Basic of Linear Inequalities Questions in English

(C) The critical points are $x = 2$ and $x = 3$.
Case $1$: $x < 2$
$|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$
$-(x - 2) - (x - 3) = 7$
$-x + 2 - x + 3 = 7$
$-2x + 5 = 7$
$-2x = 2$
$x = -1$ (which satisfies $x < 2$)
Case $2$: $2 \le x < 3$
$|x - 2| = x - 2$ and $|x - 3| = -(x - 3)$
$(x - 2) - (x - 3) = 7$
$x - 2 - x + 3 = 7$
$1 = 7$ (which is impossible)
Case $3$: $x \ge 3$
$|x - 2| = x - 2$ and $|x - 3| = x - 3$
$(x - 2) + (x - 3) = 7$
$2x - 5 = 7$
$2x = 12$
$x = 6$ (which satisfies $x \ge 3$)
Thus,the solutions are $x = 6$ or $x = -1$.

(C) Given the equation $x + y + z = a$.
This is a simple linear equation in three variables.
The value of the expression $x + y + z$ is directly given as $a$.

(A) We are given the inequality $30x < 200$.
Dividing both sides by $30$,we get $x < \frac{200}{30}$.
This simplifies to $x < 6.66...$.
Since $x$ must be a natural number,the possible values for $x$ are $1, 2, 3, 4, 5, 6$.
Thus,the solution set is $\{1, 2, 3, 4, 5, 6\}$.

(A) Given the inequality $30x < 200$.
Divide both sides by $30$:
$x < \frac{200}{30}$
$x < 6.66...$
Since $x$ is an integer,the values of $x$ must be less than $6.66...$.
Therefore,the solution set is $\{..., -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.

(A) Given inequality: $5x - 3 < 3x + 1$
Subtract $3x$ from both sides: $5x - 3x - 3 < 1$
$2x - 3 < 1$
Add $3$ to both sides: $2x < 1 + 3$
$2x < 4$
Divide by $2$: $x < 2$
Since $x$ is an integer and $x < 2$,the set of solutions is $\{..., -2, -1, 0, 1\}$.

(A) The given inequality is $24x < 100$.
Divide both sides by $24$:
$x < \frac{100}{24}$
$x < \frac{25}{6}$
$x < 4.166...$
Since $x$ is a natural number,the values of $x$ must be positive integers less than $4.166...$.
Thus,the possible values for $x$ are $1, 2, 3,$ and $4$.
Therefore,the solution set is $\{1, 2, 3, 4\}$.

(A) The given inequality is $24x < 100$.
Divide both sides by $24$:
$x < \frac{100}{24}$
$x < \frac{25}{6}$
$x < 4.166...$
Since $x$ is an integer,the values of $x$ are all integers less than $4.166...$.
Therefore,the solution set is $\{\dots, -2, -1, 0, 1, 2, 3, 4\}$.

(A) The given inequality is $-12x > 30.$
Dividing both sides by $-12,$ the inequality sign reverses:
$x < \frac{30}{-12}$
$x < -2.5$
Since $x$ must be a natural number (i.e.,$x \in \{1, 2, 3, ...\}$),and there are no natural numbers less than $-2.5$,there is no solution for the given inequality.

(A) The given inequality is $5x - 3 < 7$.
Adding $3$ to both sides:
$5x - 3 + 3 < 7 + 3$
$5x < 10$
Dividing both sides by $5$:
$x < 2$
Since $x$ is an integer,the values of $x$ that satisfy the inequality $x < 2$ are all integers less than $2$.
Therefore,the solution set is $\{..., -2, -1, 0, 1\}$.

(A) The given inequality is $5x - 3 < 7$.
Adding $3$ to both sides,we get:
$5x - 3 + 3 < 7 + 3$
$5x < 10$
Dividing both sides by $5$,we get:
$x < 2$
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) The given inequality is $3x + 8 > 2$.
Subtract $8$ from both sides:
$3x > 2 - 8$
$3x > -6$
Divide both sides by $3$:
$x > -2$
Since $x$ is an integer greater than $-2$,the set of solutions is $\{-1, 0, 1, 2, \dots\}$.

