ of $3 %$ solution be added to $460 L$ of $9 %$ solution of acid.

Then, total quantity of mixture $=(460+x) L$

Total acid content in the $(460+x) L$ of mixture

$ =460 \times \frac{9}{100}+x \times \frac{3}{100} $

It is given that acid content in the resulting mixture must be more than $5 %$ but less than $7 %$ acid.

Therefore, $\quad 5\text{of}(460+x)<560 \times \frac{9}{100}+\frac{3x}{100}<7\text{of}(460+x)$

$\Rightarrow \quad \frac{5}{100} \times(460+x)<460 \times \frac{9}{100}+\frac{3}{100} x<\frac{7}{100} \times(460+x)$

$\Rightarrow \quad 5 \times(460+x)<460 \times 9+3 x<7 \times(460+x)\text{[multiplying by 100]}$

$\begin{array}{lr} \Rightarrow & 2300+5 x<4140+3 x<3220+7 x \\ \text { Taking first two inequalities, } & 2300+5 x<4140+3 x \\ \Rightarrow & 5 x-3 x<4140-2300 \\ \Rightarrow & 2 x<1840 \\ \Rightarrow & x<\frac{1840}{2} \\ \Rightarrow & x<920\quad…(i) \end{array}$

Taking last two inequalities,

$\begin{aligned} 4140+3 x & <3220+7 x \\ \Rightarrow 3 x-7 x & <3220-4140 \\ \Rightarrow -4 x & <-920 \\ \Rightarrow 4 x & >920 \\ \Rightarrow x & >\frac{920}{4} \\ \Rightarrow x & >230 \quad …(ii) \end{aligned}$

Hence, the number of litres of the $3 %$ solution of acid must be more than $230 L$ and less than $920 L$

Solution

Let the required temperature be $x^{\circ} F$.

Given that,

$ \begin{matrix} \Rightarrow & 5 F=9 C+32 \times 5 \\ \Rightarrow & 9 C=5 F-32 \times 5 \\ \therefore & C=\frac{5 F-160}{9} \end{matrix} $

Since, temperature in degree calcius lies between $40^{\circ} C$ to $45^{\circ} C$.

$\begin{array}{lc} \text { Therefore, } & 40<\frac{5 F-160}{9}<45 \\ \Rightarrow & 40<\frac{5 x-160}{9}<45 \\ \Rightarrow & 40 \times 9<5 x-160<45 \times 9 \text{[multiplying throughout by 9]} \\ \Rightarrow & 360<5 x-160<405 \text{[adding 160 throughout]} \\ \Rightarrow & 360+160<5 x<405+160 \\ \Rightarrow & 520<5 x<565 \\ \Rightarrow & \frac{520}{5}<x<\frac{565}{5} \text{[divide throughout by 5]} \\ \Rightarrow & 104<x<113 \end{array}$

Hence, the range of temperature in degree fahrenheit is $104^{\circ} F$ to $113^{\circ} F$.

Solution

Let the length of shortest side be $x cm$.

According to the given information,

Longest side $=2 \times$ Shortest side

$ =2 x cm $

and third side $=2+$ Shortest side

$ =(2+x) cm $

Perimeter of triangle $=x+2 x+(x+2)=4 x+2$

According to the question,

Perimeter $>166 cm$

$ \begin{matrix} \Rightarrow & 4 x+2>166 \\ \Rightarrow & 4 x>166-2 \\ \Rightarrow & 4 x>164 \\ \therefore & x>\frac{164}{4}=41 cm \end{matrix} $

Hence, the minimum length of shortest side be $41 cm$.

Solution

Given that,

$ T=30+25(x-3), 3 \leq x \leq 15 $

According to the question,

$ \begin{aligned} & 155<T<205 \\ & \begin{matrix} \Rightarrow & 155<30+25(x-3)<205 \\ \Rightarrow & 155-30<25(x-3)<205-30 \end{matrix} \\ & \Rightarrow \quad 155-30<25(x-3)<205-30 \quad \text { subtracting } 30 \text { in whole } \\ & \Rightarrow \quad 125<25(x-3)<175 \\ & \Rightarrow \quad \frac{125}{25}<x-3<\frac{175}{25} \quad \text { dividing by } 25 \text { in whole } \\ & \Rightarrow \quad 5<x-3<7 \\ & \Rightarrow \quad 5+3<x<7+3 \quad \text { adding } 3 \text { in whole } \\ & \Rightarrow \quad 8<x<10 \end{aligned} $

Hence, at the depth 8 to $10 km$ temperature lies between $155^{\circ}$ to $205^{\circ} C$.

