Solve the given inequality
Question: Solve the given inequality $y-2 \leq 3 x$ graphically in two - dimensional plane. Solution: The graphical representation of $y-2 \leq 3 x$ is given by blue line in the figure below. This lines divides $x-y$ plane into two parts. Select a point (not on the line), which lies on one of the two parts, to determine whether the point satisfies the given inequality or not. We select the point as $(0,0)$ It is observed that $0-2 \leq 3 \times 0$ or $-2 \leq 0$ which is true. Therefore, the sol...
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Question: $\frac{3 t+5}{4}-1=\frac{4 t-3}{5}$ Solution: Given, $\frac{3 t+5}{4}-1=\frac{4 t-3}{5}$ $\Rightarrow$ $\frac{3 t+5-4}{4}=\frac{4 t-3}{5}$ $\Rightarrow$ $5(3) t+5-4)=4(4 t-3)$ [by cross-multiplication] $\Rightarrow \quad 5(3 t+1)=4(4 t-3)$ $\Rightarrow$ $15 t+5=16 t-12$ $\Rightarrow$ $15 t-16 t=-12-5$ [transposing 16 tto LHS and 5 to RHS] $\Rightarrow$ $-t=-17$ $\Rightarrow$ $\frac{-t}{-1}=\frac{-17}{-1}$ [dividing both sides by $-1$ ] $\therefore$ $t=17$...
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Question: Solve the given inequality $x-y \leq 3$ graphically in two - dimensional plane. Solution: The graphical representation of $x-y \leq 3$ is given by blue line in the figure below. This lines divides $x-y$ plane into two parts. Select a point (not on the line), which lies on one of the two parts, to determine whether the point satisfies the given inequality or not. We select the point as $(0,0)$ It is observed that $0-0 \leq 3$ or $0 \leq 3$ which is true. Therefore, the solution for the ...
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Question: $3 x-\frac{x-2}{3}=4-\frac{x-1}{4}$ Solution: Given, $3 x-\frac{x-2}{3}=4-\frac{x-1}{4}$ $\Rightarrow$ $\frac{9 x-(x-2)}{3}=\frac{16-(x-1)}{4}$ $\Rightarrow$ $4(9 x-x+2)=3(16-x+1)$ [by cross-multiplication] $\Rightarrow$ $4(8 x+2)=3(-x+17)$ $\Rightarrow$ $32 x+3 x=51-8 \quad$ [transposing $-3 x$ to LHS and 8 to RHS] $\Rightarrow$ $35 x=43$ $\Rightarrow$ $\frac{35 x}{35}=\frac{43}{35}$ [dividing both sides by 35 ] $\therefore$ $x=\frac{43}{35}$...
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Question: Solve the given inequality $x+y \geq 4$ graphically in two - dimensional plane. Solution: The graphical representation of $x+y \geq 4$ is given by blue line in the figure below. This lines divides $x-y$ plane into two parts Select a point (not on the line),which lies on one of the two parts, to determine whether the point satisfies the given inequality or not. We select the point as $(0,0)$ It is observed that $0+0 \geq 4$ or $0 \geq 4$ which is false. Therefore, the solution for the g...
Read More →1-(x-2)-[(x-3)-(x-1)]=0
Question: 1-(x-2)-[(x-3)-(x-1)]=0 Solution: Given, $\quad 1-(x-2)-[(x-3)-(x-1)]=0$ $\Rightarrow$ $1-x+2-[x-3-x+1]=0$ $\Rightarrow \quad 3-x-[-2]=0$ $\Rightarrow \quad 3-x+2=0$ $\Rightarrow \quad-x+5=0$ $\Rightarrow$ $-x+5=0$ $\Rightarrow$ $-x=-5$[transposing 5 to RHS] $\Rightarrow$ $\frac{-x}{-1}=\frac{-5}{-1}$ [dividing both sides by -1] $\therefore$ $x=5$...
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Question: $\frac{2 x-1}{5}=\frac{3 x+1}{3}$ Solution: Given $\frac{2 x-1}{5}=\frac{3 x+1}{3}$ $\Rightarrow$ $3(2 x-1)=5(3 x+1)$ [by cross-multiplication] $\Rightarrow$ $6 x-3=15 x+5$ $\Rightarrow$ $6 x-15 x=3+5$ [transposing $15 x$ to LHS and $-3$ to RHS] $\Rightarrow$ $-9 x=8$ $\Rightarrow$ $\frac{-9 x}{-9}=\frac{8}{-9}$ [dividing both sides by $-9$ ] $\therefore$$x=\frac{-8}{9}$...
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Question: Solve $|x|4$, when $x \in R$. Solution: $|x|4$ Square $\Rightarrow x^{2}16$ $\Rightarrow x^{2}-160$ $\Rightarrow x^{2}-4^{2}0$ $\Rightarrow(x+4)(x-4)0$ Observe that when x is greater than 4, (x + 4)(x 4) is positive And for each root the sign changes hence We want greater than 0 that is positive part Hence $x$ should be less than $-4$ and greater than 4 for $(x+4)(x-4)$ to be positive $x$ less than $-4$ means $x$ is from negative infinity to $-4$ and $x$ greater than 4 means $x$ is fro...