(A) $4x + 3 < 5x + 7$
Subtract $3$ from both sides:
$4x < 5x + 4$
Subtract $5x$ from both sides:
$4x - 5x < 4$
$-x < 4$
Multiply by $-1$ and reverse the inequality sign:
$x > -4$
Thus,the solution set is $(-4, \infty)$.

(A) Given inequality: $3x - 7 > 5x - 1$
Subtracting $5x$ from both sides:
$3x - 5x - 7 > -1$
$-2x - 7 > -1$
Adding $7$ to both sides:
$-2x > -1 + 7$
$-2x > 6$
Dividing by $-2$ (note that dividing by a negative number reverses the inequality sign):
$x < \frac{6}{-2}$
$x < -3$
Thus,the solution set is $(-\infty, -3)$.

(A) Given inequality: $3(x-1) \leq 2(x-3)$
Expand both sides: $3x - 3 \leq 2x - 6$
Subtract $2x$ from both sides: $3x - 2x - 3 \leq -6$
Simplify: $x - 3 \leq -6$
Add $3$ to both sides: $x \leq -6 + 3$
Result: $x \leq -3$
Thus,the solution set is $(-\infty, -3]$.

(A) Given inequality: $x+\frac{x}{2}+\frac{x}{3} < 11$
Taking $x$ as a common factor:
$x(1+\frac{1}{2}+\frac{1}{3}) < 11$
Finding the common denominator for the terms inside the bracket:
$x(\frac{6+3+2}{6}) < 11$
$x(\frac{11}{6}) < 11$
Multiply both sides by $\frac{6}{11}$:
$x < 11 \times \frac{6}{11}$
$x < 6$
Thus,all real numbers $x$ which are less than $6$ are the solutions of the given inequality.
Hence,the solution set is $(-\infty, 6)$.

(A) Given inequality: $\frac{x}{3} > \frac{x}{2} + 1$
Subtract $\frac{x}{2}$ from both sides:
$\frac{x}{3} - \frac{x}{2} > 1$
Find a common denominator $(6)$:
$\frac{2x - 3x}{6} > 1$
Simplify the numerator:
$-\frac{x}{6} > 1$
Multiply both sides by $6$:
$-x > 6$
Multiply by $-1$ and reverse the inequality sign:
$x < -6$
Thus,the solution set is $(-\infty, -6)$.

(A) Given inequality: $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$
Multiply both sides by $15$ (the $LCM$ of $3$ and $5$):
$9(x-2) \leq 25(2-x)$
Expand the terms:
$9x - 18 \leq 50 - 25x$
Add $25x$ to both sides:
$9x + 25x - 18 \leq 50$
$34x - 18 \leq 50$
Add $18$ to both sides:
$34x \leq 50 + 18$
$34x \leq 68$
Divide by $34$:
$x \leq 2$
Thus,the solution set is $(-\infty, 2]$.

(A) Given inequality: $\frac{1}{2}\left(\frac{3x}{5}+4\right) \geq \frac{1}{3}(x-6)$
Multiply both sides by $6$ to clear the denominators:
$3\left(\frac{3x}{5}+4\right) \geq 2(x-6)$
Expand the terms:
$\frac{9x}{5} + 12 \geq 2x - 12$
Rearrange the terms to isolate $x$:
$12 + 12 \geq 2x - \frac{9x}{5}$
$24 \geq \frac{10x - 9x}{5}$
$24 \geq \frac{x}{5}$
Multiply by $5$:
$120 \geq x$
Thus,all real numbers $x$ such that $x \leq 120$ are solutions.
The solution set is $(-\infty, 120]$.

(A) Given inequality: $2(2x + 3) - 10 < 6(x - 2)$
Expand the terms: $4x + 6 - 10 < 6x - 12$
Simplify: $4x - 4 < 6x - 12$
Rearrange the terms to isolate $x$: $-4 + 12 < 6x - 4x$
$8 < 2x$
Divide by $2$: $4 < x$ or $x > 4$
Thus,all real numbers $x$ greater than $4$ are the solutions.
Hence,the solution set is $(4, \infty)$.

(A) Given inequality: $37-(3x+5) \geq 9x-8(x-3)$
Step $1$: Simplify both sides of the inequality.
$37-3x-5 \geq 9x-8x+24$
$32-3x \geq x+24$
Step $2$: Collect the terms involving $x$ on one side and constants on the other.
$32-24 \geq x+3x$
$8 \geq 4x$
Step $3$: Divide by $4$.
$2 \geq x$ or $x \leq 2$
Thus,all real numbers $x$ which are less than or equal to $2$ are the solutions.
Hence,the solution set is $(-\infty, 2]$.

