CBSE Questions for Class 6 Maths Algebra Quiz 3 - MCQExams.com

$${\textbf{Step -1: Considering option A.}}$$

               $${\text{We have given,}}$$ $$x - 5 = 1$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$ - 1 - 5 = 6$$

               $$ \Rightarrow  - 6 = 1$$ $${\text{Which is not possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is not a solution of }}x - 5 = 6.$$

$${\textbf{Step -2: Considering option B.}}$$

               $${\text{We have given,}}$$ $$2x + 5 = 7$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$2\left( { - 1} \right) + 5 = 7$$

               $$ \Rightarrow  - 2 + 5 = 7$$

               $$ \Rightarrow 3 = 7$$ $${\text{Which is not possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S }} \ne {\text{ R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is not a solution of 2}}x + 5 = 7.$$

$${\textbf{Step -3: Considering option C.}}$$

               $${\text{We have given,}}$$ $$2\left( {x - 2} \right) + 6 = 0$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$2\left( { - 1 - 2} \right) + 6 = 0$$

               $$ \Rightarrow 2\left( { - 3} \right) + 6 = 0$$

               $$ \Rightarrow  - 6 + 6 = 0$$

               $$ \Rightarrow 0 = 0$$ $${\text{Which is possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is a solution of 2}}\left( {x - 2} \right) + 6 = 0.$$

$${\textbf{Step -4: Considering option D.}}$$

               $${\text{We have given,}}$$ $$3x + 5 = 4$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$3\left( { - 1} \right) + 5 = 4$$

               $$ \Rightarrow  - 3 + 5 = 4$$

               $$ \Rightarrow 2= 4$$ $${\text{Which is not possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is not a solution of }}3x + 5 = 4.$$

$${\textbf{ Hence, option C}}{\textbf{. 2}}\left(\mathbf {x - 2} \right)\mathbf{ + 6 = 0}{\textbf{ is correct answer.}}$$

$${\textbf{Step -1: Considering option A.}}$$

               $${\text{We have given,}}$$ $$x +1 = 0$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$ - 1 +1 = 0$$

               $$ \Rightarrow  0 = 0$$ $${\text{Which is possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is a solution of }}x +1 = 0.$$

$${\textbf{Step -2: Considering option B.}}$$

               $${\text{We have given,}}$$ $$3x + 4 = 1$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$3\left( { - 1} \right) + 4 = 1$$

               $$ \Rightarrow  - 3 + 4 = 1$$

               $$ \Rightarrow 1 = 1$$ $${\text{Which is possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S }} = {\text{ R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is a solution of 3}}x + 4 = 1.$$

$${\textbf{Step -3: Considering option C.}}$$

               $${\text{We have given,}}$$ $$5x + 7 = 2$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$5\left( { - 1 } \right) + 7 = 2$$

               $$ \Rightarrow -5+ 7 = 2$$

               $$ \Rightarrow  2 = 2$$ $${\text{Which is possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is a solution of }}5x+7 =2.$$

$${\textbf{Step -4: Considering option D.}}$$

               $${\text{We have given,}}$$ $$x -1 = 2$$

               $${\text{Putting }}x =  - 1{\text{ we get,}}$$

               $$-1-1= 2$$

               $$ \Rightarrow  - 2 = 2$$ $${\text{Which is not possible}}{\text{.}}$$

               $${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$$

               $${\text{So,  - 1 is not a solution of }}x -1 = 2.$$

$${\textbf{ Hence, option D. }}\mathbf{ x-1 =2}{\textbf{ is correct answer.}}$$