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Question: Solve $|x|4$, when $x \in R$. Solution: $|x|4$ Square $\Rightarrow x^{2}16$ $\Rightarrow x^{2}-160$ $\Rightarrow x^{2}-4^{2}0$ $\Rightarrow(x+4)(x-4)0$ Observe that when x is greater than 4, (x + 4)(x 4) is positive And for each root the sign changes hence We want less than 0 that is negative part Hence $x$ should be between $-4$ and 4 for $(x+4)(x-4)$ to be negative Hence $x \in(-4,4)$ Hence the solution set of $|x|4$ is $(-4,4)$...
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Question: $\frac{x+1}{4}=\frac{x-2}{3}$ Solution: Given, $\frac{x+1}{4}=\frac{x-2}{3}$ $\Rightarrow \quad 3(x+1)=4(x-2)$ [by cross-multiplication] $\Rightarrow \quad 3 x+3=4 x-8$ $\Rightarrow$ $3 x-4 x=-8-3$ [transposing $4 x$ to LHS and 3 to RHS] $\Rightarrow$ $-x=-11$ [dividing both sides by $-1$ ] $\Rightarrow$ $\frac{-x}{-1}=\frac{-11}{-1}$ $\therefore$ $x=11$...
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Question: Solve $\frac{x}{x-5}\frac{1}{2}$ when $x \in \mathbf{R}$. Solution: $\frac{x}{x-5}\frac{1}{2}$ $\Rightarrow \frac{x}{x-5}-\frac{1}{2}0$ $\Rightarrow \frac{2 x-x+5}{2(x-5)}0$ $\Rightarrow \frac{x+5}{x-5}0$ Observe that $\frac{x+5}{x-5}$ is zero at $\mathrm{x}=-5$ and not defined at $\mathrm{x}=5$ Hence plotting these two points on number line Now for $x5, \frac{x+5}{x-5}$ is positive For every root and not defined value of $\frac{x+5}{x-5}$ the sign will change We want greater than zero...
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Question: $\frac{1}{2}(x+1)+\frac{1}{3}(x-1)=\frac{5}{12}(x-2)$ Solution: Given, $\frac{1}{2}(x+1)+\frac{1}{3}(x-1)=\frac{5}{12}(x-2)$ $\Rightarrow \quad \frac{x}{2}+\frac{1}{2}+\frac{x}{3}-\frac{1}{3}=\frac{5 x}{12}-\frac{5}{6}$ $\Rightarrow \quad \frac{x}{2}+\frac{x}{3}-\frac{5 x}{12}=\frac{1}{3}-\frac{1}{2}-\frac{5}{6} \quad\left[\right.$ transposing $\frac{1}{2}, \frac{1}{3}$ to RHS and $\frac{5 x}{12}$ to LHS $\Rightarrow$ $\frac{6 x+4 x-5 x}{12}=\frac{2-3-5}{6}$ $\Rightarrow$ $\frac{5 x}{1...
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Question: Solve $\frac{3}{x-2}1$, when $x \in R$ Solution: $\frac{3}{x-2}1$ $\Rightarrow \frac{3}{x-2}-10$ $\Rightarrow \frac{3-x+2}{x-2}0$ $\Rightarrow \frac{5-x}{x-2}0$ Observe that $\frac{5-x}{x-2}$ is zero at $\mathrm{x}=5$ and not defined at $\mathrm{x}=2$ Hence plotting these two points on number line Now for $x5, \frac{5-x}{x-2}$ is negative For every root and not defined value of $\frac{5-x}{x-2}$ the sign will change We want the negative part hence x 2 and x 5 x 2 means x is from negati...
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Question: $\frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12}$ Solution: Given, $\frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12}$ $\Rightarrow \quad \frac{x}{2}-\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{1}{6}+\frac{1}{12}$ $\Rightarrow \quad \frac{2 x-x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{2+1}{12}$ $\Rightarrow \quad \frac{2 x-x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{2+1}{12}$ $\Rightarrow$ $\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{3}{12...
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Question: Solve $x+54 x-10$, when $x \in R$ Solution: $x+54 x-10$ $\Rightarrow 5+104 x-x$ $\Rightarrow 153 x$ Divide by 3 $\Rightarrow 5x$ $\Rightarrow x5$ $x5$ means $x$ is from $-\infty$ to 5 that is $x \in(-\infty, 5)$ Hence solution of $x+54 x-10$ is $x \in(-\infty, 5)$...
Read More →5(x-1)-2(x+8)=0
Question: 5(x-1)-2(x+8)=0 Solution: Given, $5(x-1)-2(x+8)=0$ $\Rightarrow \quad 5 x-5-2 x-16=0$ $\Rightarrow \quad 3 x-21=0$ $\Rightarrow$ $3 x=21$ [transposing $-21$ to RHS] $\Rightarrow$ $\frac{3 x}{3}=\frac{21}{3}$ [dividing both sides by 3 ] $\therefore$ $x=7$...