(A) Given inequality: $\frac{x}{4} < \frac{5x-2}{3} - \frac{7x-3}{5}$
Taking the $LCM$ of the denominators on the right side:
$\frac{x}{4} < \frac{5(5x-2) - 3(7x-3)}{15}$
$\frac{x}{4} < \frac{25x - 10 - 21x + 9}{15}$
$\frac{x}{4} < \frac{4x - 1}{15}$
Cross-multiplying:
$15x < 4(4x - 1)$
$15x < 16x - 4$
Rearranging the terms:
$4 < 16x - 15x$
$4 < x$ or $x > 4$
Thus,all real numbers $x$ greater than $4$ are solutions.
The solution set is $(4, \infty)$.

(A) Given inequality: $\frac{(2x-1)}{3} \geq \frac{(3x-2)}{4} - \frac{(2-x)}{5}$
Taking the $LCM$ of the denominators on the $RHS$: $\frac{(2x-1)}{3} \geq \frac{5(3x-2) - 4(2-x)}{20}$
Expanding the terms: $\frac{(2x-1)}{3} \geq \frac{15x - 10 - 8 + 4x}{20}$
Simplifying: $\frac{(2x-1)}{3} \geq \frac{19x - 18}{20}$
Cross-multiplying: $20(2x - 1) \geq 3(19x - 18)$
$40x - 20 \geq 57x - 54$
Rearranging terms: $-20 + 54 \geq 57x - 40x$
$34 \geq 17x$
Dividing by $17$: $2 \geq x$ or $x \leq 2$
Thus,the solution set is $(-\infty, 2]$.

(N/A) $3x - 2 < 2x + 1$
$\Rightarrow 3x - 2x < 1 + 2$
$\Rightarrow x < 3$
The graphical representation of the solution $x < 3$ on the number line is shown below:
(The graph consists of a number line with an open circle at $3$,and the region to the left of $3$ is shaded to represent all values less than $3$.)

(N/A) $5x - 3 \geq 3x - 5$
$\Rightarrow 5x - 3x \geq -5 + 3$
$\Rightarrow 2x \geq -2$
$\Rightarrow \frac{2x}{2} \geq \frac{-2}{2}$
$\Rightarrow x \geq -1$
The graphical representation of the solution $x \geq -1$ on a number line is shown below. The point $-1$ is included (represented by a solid circle),and all values to the right of $-1$ are shaded.

(N/A) $3(1-x) < 2(x+4)$
$\Rightarrow 3-3x < 2x+8$
$\Rightarrow 3-8 < 2x+3x$
$\Rightarrow -5 < 5x$
$\Rightarrow \frac{-5}{5} < \frac{5x}{5}$
$\Rightarrow -1 < x$
The graphical representation of the solution $x > -1$ on the number line is shown by an open circle at $-1$ with the line extending to the right.

(N/A) Given inequality: $\frac{x}{2} \geq \frac{5x-2}{3} - \frac{7x-3}{5}$
$\Rightarrow \frac{x}{2} \geq \frac{5(5x-2) - 3(7x-3)}{15}$
$\Rightarrow \frac{x}{2} \geq \frac{25x - 10 - 21x + 9}{15}$
$\Rightarrow \frac{x}{2} \geq \frac{4x - 1}{15}$
Multiplying both sides by $30$:
$\Rightarrow 15x \geq 2(4x - 1)$
$\Rightarrow 15x \geq 8x - 2$
$\Rightarrow 15x - 8x \geq -2$
$\Rightarrow 7x \geq -2$
$\Rightarrow x \geq -\frac{2}{7}$
The solution set is $[-\frac{2}{7}, \infty)$. The graphical representation is a solid circle at $-\frac{2}{7}$ with a line extending to the right.

