Linear inequalities for Single Line Questions in English

(B) Given the inequality: $5x - 3 < 3x + 1$
Add $3$ to both sides: $5x < 3x + 4$
Subtract $3x$ from both sides: $2x < 4$
Divide by $2$: $x < 2$
Since $x$ is a real number,the solution set is all real numbers less than $2$.
Thus,the solution set is $x \in (-\infty, 2)$.

(A) Given inequality: $4x + 3 < 6x + 7$
Subtract $6x$ from both sides: $4x - 6x + 3 < 7$
$-2x + 3 < 7$
Subtract $3$ from both sides: $-2x < 4$
Divide by $-2$ (remember to reverse the inequality sign when dividing by a negative number): $x > -2$
Thus,the solution set is $(-2, \infty)$.

(N/A) We have $\frac{5-2x}{3} \leq \frac{x}{6}-5$.
Multiply both sides by $6$ to clear the denominators:
$2(5-2x) \leq x - 30$
$10 - 4x \leq x - 30$
Subtract $x$ from both sides:
$10 - 5x \leq -30$
Subtract $10$ from both sides:
$-5x \leq -40$
Divide by $-5$ (remembering to reverse the inequality sign when dividing by a negative number):
$x \geq 8$
Thus,the solution set is $x \in [8, \infty)$.

(N/A) Given inequality: $7x + 3 < 5x + 9$
Subtract $5x$ from both sides: $2x + 3 < 9$
Subtract $3$ from both sides: $2x < 6$
Divide by $2$: $x < 3$
The solution set is all real numbers less than $3$. The graphical representation on the number line is a ray starting from $3$ (with an open circle at $3$ to indicate that $3$ is not included) extending to the left towards negative infinity.

(N/A) We have $\frac{3x-4}{2} \geq \frac{x+1}{4}-1$.
Multiplying both sides by $4$ to clear the denominators:
$2(3x-4) \geq (x+1) - 4$
$6x - 8 \geq x - 3$
Subtract $x$ from both sides:
$5x - 8 \geq -3$
Add $8$ to both sides:
$5x \geq 5$
Divide by $5$:
$x \geq 1$.
The solution set is $[1, \infty)$. The graphical representation is a solid circle at $1$ with a ray extending to the right.

(A) The given inequality is $-12x > 30.$
Dividing both sides by $-12,$ the inequality sign reverses:
$x < \frac{30}{-12}$
$x < -2.5$
The integers less than $-2.5$ are $....., -5, -4, -3.$
Thus,the solution set is $\{....., -5, -4, -3\}.$

(A) The given inequality is $3x + 8 > 2$.
Subtract $8$ from both sides:
$3x + 8 - 8 > 2 - 8$
$3x > -6$
Divide both sides by $3$:
$\frac{3x}{3} > \frac{-6}{3}$
$x > -2$
When $x$ is a real number,the solution set consists of all real numbers greater than $-2$.
Thus,the solution set is $(-2, \infty)$.

(A) Given inequality: $3(2-x) \geq 2(1-x)$
Expanding both sides:
$6 - 3x \geq 2 - 2x$
Adding $3x$ to both sides:
$6 \geq 2 + x$
Subtracting $2$ from both sides:
$4 \geq x$ or $x \leq 4$
Thus,the solution set for the given inequality is $(-\infty, 4]$.

(N/A) The graph of $3x + 2y = 6$ is represented by a dotted line in the figure.
This line divides the $xy$-plane into two half-planes,$I$ and $II$. We select a test point not on the line,such as $(0, 0)$,which lies in half-plane $I$,and determine if it satisfies the given inequality:
$3(0) + 2(0) > 6$
$0 > 6$,which is false.
Hence,half-plane $I$ is not the solution region. Since the inequality is strict $(>)$,points on the line are not included. Therefore,the shaded half-plane $II$ represents the solution region of the inequality.