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Question: Solve $-4 x16$, when $x \in Z$ Solution: We have to find integer values of x for which -4x 16 (why only integer values because it is given that $x \in Z$ that is set of integers) $-4 x16$ $\Rightarrow-x4$ $\Rightarrow x-4$ The integers less than $-4$ are $-5,-6,-7,-8, \ldots$ Generalizing the solution in terms of n $x=-(4+n)$ where $n$ is integers from 1 to infinity Hence solution of $-4 x16$ is $x=-(4+n) \forall n=\{1,2,3,4 \ldots\}$...
Read More →4t-3-(3t+1)=5t-4
Question: 4t-3-(3t+1)=5t-4 Solution: Given, $4 t-3-(3 t+1)=5 t-4$ $\Rightarrow \quad 4 t-3-3 t-1=5 t-4$ $\Rightarrow \quad t-4=5 t-4$ $\Rightarrow \quad t-5 t=-4+4$ [transposing $5 t$ to LHS and $-4$ to RHS] $\Rightarrow \quad-4 t=0$ $\Rightarrow$ $\frac{-4 t}{-4}=\frac{0}{-4}$ [dividing both sides by $-4$ ] $\therefore$ $t=0$...
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Question: Solve the system of in equation $x-2 \geq 0,2 x-5 \leq 3$ Solution: We have to find values of x for which both the equations hold true $x-2 \geq 0$ and $2 x-5 \leq 3$ We will solve both the equations separately and then their intersection set will be solution of the system $x-2 \geq 0$ $\Rightarrow x \geq 2$ Hence $x \in(2, \infty)$ Now, $2 x-5 \leq 3$ $\Rightarrow 2 x \leq 3+5$ $\Rightarrow 2 x \leq 8$ $\Rightarrow x \leq 4$ Hence $x \in(-\infty, 4)$ The intersection of set $(2, \inft...
Read More →10x-5-7x=5x+15-8
Question: 10x-5-7x=5x+15-8 Solution: Given, $10 x-5-7 x=5 x+15-8$ $\Rightarrow \quad 10 x-7 x-5 x=5+15-8$ [transposing $5 x$ to LHS and $-5$ to RHS] $\Rightarrow \quad-2 x=12$ $\Rightarrow \quad \frac{-2 x}{-2}=\frac{12}{-2}$ [dividing both sides by -2] $\therefore \quad x=-6$...
Read More →Find the solution set of the in equation
Question: Find the solution set of the in equation $\frac{x+1}{x+2}1$. Solution: $\frac{x+1}{x+2}1$ $\Rightarrow \frac{x+1}{x+2}-10$ $\Rightarrow \frac{x+1-x-2}{x+2}0$ $\Rightarrow \frac{-1}{x+2}0$ We have to find values of $x$ for which $\frac{-1}{x+2}0$ that is $\frac{-1}{x+2}$ is negative The numerator of $\frac{-1}{x+2}$ is $-1$ which is negative hence for $\frac{-1}{x+2}$ to be negative $x+2$ must be positive (otherwise if $x+2$ is negative then negative upon negative will be positive) That...
Read More →8x-7-3x=6x-2x-3
Question: 8x-7-3x=6x-2x-3 Solution: Given, $\quad 8 x-7-3 x=6 x-2 x-3$ $\Rightarrow 8 x-3 x-6 x+2 x=-3+7$ [transposing $6 x,-2 x$ to LHS and $-7$ to RHS] $\therefore$ $x=4$...
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Question: 0.4(3x-1)=0.5x +1 Solution: Given, $0.4(3 x-1)=0.5 x+1$ $\Rightarrow$ $1.2 x-0.4=0.5 x+1$ $\Rightarrow$ $1.2 x-0.5 x=1+0.4$ [transposing $0.5 x$ to LHS and $-0.4$ to RHS] $\Rightarrow$$0.7 x=1.4$ $\Rightarrow$ $\frac{0.7 x}{0.7}=\frac{1.4}{0.7}$ [dividing both sides by $0.7$ ] $\therefore$ $x=2$...
Read More →Find the solution set of the in equation
Question: Find the solution set of the in equation $\frac{|x-2|}{(x-2)}0 . x \neq 2$ Solution: $\frac{|x-2|}{(x-2)}0$ means we have to find values of x for which $\frac{|x-2|}{(x-2)}$ is negative Observe that the numerator $|x-2|$ is always positive because of mod, hence for $^{\frac{|x-2|}{(x-2)}}$ to be a negative quantity the denominator $(x-2)$ has to be negative That is $x-2$ should be less than 0 $\Rightarrow x-20$ $\Rightarrow x2$ Hence $x$ should be less than 2 for $\frac{|x-2|}{(x-2)}0$...
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Question: $\frac{x}{5}=\frac{x-1}{6}$ Solution: Given, $\frac{x}{5}=\frac{x-1}{6}$ $\Rightarrow$ $6 x=5(x-1)$ [by cross-multiplication] $\Rightarrow$ $6 x=5 x-5$ $\Rightarrow$ $6 x-5 x=-5$ [transposing $5 x$ to LHS] $\therefore$ $x=-5$...
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