Question

Using the trial and error method, find the solution of the equation a /2 – 3 = 1

Answer 1

Given :-

$\sf \bullet \; \; \dfrac{a}{2} -3 = 1$

To Find :-  

• Value of the variable a by trial and error method  

Solution :-  

By trial and error method  

$\sf \star\;\;Let\;the\;a\;be\;5:$

$\sf \implies \dfrac{5}{2} -3 = 1$

$\sf \implies \dfrac{5-6}{2} = 1$

$\sf \implies \dfrac{-1}{2} = 1$

LHS ≠  RHS  

∴ This is not the valid answer  

_______________

$\sf \star\;\;Let\;the\;a\;be\;6$  

$\sf \implies \dfrac{6}{2} -3 = 1$

$\sf \implies 3 -3 = 1$

$\sf \implies 0 = 1$

LHS ≠ RHS  

∴This is not the valid answer  

_______________

$\sf \star\;\;Let\;the\;a\;be\;7:$

$\sf \implies \dfrac{7}{2} -3 = 1$

 

$\sf \implies \dfrac{7-6}{2} = 1$

$\sf \implies \dfrac{1}{2} = 1$

LHS ≠ RHS  

∴ This is not the valid answer  

_______________

$\sf \star\;\;Let\;the\;a\;be\;8:$  

$\sf \implies \dfrac{8}{2} -3 = 1$

$\sf \implies 4 -3 = 1$

$\sf \implies 1 = 1$

LHS = RHS  

∴ This is the valid answer  

_______________

Therefore,

• The value of ‘a’ is 8  

_______________

Answer 2

$\large\sf\underline{Given}$

• $\sf\:\frac{a}{2}-3=1$

‎

$\large\sf\underline{To\:Find}$

• Value of a .

‎

$\large\sf\underline{Solution}$

‎

$\sf\:\frac{a}{2}-3=1$

‎

• LCM of 2 and 1

‎

$\sf➞\:\frac{a-(3 \times 2)}{2}=1$

‎

$\sf➞\:\frac{a-6}{2}=1$

‎

• Cross multiplying

‎

$\sf➞\:a-6=2 \times 1$

‎

$\sf➞\:a-6=2$

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• Transporting 6 to RHS

‎

$\sf➞\:a=2+6$

‎

${\sf{{\purple{➞\:a=8}}}}$

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!! Hope it helps !!