(N/A) Given the inequality: $-5 \leq \frac{5-3x}{2} \leq 8$
Multiply all parts by $2$:
$-10 \leq 5-3x \leq 16$
Subtract $5$ from all parts:
$-15 \leq -3x \leq 11$
Divide by $-3$ (note that the inequality signs reverse when dividing by a negative number):
$5 \geq x \geq -\frac{11}{3}$
Rearranging the inequality,we get:
$-\frac{11}{3} \leq x \leq 5$

(N/A) From inequality $(1)$,we have:
$3x - 7 < 5 + x$
$2x < 12$
$x < 6$ ..... $(3)$
From inequality $(2)$,we have:
$11 - 5x \leqslant 1$
$-5x \leqslant 1 - 11$
$-5x \leqslant -10$
Dividing by $-5$ reverses the inequality sign:
$x \geqslant 2$ ..... $(4)$
Combining $(3)$ and $(4)$,the solution is the set of all real numbers $x$ such that $2 \leqslant x < 6$.

(A) The given set is ${ x:x \in R, -12 < x < -10 }$.
Since the inequality is strict $( < )$,the endpoints are not included.
Therefore,the interval is $(-12, -10)$.

(A) Given inequality: $6 \leq -3(2x - 4) < 12$
Divide the entire inequality by $-3$. Remember that dividing by a negative number reverses the inequality signs:
$-2 \geq 2x - 4 > -4$
Add $4$ to all parts of the inequality:
$-2 + 4 \geq 2x > -4 + 4$
$2 \geq 2x > 0$
Divide by $2$:
$1 \geq x > 0$
Thus,the solution set is $(0, 1]$.

(A) Given inequality: $-3 \leq 4 - \frac{7x}{2} \leq 18$
Subtract $4$ from all parts:
$-3 - 4 \leq -\frac{7x}{2} \leq 18 - 4$
$-7 \leq -\frac{7x}{2} \leq 14$
Multiply by $-1$ and reverse the inequality signs:
$7 \geq \frac{7x}{2} \geq -14$
Divide by $7$:
$1 \geq \frac{x}{2} \geq -2$
Multiply by $2$:
$2 \geq x \geq -4$
Thus,the solution set is $[-4, 2]$.

(A) Given the inequality: $-15 < \frac{3(x-2)}{5} \leq 0$
Multiply all parts by $5$:
$-75 < 3(x-2) \leq 0$
Divide all parts by $3$:
$-25 < x-2 \leq 0$
Add $2$ to all parts:
$-25 + 2 < x \leq 0 + 2$
$-23 < x \leq 2$
Thus,the solution set is $(-23, 2]$.

(A) Given the inequality: $7 \leq \frac{3x+11}{2} \leq 11$
Multiply all parts by $2$:
$14 \leq 3x+11 \leq 22$
Subtract $11$ from all parts:
$14 - 11 \leq 3x \leq 22 - 11$
$3 \leq 3x \leq 11$
Divide by $3$:
$1 \leq x \leq \frac{11}{3}$
Thus,the solution set is $[1, 11/3]$.

(N/A) $5x + 1 > -24 \Rightarrow 5x > -25$
$\Rightarrow x > -5$ ..... $(1)$
$5x - 1 < 24 \Rightarrow 5x < 25$
$\Rightarrow x < 5$ ..... $(2)$
From $(1)$ and $(2)$,it can be concluded that the solution set for the given system of inequalities is $(-5, 5)$. The solution of the given system of inequalities can be represented on a number line as shown below:
(The number line shows an open interval between $-5$ and $5$,with open circles at $-5$ and $5$ indicating that these points are not included in the solution set.)

(N/A) First,solve the first inequality:
$2(x-1) < x+5$
$2x - 2 < x + 5$
$2x - x < 5 + 2$
$x < 7$ ..... $(1)$
Next,solve the second inequality:
$3(x+2) > 2-x$
$3x + 6 > 2 - x$
$3x + x > 2 - 6$
$4x > -4$
$x > -1$ ..... $(2)$
From $(1)$ and $(2)$,the solution set is the intersection of $x < 7$ and $x > -1$,which is $(-1, 7)$.
The solution set $(-1, 7)$ is represented on the number line as an open interval between $-1$ and $7$.