Solution

The given system of inequations is

$ \begin{aligned} & \frac{2 x+1}{7 x-1}>5 \\ & \text { and } \quad \frac{x+7}{x-8}>2 \\ & \text { Now, } \quad \frac{2 x+1}{7 x-1}-5>0 \\ & \Rightarrow \quad \frac{(2 x+1)-5(7 x-1)}{7 x-1}>0 \\ & \Rightarrow \quad \frac{2 x+1-35 x+5}{7 x-1}>0 \\ & \Rightarrow \quad \frac{-33 x+6}{7 x-1}>0 \Rightarrow \frac{33 x-6}{7 x-1}<0 \\ & \Rightarrow \quad x \in \frac{1}{7}, \frac{6}{33} \\ \end{aligned} $

$ \frac{x+7}{x-8}>2 \Rightarrow \frac{x+7}{x-8}-2>0 \\ \frac{x+7-2(x-8)}{x-8}>0 \Rightarrow \frac{x+7-2 x+16}{x-8}>0 \\ \Rightarrow \quad \frac{-x+23}{x-8}>0 \Rightarrow \frac{x-23}{x-8}<0 \\ $

$x \in(8,23) $

Since, the intersection of Eqs. (iii) and (iv) is the null set. Hence, the given system of equation has no solution.

Solution

Consider the line $3 x+2 y=48$, we observe that the shaded region and the origin are on the same side of the line $3 x+2 y=48$ and $(0,0)$ satisfy the linear constraint $3 x+2 y \leq 48$. So, we must have one inequation as $3 x+2 y \leq 48$.

Now, consider the line $x+y=20$. We find that the shaded region and the origin are on the same side of the line $x+y=20$ and $(0,0)$ satisfy the constraints $x+y \leq 20$. So, the second inequation is $x+y \leq 20$.

We also notice that the shaded region is above $X$-axis and is on the right side of $Y$-axis, so we must have $x \geq 0, y \geq 0$.

Thus, the linear inequations corresponding to the given solution set are $3 x+2 y \leq 48$, $x+y \leq 20$ and $x \geq 0, y \geq 0$.

Solution

Consider the line $x+y=4$.

We observe that the shaded region and the origin lie on the opposite side of this line and $(0,0)$ satisfies $x+y \leq 4$. Therefore, we must have $x+y \geq 4$ as the linear inequation corresponding to the line $x+y=4$.

Consider the line $x+y=8$, clearly the shaded region and origin lie on the same side of this line and $(0,0)$ satisfies the constraints $x+y \leq 8$. Therefore, we must have $x+y \leq 8$, as the linear inequation corresponding to the line $x+y=8$.

Consider the line $x=5$. It is clear from the graph that the shaded region and origin are on the left of this line and $(0,0)$ satisfy the constraint $x \leq 5$.

Hence, $x \leq 5$ is the linear inequation corresponding to $x=5$.

Consider the line $y=5$, clearly the shaded region and origin are on the same side (below) of the line and $(0,0)$ satisfy the constrain $y \leq 5$.

Therefore, $y \leq 5$ is an inequation corresponding to the line $y=5$.

We also notice that the shaded region is above the $X$-axis and on the right of the $Y$-axis i.e., shaded region is in first quadrant. So, we must have $x \geq 0, y \geq 0$.

Thus, the linear inequations comprising the given solution set are

$ x+y \geq 4 ; x+y \leq 8 ; x \leq 5 ; y \leq 5 ; x \geq 0 \text { and } y \geq 0 $

Solution

Consider the inequation $x+2 y \leq 3$ as an equation, we have

$x+2y=3$

$\Rightarrow x=3-2y$

$\Rightarrow 2y=3-x$

Now, $(0,0)$ satisfy the inequation $x+2 y \leq 3$.

So, half plane contains $(0,0)$ as the solution and the line $x+2 y=3$ intersect the coordinate axis at $(3,0)$ and $(0,3 / 2)$.

