7th Class Mathematics Simple Equations

  • question_answer 1)
    Give first the step you will use to separate the variable and then solve the equation (a) \[x-1=0\]                       (b) \[x+1=0\]                     (c) \[x-1=5\]                       (d) \[x+6=2\] (e) \[y-4=-7\]                     (f) \[y-4=4\]                       (g) \[y+4=4\]                      (h) \[y+4=-4\]                

    Answer:

                    (a) The given equation is \[x-1=0\] Add 1 to both sides, \[x-1+1=0+1\] \[\Rightarrow \]               \[x=1\] It is the required solution. Check. Put the solution \[x=1\] back into the equation. \[\text{L}\text{.H}\text{.S}=x-1=1-1=0=\text{R}\text{.H}\text{.S}\] The solution is thus checked for its correctness. (b) The given equation is \[x+1=0\] Subtracting 1 from both sides, \[x+1-1=0-1\] \[\Rightarrow \]               \[x=-1\] It is the required solution. Check. Put the solution \[x=-1\] back into the equation. \[\text{L}\text{.H}\text{.S}\text{.}=x+1=(-1)+1=0=\text{R}\text{.H}\text{.S}\text{.}\] The solution is thus checked for its correctness. (c) The given equation is \[x-1=5\] Add 1 to both sides, \[x+1-1=5+1\text{ }\Rightarrow \text{ }x=6\] It is the required solution Check. Put the solution \[x=6\] back into the equation. \[\text{L}.\text{H}.\text{S}.=x-1=6-1=5=\text{R}.\text{H}.\text{S}.\] The solution is thus checked for its correctness. (d) The given equation is \[x+6=2\] Subtracting 6 from both sides, \[x+6-6=2-6\text{ }\Rightarrow \text{ }x=-4\] It is the required solution. Check. Put the solution \[x=-\text{ }4\] back into the equation. \[\text{L}.\text{H}.\text{S}.=x+\text{6}=-\text{4}+\text{6}=\text{2}=\text{R}.\text{H}.\text{S}.\] The solution is thus checked for its correctness. (e) The given equation is \[y-4=-7\] Add 4 to both side, \[y-4+4=-7+4\]\[\Rightarrow \] \[y=-3\] It is the required solution, Check. Put the solution \[y=-\text{ }3\] back into the equation. \[\text{L}.\text{H}.\text{S}.=y-\text{4}=-\text{3}-\text{4}=-\text{7}=\text{ R}.\text{H}.\text{S}.\] The solution is thus checked for its correctness. (f) The given equation is \[y-4=4\] Add 4 to both sides, \[y-4+4=4+4\]\[\Rightarrow \]\[y=8\] It is the required solution. Check. Put the solution \[y=8\] back into the equation. \[\text{L}\text{.H}\text{.S}\text{.}=y-4=8-4=4=\text{R}\text{.H}\text{.S}\text{.}\] The solution is thus checked for its correctness. (g) The given equation is \[y+4=4\] Subtracting 4 from both sides, \[y+\text{4}-\text{4}=\text{ 4}-\text{4 }\Rightarrow \text{ }y=0\] It is the required solution. Check. Put the solution \[y=0\] back into the equation. \[\text{L}.\text{H}.\text{S}.\text{ }=y+\text{4}=0+\text{4}=\text{4}=\text{R}.\text{H}.\text{S}.\] The solution is thus checked for its correctness. (h) The given equation is \[y+4=-4\] Subtracting 4 form both sides, \[y+4-4=-4-4\]    \[\Rightarrow \] \[y=-8\] It is the required solution. Check. Put the solution \[y=-\text{ }8\] back into the equation. \[\text{L}.\text{H}.\text{S}.\text{ }=y+\text{4}=-\text{8}+\text{4}=-\text{4}=\text{ R}.\text{H}.\text{S}.\] The solution is thus checked for its correctness.


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