(N/A) First,solve the first inequality:
$3x - 7 > 2(x - 6)$
$3x - 7 > 2x - 12$
$3x - 2x > -12 + 7$
$x > -5$ ..... $(1)$
Next,solve the second inequality:
$6 - x > 11 - 2x$
$-x + 2x > 11 - 6$
$x > 5$ ..... $(2)$
From $(1)$ and $(2)$,the common solution is the intersection of the two intervals,which is $x > 5$.
Thus,the solution set is $(5, \infty)$.
The graphical representation on the number line is shown below:

(N/A) Given the inequality: $x + 2 < -8$.
Subtracting $2$ from both sides: $x < -8 - 2$,which simplifies to $x < -10$.
$(1)$ For $x \in N$ (natural numbers): Since there are no natural numbers less than $-10$,the solution set is $\phi$ (the empty set).
$(2)$ For $x \in Z$ (integers): The integers less than $-10$ are $\{\ldots, -13, -12, -11\}$.
$(3)$ For $x \in R$ (real numbers): The solution set is the interval $(-\infty, -10)$.

(N/A) Given inequality: $-6x \leq 18$
Divide both sides by $-6$. Remember that dividing by a negative number reverses the inequality sign:
$x \geq \frac{18}{-6}$
$x \geq -3$
$(1)$ If $x \in N$ (Natural numbers: $\{1, 2, 3, \ldots\}$),then $x \in \{1, 2, 3, \ldots\}$.
$(2)$ If $x \in Z$ (Integers: $\{\ldots, -2, -1, 0, 1, 2, \ldots\}$),then $x \in \{-3, -2, -1, 0, 1, 2, \ldots\}$.
$(3)$ If $x \in R$ (Real numbers),then $x \in [-3, \infty)$.

(A) Given the inequality: $3x - 22 \geq 5$.
Add $22$ to both sides: $3x \geq 5 + 22$.
$3x \geq 27$.
Divide both sides by $3$: $x \geq 9$.
Since $x \in R$,the solution set is $[9, \infty)$.

(A) Given inequality: $3(2x - 1) + 5 \leq \frac{1}{2}(x + 15)$
Multiply both sides by $2$ to clear the fraction: $6(2x - 1) + 10 \leq x + 15$
Expand the terms: $12x - 6 + 10 \leq x + 15$
Simplify: $12x + 4 \leq x + 15$
Subtract $x$ from both sides: $11x + 4 \leq 15$
Subtract $4$ from both sides: $11x \leq 11$
Divide by $11$: $x \leq 1$
Thus,the solution set is $x \in (-\infty, 1]$.

(A) Given the inequality: $\frac{3-2x}{5} < \frac{x}{3} + 2$
Multiply the entire inequality by $15$ (the least common multiple of $5$ and $3$) to clear the denominators:
$15 \times \frac{3-2x}{5} < 15 \times (\frac{x}{3} + 2)$
$3(3-2x) < 5x + 30$
$9 - 6x < 5x + 30$
Subtract $5x$ from both sides:
$9 - 11x < 30$
Subtract $9$ from both sides:
$-11x < 21$
Divide by $-11$ and reverse the inequality sign:
$x > -\frac{21}{11}$
Thus,the solution set is $x \in (-\frac{21}{11}, \infty)$.

(A) Given inequality: $\frac{x+1}{2} > 6(x+2)$
Multiply both sides by $2$: $x+1 > 12(x+2)$
Expand the right side: $x+1 > 12x + 24$
Subtract $x$ from both sides: $1 > 11x + 24$
Subtract $24$ from both sides: $-23 > 11x$
Divide by $11$: $-\frac{23}{11} > x$
Thus,the solution set is $x \in (-\infty, -\frac{23}{11})$.

(A) Given inequality: $3(x-1)+2(x-2)<5(x+2)$
Expand the terms: $3x-3+2x-4<5x+10$
Combine like terms on the left side: $5x-7<5x+10$
Subtract $5x$ from both sides: $-7<10$
Since $-7<10$ is a true statement for all real values of $x$,the solution set is the set of all real numbers,$x \in \mathbb{R}$.

Given inequality: $\frac{2x-1}{3} + 5 < \frac{3x-1}{2} - 2$
Multiply the entire inequality by $6$ to clear the denominators:
$2(2x-1) + 30 < 3(3x-1) - 12$
$4x - 2 + 30 < 9x - 3 - 12$
$4x + 28 < 9x - 15$
Subtract $4x$ from both sides:
$28 < 5x - 15$
Add $15$ to both sides:
$43 < 5x$
Divide by $5$:
$x > \frac{43}{5}$ or $x > 8.6$
The solution set is $(8.6, \infty)$.
On the number line,this is represented by an open circle at $8.6$ with a line extending to the right.