(N/A) First,consider the corresponding equation $3x - 6 = 0$,which simplifies to $x = 2$.
Draw the line $x = 2$ in the Cartesian plane. Since the inequality is $\geq$,the line $x = 2$ is included in the solution region (represented by a solid line).
To determine the solution region,test a point not on the line,such as $(0, 0)$.
Substituting $(0, 0)$ into the inequality $3x - 6 \geq 0$ gives $3(0) - 6 \geq 0$,which simplifies to $-6 \geq 0$.
Since $-6 \geq 0$ is false,the origin $(0, 0)$ does not lie in the solution region.
Therefore,the solution region is the half-plane to the right of the line $x = 2$,including the line itself.

(N/A) The graph of $y = 2$ is shown in the figure.
Let us select a point $(0, 0)$ in the lower half-plane $I$ and substitute $y = 0$ into the given inequality:
$0 < 2$,which is true.
Thus,the solution region is the shaded region below the line $y = 2$. Hence,every point below the line (excluding all the points on the line,as indicated by the dashed line) represents the solution of the given inequality.

(N/A) The graphical representation of $x+y=5$ is given as a dotted line in the figure below.
This line divides the $xy$-plane into two half-planes,$I$ and $II$.
Select a point (not on the line),which lies in one of the half-planes,to determine whether the point satisfies the given inequality or not.
We select the point as $(0,0)$.
It is observed that,
$0+0 < 5$ or $0 < 5$,which is true.
Therefore,the half-plane $I$ is the solution region of the given inequality.
Also,it is evident that any point on the line does not satisfy the given strict inequality.
Thus,the solution region of the given inequality is the shaded half-plane $I$ excluding the points on the line.

(N/A) The graphical representation of the line $2x + y = 6$ is shown in the figure.
This line divides the $xy$-plane into two half-planes,$I$ and $II$.
Select a point (not on the line),which lies in one of the half-planes,to determine whether the point satisfies the given inequality.
We select the point $(0, 0)$.
It is observed that,
$2(0) + 0 \geq 6$ or $0 \geq 6$,which is false.
Therefore,half-plane $I$ is not the solution region of the given inequality. Also,it is evident that any point on the line satisfies the given inequality.
Thus,the solution region of the given inequality is the shaded half-plane $II$ including the points on the line.

(N/A) To solve the inequality $3x + 4y \leq 12$ graphically,we first consider the corresponding equation $3x + 4y = 12$.
Find the intercepts of the line $3x + 4y = 12$:
If $x = 0$,then $4y = 12 \implies y = 3$. So,the point is $(0, 3)$.
If $y = 0$,then $3x = 12 \implies x = 4$. So,the point is $(4, 0)$.
Draw the line passing through $(0, 3)$ and $(4, 0)$. Since the inequality is $\leq$,the line is solid.
To determine the solution region,test the origin $(0, 0)$ in the inequality $3x + 4y \leq 12$:
$3(0) + 4(0) = 0 \leq 12$,which is true.
Therefore,the solution region is the half-plane containing the origin $(0, 0)$,which is the shaded region below the line $3x + 4y = 12$.

The graphical representation of the line $y+8=2x$ is shown in the figure.
This line divides the $xy$-plane into two half-planes.
To determine the solution region,we select a test point not on the line. Let us choose the point $(0,0)$.
Substituting $(0,0)$ into the inequality $y+8 \geq 2x$:
$0+8 \geq 2(0)$
$8 \geq 0$
Since $8 \geq 0$ is a true statement,the solution region is the half-plane that contains the point $(0,0)$.
Because the inequality is $\geq$,the line $y+8=2x$ is included in the solution region. The shaded area in the figure represents the solution set.

(N/A) The graphical representation of the line $x-y=2$ is shown in the figure.
This line divides the $xy$-plane into two half-planes.
To determine the solution region,we select a test point not on the line,such as $(0,0)$.
Substituting $(0,0)$ into the inequality $x-y \leq 2$,we get:
$0-0 \leq 2 \implies 0 \leq 2$,which is a true statement.
Since the inequality holds true for the point $(0,0)$,the solution region is the half-plane containing the origin $(0,0)$.
Because the inequality is $\leq$,the line $x-y=2$ is included in the solution region (represented by a solid line).
The solution region is the shaded area shown in the figure.