Consider the inequation $3 x+4 y \geq 12$ as an equation, we have $3 x+4 y=12$ $\Rightarrow$ $4 y=12-3 x$

Thus, coordinate axis intersected by the line $3 x+4 y=12$ at points $(4,0)$ and $(0,3)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+4 y=12$.

Therefore, half plane of the solution does not contained $(0,0)$.

Consider the inequation $y \geq 1$ as an equation, we have

$ y=1 $

It represents a straight line parallel to $X$-axis passing through point $(0,1)$.

Now, $(0,0)$ does not satisfy the inequation $y \geq 1$.

Therefore, half plane of the solution does not contains $(0,0)$.

Clearly $x \geq 0$ represents the region lying on the right side of $Y$-axis.

The solution set of the given linear constraints will be the intersection of the above region.

It is clear from the graph the shaded portions do not have common region.

So, solution set is null set.

Solution

Consider the inequation $3 x+2 y \geq 24$ as an equation, we have $3 x+2 y=24$. $\Rightarrow$

$ 2 y=24-3 x $

Hence, line $3 x+y=24$ intersect coordinate axes at points $(8,0)$ and $(0,12)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+2 y \geq 24$.

Therefore, half plane of the solution set does not contains $(0,0)$.

Consider the inequation $3 x+y \leq 15$ as an equation, we have

$ 3 x+y=15 $

$ y=15-3 x $

Line $3 x+y=15$ intersects coordinate axes at points $(5,0)$ and $(0,15)$.

Now, point $(0,0)$ satisfy the inequation $3 x+y \leq 15$.

Therefore, the half plane of the solution contain origin.

Consider the inequality $x \geq 4$ as an equation, we have

$ x=4 $

It represents a straight line parallel to $Y$-axis passing through $(4,0)$. Now, point $(0,0)$ does not satisfy the inequation $x \geq 4$.

Therefore, half plane does not contains $(0,0)$,

The graph of the above inequations is given below.

It is clear from the graph that there is no common region corresponding to these inequality. Hence, the given system of inequalities have no solution.

Solution

Consider the inequation $2 x+y \geq 8$ as an equation, we have

2x + y = 8

y = 8-2x

The line $2 x+y=8$ intersects coordinate axes at $(4,0)$ and $(0,8)$. Now, point $(0,0)$ does not satisfy the inequation $2 x+y \geq 8$. Therefore, half plane does not contain origin.

Consider the inequation $x+2 y \geq 10$, as an equation, we have

$\Rightarrow$

$ x+2 y=10 $

$ 2 y=10-x $

The line $2 x+y=8$ intersects the coordinate axes at $(10,0)$ and $(0,5)$.

Now, point $(0,0)$ does not satisfy the inequation $x+2 y \geq 10$.

Therefore, half plane does not contain $(0,0)$.

Consider the inequation $x \geq 0$ and $y \geq 0$ clearly, it represents the region in first quadrant. The graph of the above inequations is given below

It is clear from the graph that common shaded portion is unbounded.

Solution

(c) If $x<5$, then $-x>-5$

[if we multiply by negative numbers, then inequality get reversed]

Solution

(c) It is given that,

$ x<y, b<0 $

$\Rightarrow \quad \frac{x}{b}>\frac{y}{b}$

$[\because b<0]$

Solution

(a) Given that, $-3 x+17<-13$

Solution

(b) Given,

$ |x|<3 $

$ \Rightarrow \quad-3<x<3 \quad[\because|x|<a \Rightarrow-a<x<a] $

Solution

(d) Given,

$ \begin{aligned} |x| & >b \text { and } b>0 \\ x & <-b \text { or } x>b \\ x & \in(-\infty,-b) \cup(b, \infty) \end{aligned} $

$ \Rightarrow $

Solution

(c) Given,

$ \begin{aligned} |x-1| & >5 \\ (x-1) & <-5 \text { or }(x-1)>5 \quad[\because|x|>a \Rightarrow x<a \text { or } x>a] \\ x & <-4 \text { or } x>6 \\ x & \in(-\infty,-4) \cup(6, \infty) \end{aligned} $

Solution

(b) Given,

$ \begin{aligned} |x+2| & \leq 9, \\ -9 & \leq x+2 \leq \\ -9-2 & \leq x \leq 9 \\ -11 & \leq x \leq 7 \\ x & \in-11,7 \end{aligned} $

$ \begin{matrix} \Rightarrow & -9 \leq x+2 \leq 9 & {[\because|x| \leq a \Rightarrow-a \leq x \leq a]} \\ \Rightarrow & -9-2 \leq x \leq 9-2 & \text { [subtracting 2 througout] } \\ \Rightarrow & -11 \leq x \leq 7 & \end{matrix} $

$\Rightarrow$

The inequality representing the following graphs is

Solution

(a) The given graph represent $x>-5$ and $x<5$.