(N/A) Given inequality: $\frac{3x}{2} + 15 \leq \frac{2x}{3} + 6$
Subtract $15$ from both sides: $\frac{3x}{2} \leq \frac{2x}{3} - 9$
Subtract $\frac{2x}{3}$ from both sides: $\frac{3x}{2} - \frac{2x}{3} \leq -9$
Find a common denominator $(6)$: $\frac{9x - 4x}{6} \leq -9$
$\frac{5x}{6} \leq -9$
Multiply both sides by $6$: $5x \leq -54$
Divide by $5$: $x \leq -10.8$
The solution set is $(-\infty, -10.8]$.
On the number line,this is represented by a solid circle at $-10.8$ with a line extending to the left.

(N/A) Given inequality: $\frac{4x+1}{9} > \frac{9x+1}{4} - 2$
Multiply both sides by $36$ (the $LCM$ of $9$ and $4$):
$4(4x+1) > 9(9x+1) - 72$
$16x + 4 > 81x + 9 - 72$
$16x + 4 > 81x - 63$
$4 + 63 > 81x - 16x$
$67 > 65x$
$x < \frac{67}{65}$
Thus,the solution set is $\left(-\infty, \frac{67}{65}\right)$.
The number line representation shows all values to the left of $\frac{67}{65}$ with an open circle at $\frac{67}{65}$.

(N/A) Given inequality: $\frac{x-1}{2}+5 \geq \frac{2x-1}{3}+15$
Subtract $5$ from both sides: $\frac{x-1}{2} \geq \frac{2x-1}{3}+10$
Multiply by $6$ to clear denominators: $3(x-1) \geq 2(2x-1) + 60$
Expand: $3x-3 \geq 4x-2+60$
Simplify: $3x-3 \geq 4x+58$
Rearrange terms: $-3-58 \geq 4x-3x$
Result: $-61 \geq x$ or $x \leq -61$
The solution set is $(-\infty, -61]$.
The number line representation shows a solid circle at $-61$ with a line extending to the left.

(N/A) Given inequality: $\frac{x}{3} + 5 \geq \frac{x}{2} + 7$
Subtract $5$ from both sides:
$\frac{x}{3} \geq \frac{x}{2} + 2$
Subtract $\frac{x}{2}$ from both sides:
$\frac{x}{3} - \frac{x}{2} \geq 2$
Find a common denominator $(6)$:
$\frac{2x - 3x}{6} \geq 2$
$-\frac{x}{6} \geq 2$
Multiply both sides by $-6$ (remember to reverse the inequality sign when multiplying by a negative number):
$x \leq -12$
The solution set is $(-\infty, -12]$.
This is represented on the number line by a solid circle at $-12$ and a shaded line extending to the left.

(D) The given inequality is $|x-2| \geq 8.$
By the property of absolute value inequalities,$|u| \geq a$ implies $u \leq -a$ or $u \geq a.$
Applying this to the given expression:
$x-2 \leq -8$ or $x-2 \geq 8.$
Solving the first part:
$x \leq -8 + 2$
$x \leq -6.$
Solving the second part:
$x \geq 8 + 2$
$x \geq 10.$
Combining these intervals,we get $x \in (-\infty, -6] \cup [10, \infty).$
Thus,the correct option is $D.$

(A) Given the inequality: $\left|\frac{3x-4}{2}\right| \leq \frac{5}{12}$
This is equivalent to: $-\frac{5}{12} \leq \frac{3x-4}{2} \leq \frac{5}{12}$
Multiply the entire inequality by $2$: $-\frac{5}{6} \leq 3x-4 \leq \frac{5}{6}$
Add $4$ to all parts: $4 - \frac{5}{6} \leq 3x \leq 4 + \frac{5}{6}$
$\frac{24-5}{6} \leq 3x \leq \frac{24+5}{6}$
$\frac{19}{6} \leq 3x \leq \frac{29}{6}$
Divide by $3$: $\frac{19}{18} \leq x \leq \frac{29}{18}$
Thus,the solution set is $[\frac{19}{18}, \frac{29}{18}]$.