(N/A) The graphical representation of $2x - 3y = 6$ is given as a dotted line in the figure below.
This line divides the $xy$-plane into two half-planes.
Select a point (not on the line),which lies in one of the half-planes,to determine whether the point satisfies the given inequality or not.
We select the point as $(0, 0)$.
It is observed that,
$2(0) - 3(0) > 6$ or $0 > 6$,which is false.
Therefore,the upper half-plane is not the solution region of the given inequality. Also,it is clear that any point on the line does not satisfy the given inequality because the inequality is strict $(>)$.
Thus,the solution region of the given inequality is the half-plane that does not contain the point $(0, 0)$,excluding the line itself.
The solution region is represented by the shaded region in the figure.

(N/A) The graphical representation of the line $-3x + 2y = -6$ is shown in the figure.
This line divides the $xy$-plane into two half-planes.
To determine the solution region,we select a test point not on the line,such as $(0, 0)$.
Substituting $(0, 0)$ into the inequality:
$-3(0) + 2(0) \geq -6$
$0 \geq -6$,which is a true statement.
Since the point $(0, 0)$ satisfies the inequality,the solution region is the half-plane containing the origin $(0, 0)$.
Because the inequality is $\geq$,the line itself is included in the solution region. The shaded region in the figure represents the solution set.

(N/A) The graphical representation of the line $3y - 5x = 30$ is drawn as a dotted line because the inequality is strict $( < )$.
This line divides the $xy$-plane into two half-planes.
To determine the solution region,we test a point not on the line,such as the origin $(0, 0)$.
Substituting $(0, 0)$ into the inequality:
$3(0) - 5(0) < 30$
$0 < 30$
Since $0 < 30$ is a true statement,the half-plane containing the origin $(0, 0)$ is the solution region.
Thus,the solution region is the half-plane containing the origin,excluding the line $3y - 5x = 30$ itself.

(N/A) The graphical representation of $y = -2$ is given as a dotted line in the figure. This line divides the $xy$-plane into two half-planes.
Select a point (not on the line),which lies in one of the half-planes,to determine whether the point satisfies the given inequality or not.
We select the point $(0, 0)$.
It is observed that $0 < -2$,which is false.
Also,it is evident that any point on the line does not satisfy the given inequality.
Hence,every point below the line $y = -2$ (excluding all the points on the line) determines the solution of the given inequality.
The solution region is represented by the shaded region in the figure.

The graphical representation of $x = -3$ is given as a dotted line in the figure. This line divides the $xy$-plane into two half-planes.
Select a point (not on the line),which lies in one of the half-planes,to determine whether the point satisfies the given inequality or not.
We select the point $(0, 0)$.
It is observed that $0 > -3$,which is true.
Also,it is evident that any point on the line does not satisfy the given inequality.
Hence,every point on the right side of the line $x = -3$ (excluding all the points on the line) determines the solution of the given inequality.
The solution region is represented by the shaded region in the figure.

(N/A) First,we draw the graph of the line $8x + 3y = 100$.
To find the intercepts,if $x = 0$,then $3y = 100 \implies y = 33.33$. If $y = 0$,then $8x = 100 \implies x = 12.5$.
The inequality $8x + 3y \leq 100$ represents the shaded region below the line,including the points on the line $8x + 3y = 100$.
Since $x \geq 0$ and $y \geq 0$,the solution is restricted to the first quadrant.
Thus,every point in the shaded region in the first quadrant,including the points on the line and the axes,represents the solution of the given system of inequalities.