On combining these two result, we get

$ |x|<5 $

Solution of a linear inequality in variable $x$ is represented on number line in following questions.

Solution

(d) The given graph represents all the values greater than 5 except $x=5$ on the real line So,

$x \in(5, \infty)$.

Solution

(b) The given graph represents all the values greater than $\frac{9}{2}$ including $\frac{9}{2}$ as the real line.

$ x \in \frac{9}{2}, \infty $

Solution

(a) The given graph represents all the values less than $\frac{7}{2}$ on the real line.

$\Rightarrow$

$ x \in-\infty, \frac{7}{2} $

Solution

(b) The given graph represents all values less than -2 including -2 .

$ x \in(-\infty,-2] $

Solution

(i) If $x<y$ and $b<0$ $\Rightarrow$ $\frac{x}{b}>\frac{y}{b}$

Hence, statement (i) is false.

(ii) If $x y>0$, then, $x>0, y>0$ or $x<0, y<0$.

Hence, statement (ii) is true.

(iii) If $x y>0$, then $x<0$ and $y<0$.

Hence, statement (iii) is true. (iv) If $x y<0 \Rightarrow x<0, y>0$ or $x>0, y<0$. Hence, statement (iv) is false.

(v) If $x<-5$ and $x<-2$, then

$ x \in(-\infty,-5) $

Hence, statement (v) is true.

(vi) If $x<-5$ and $x>2$, then $x$ have no value.

Hence, statement (vi) is false.

(vii) If $x>-2$ and $x<9$, then $x \in(-2,9)$.

Hence, statement (vii) is true.

(viii) If $|x|>5$, then either $x<-5$ or $x>5$.

$ \Rightarrow \quad x \in(-\infty,-5) \cup(5, \infty) $

Hence, statement (viii) is false.

(ix) If $|x| \leq 4$, then

$-4 \leq x \leq 4 \\ $

$\Rightarrow x \in-4,4 $

Hence, statement (ix) is true.

(x) The given graph represents $x \leq 3$.

Hence, statement $(x)$ is false.

(xi) The given graph represents $x \geq 0$.

Hence, statement (xi) is true.

(xii) The given graph represent $y \geq 0$.

Hence, statement (xii) is false.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ is

Hence, statement (xiii) is false.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ is

Hence, statement (xiv) is false.

(xv) The given graph represents $x+y \geq 0$.

Hence, statement (xv) is correct.

Solution

(i) If $-4 x \geq 12 \Rightarrow x \leq-3$

(ii) If $\frac{-3}{4} x \leq-3$

$ \Rightarrow \quad x \geq(-3) \times \frac{4}{-3} \Rightarrow x \geq 4 $

(iii) If $\frac{2}{x+2}>0$

$ \Rightarrow $

(iv) If $x>-5 \Rightarrow 4 x>-20$

(v) If $x>y$ and $z<0$, then

$ \Rightarrow \quad-x z>-y z $

(vi) If $p>0$ and $q<0$,

then

e.g., consider $2>0$ and $-3<0$.

$ p-q>p $

$\Rightarrow$

$ 2-(-3)=2+3=5>2 $

(vii) If $|x+2|>5$, then

$ x+2<-5 \text { or } x+2>5 \\ $

$\Rightarrow x<-5-2 \text { or } x>5-2 \\ $

$\Rightarrow x<-7 \text { or } x>3 \\ $

$\Rightarrow x \in-\infty,-7 \cup(3, \infty) $

(viii) If $-2 x+1 \geq 9$, then

$ \begin{aligned} -2 x \geq 9-1 & \Rightarrow-2 x \geq 8 \\ 2 x \leq-8 & \Rightarrow x \leq-4 \end{aligned} $