(A) We are given the compound inequality: $-8 \leq 5x - 3 < 7$.
First,add $3$ to all parts of the inequality:
$-8 + 3 \leq 5x - 3 + 3 < 7 + 3$
$-5 \leq 5x < 10$.
Next,divide all parts by $5$:
$\frac{-5}{5} \leq \frac{5x}{5} < \frac{10}{5}$
$-1 \leq x < 2$.
Thus,the solution in interval notation is $[-1, 2)$.

(A) Given the inequality: $2 \leq 3x - 4 \leq 5$.
Add $4$ to all parts of the inequality:
$2 + 4 \leq 3x - 4 + 4 \leq 5 + 4$
This simplifies to:
$6 \leq 3x \leq 9$
Divide all parts by $3$:
$\frac{6}{3} \leq \frac{3x}{3} \leq \frac{9}{3}$
Resulting in:
$2 \leq x \leq 3$
Thus,the solution set for the given inequality is $[2, 3]$.

(A) Given inequality: $-12 < 4 - \frac{3x}{-5} \leq 2$
Simplify the term: $-12 < 4 + \frac{3x}{5} \leq 2$
Subtract $4$ from all parts: $-12 - 4 < \frac{3x}{5} \leq 2 - 4$
$-16 < \frac{3x}{5} \leq -2$
Multiply all parts by $5$: $-80 < 3x \leq -10$
Divide all parts by $3$: $-\frac{80}{3} < x \leq -\frac{10}{3}$
Thus,the solution set is $\left(\frac{-80}{3}, \frac{-10}{3}\right]$.

(N/A) First,solve the first inequality:
$5(2x - 7) - 3(2x + 3) \leq 0$
$10x - 35 - 6x - 9 \leq 0$
$4x - 44 \leq 0$
$4x \leq 44$
$x \leq 11$ ..... $(1)$
Next,solve the second inequality:
$2x + 19 \leq 6x + 47$
$19 - 47 \leq 6x - 2x$
$-28 \leq 4x$
$-7 \leq x$ ..... $(2)$
Combining $(1)$ and $(2)$,the solution set is $[-7, 11]$.
The solution on the number line is represented by shading the interval between $-7$ and $11$,including the endpoints.

(A) Given inequality: $\frac{5-3x}{3} \leq \frac{x}{6}-5$.
Multiply the entire inequality by $6$ to clear the denominators:
$2(5-3x) \leq x - 30$.
Expand the terms:
$10 - 6x \leq x - 30$.
Rearrange the terms to isolate $x$:
$10 + 30 \leq x + 6x$.
$40 \leq 7x$.
Divide by $7$:
$x \geq \frac{40}{7}$.
Thus,the solution set is $x \in [\frac{40}{7}, \infty)$.

(B) Step $1$: Consider the corresponding equation $4x - y = 0$,which simplifies to $y = 4x$.
Step $2$: Draw the line $y = 4x$ as a dashed line because the inequality is strict $(>)$.
Step $3$: Test a point not on the line,such as $(1, 0)$.
Step $4$: Substitute $(1, 0)$ into the inequality: $4(1) - 0 = 4 > 0$. This is true.
Step $5$: Since the point $(1, 0)$ satisfies the inequality,the solution region is the half-plane containing $(1, 0)$,which is the region below the line $y = 4x$.

(N/A) Step $1$: Consider the corresponding equation $5x + 2y = 10$.
Step $2$: Find the intercepts of the line. If $x = 0$,then $2y = 10$,so $y = 5$. The point is $(0, 5)$. If $y = 0$,then $5x = 10$,so $x = 2$. The point is $(2, 0)$.
Step $3$: Draw the line passing through $(0, 5)$ and $(2, 0)$. Since the inequality is $\leq$,the line is solid.
Step $4$: Test the origin $(0, 0)$. Substituting into the inequality: $5(0) + 2(0) \leq 10$,which is $0 \leq 10$. This is true.
Step $5$: Since the statement is true,the solution region is the half-plane containing the origin.

(A) Given inequality is $4x - 6 \geq 0$.
First,solve for $x$:
$4x \geq 6$
$x \geq \frac{6}{4}$
$x \geq 1.5$.
To represent this graphically in a two-dimensional plane,draw the line $x = 1.5$,which is a vertical line passing through $(1.5, 0)$.
Since the inequality is $\geq$,use a solid line to indicate that points on the line are included.
Shade the region to the right of the line $x = 1.5$ because all $x$-values greater than or equal to $1.5$ satisfy the inequality.

(A) Given inequality: $5y - 3 \leq 12$.
Step $1$: Simplify the inequality.
$5y \leq 12 + 3$
$5y \leq 15$
$y \leq 3$.
Step $2$: Draw the line $y = 3$ in the Cartesian plane. Since the inequality is $\leq$,the line is solid.
Step $3$: Test a point $(0, 0)$. Since $0 \leq 3$ is true,the solution region is the half-plane containing the origin.
Thus,the solution is the region below the line $y = 3$ including the line itself.

(A) Step $1$: Consider the corresponding equation $2x + y = 3$.
Step $2$: Find the intercepts to plot the line. If $x = 0$,$y = 3$. If $y = 0$,$x = 1.5$. The line passes through $(0, 3)$ and $(1.5, 0)$.
Step $3$: Since the inequality is strictly greater than $(>)$,draw the line as a dashed line to indicate that points on the line are not included in the solution.
Step $4$: Test a point not on the line,such as $(0, 0)$. Substituting into the inequality: $2(0) + 0 > 3$,which simplifies to $0 > 3$. This is false.
Step $5$: Since the test point $(0, 0)$ does not satisfy the inequality,the solution region is the half-plane that does not contain the origin. Thus,the solution is the region above the line $2x + y = 3$.

(A) Given inequality: $y + 8 > 2x$.
First,we draw the boundary line $y = 2x - 8$.
To draw this line,we find two points:
If $x = 0$,$y = -8$. Point: $(0, -8)$.
If $y = 0$,$2x = 8 \implies x = 4$. Point: $(4, 0)$.
Since the inequality is strict $(>)$,we draw the line as a dashed line.
Now,test the origin $(0, 0)$ in the inequality: $0 + 8 > 2(0) \implies 8 > 0$,which is true.
Therefore,the solution region is the half-plane containing the origin $(0, 0)$,which is the region above the line $y = 2x - 8$.

(B) From the figure,we observe that the shaded region lies between the vertical lines $x = -2$ and $x = 2$.
Since the lines $x = -2$ and $x = 2$ are solid (not dashed),the boundary values are included in the region.
Therefore,the inequality is $-2 \leq x \leq 2$.
This can be written in terms of the absolute value as $|x| \leq 2$.

(D) The coloured region lies to the left of the $Y$-axis.
In the Cartesian plane,the $Y$-axis is represented by the equation $x = 0$.
Points to the left of the $Y$-axis have $x$-coordinates that are less than or equal to $0$.
Therefore,the inequality representing the coloured region is $x \leq 0$.

(B) Given the inequalities $x < 5$ and $x \geq 2$.
Combining these two inequalities,we get $2 \leq x < 5$.
In interval notation,this is represented as $[2, 5)$,where the square bracket indicates that $2$ is included and the parenthesis indicates that $5$ is excluded.

(C) The given inequality is $x - y \geq 0$,which can be rewritten as $x \geq y$ or $y \leq x$.
To graph this,we first consider the boundary line $x - y = 0$,which is the line $y = x$.
Since the inequality is $\geq$ (greater than or equal to),the boundary line $y = x$ must be included in the solution set,so we draw it as a solid line.
Next,we test a point not on the line,such as $(1, 0)$.
Substituting $(1, 0)$ into the inequality: $1 - 0 = 1$,and $1 \geq 0$ is true.
Therefore,the region containing the point $(1, 0)$ is the solution set.
This corresponds to the region below the line $y = x$ (where $x$ values are greater than $y$ values).
Looking at the options,the graph in Option $C$ shows the solid line $x - y = 0$ with the shaded region below it.

(B) To find the solution set of the inequality $2x + y > 5$,we first consider the boundary line $2x + y = 5$.
Since the inequality is strict $(>)$,the points on the line $2x + y = 5$ are not included in the solution set.
Now,we test the origin $(0, 0)$ in the inequality:
$2(0) + 0 > 5 \implies 0 > 5$,which is false.
Since the origin does not satisfy the inequality,the solution set is the open half-plane that does not contain the origin.

(B) To determine if a point $(x, y)$ lies in the region defined by the inequality $2y - 3x < 5$,we substitute the coordinates of the point into the expression $2y - 3x$ and check if the result is less than $5$.
For point $O(0, 0)$:
$2(0) - 3(0) = 0 - 0 = 0$.
Since $0 < 5$,the point $O(0, 0)$ lies inside the region.
For point $P(2, -1)$:
$2(-1) - 3(2) = -2 - 6 = -8$.
Since $-8 < 5$,the point $P(2, -1)$ also lies inside the region.
Therefore,both points $O$ and $P$ are inside the region.

(B) To determine which points satisfy the inequality $2x - 3y > -5$,we substitute each point $(x, y)$ into the expression $2x - 3y$ and check if the result is greater than $-5$:
$1. (1, 1): 2(1) - 3(1) = 2 - 3 = -1$. Since $-1 > -5$,this point satisfies the inequality.
$2. (-1, 1): 2(-1) - 3(1) = -2 - 3 = -5$. Since $-5$ is not greater than $-5$,this point does not satisfy it.
$3. (1, -1): 2(1) - 3(-1) = 2 + 3 = 5$. Since $5 > -5$,this point satisfies the inequality.
$4. (-1, -1): 2(-1) - 3(-1) = -2 + 3 = 1$. Since $1 > -5$,this point satisfies the inequality.
$5. (-2, 1): 2(-2) - 3(1) = -4 - 3 = -7$. Since $-7$ is not greater than $-5$,this point does not satisfy it.
$6. (2, -1): 2(2) - 3(-1) = 4 + 3 = 7$. Since $7 > -5$,this point satisfies the inequality.
$7. (-1, 2): 2(-1) - 3(2) = -2 - 6 = -8$. Since $-8$ is not greater than $-5$,this point does not satisfy it.
$8. (-2, -1): 2(-2) - 3(-1) = -4 + 3 = -1$. Since $-1 > -5$,this point satisfies the inequality.
The points that satisfy the inequality are $(1, 1), (1, -1), (-1, -1), (2, -1)$,and $(-2, -1)$.
Therefore,there are $5$ such points.

(B) From the figure,the shaded region is bounded by the vertical lines $x = -3$ and $x = 3$.
Since the lines $x = -3$ and $x = 3$ are solid lines,the points on these lines are included in the region.
Thus,the region represents all $x$ such that $-3 \leq x \leq 3$.
This inequality can be written in terms of the absolute value as $|x| \leq 3$.

(A) For the first inequality: $3x + 8 < 17$ $\Rightarrow 3x < 9$ $\Rightarrow x < 3$.
For the second inequality: $2x + 8 \geq 12$ $\Rightarrow 2x \geq 4$ $\Rightarrow x \geq 2$.
Thus,Statement $(I)$ is true.
The common set of solutions is the intersection of $x < 3$ and $x \geq 2$,which is the interval $[2, 3)$.
Since Statement $(II)$ claims the set is $(2, 3)$,which excludes $2$,Statement $(II)$ is false.

(B) Given the inequality: $(8-t)^2 < t^2-3t-10$
Expanding the left side: $64-16t+t^2 < t^2-3t-10$
Subtracting $t^2$ from both sides: $64-16t < -3t-10$
Rearranging the terms: $64+10 < 16t-3t$
$74 < 13t$
$t > \frac{74}{13}$
Thus,the solution set is $t \in (\frac{74}{13}, \infty)$.