Question

(z-1) /3 = 1 + (z-2) /4; solve the equation and verify​

Answer 1

(z-1)/3= 1+ (z-2)/4

=> (z-1)/3 - (z-2)/4 = 1

=> [4(z-1) - 3(z-2)]/12=1

=> 4(z-1) -3(z-2)= 12

=> 4z-4-3z+6=12

=>z+2=12

=>z=10

Verification: put z=10 in the equation and see if LHS=RHS

(10-1)/3= 1+ (10-2)/4

9/3= 1+ 8/4

3= 1+2

3=3

LHS=RHS

Verified.

Answer 2

$\frac{z - 1}{3} = 1 + \frac{z - 2}{4} \\ \frac{z - 1}{3} = \frac{4 + z - 2}{4} \\ \frac{z - 1}{3} = \frac{z + 2}{4} \\ 4(z - 1) = 3(z + 2) \\ 4z - 4 = 3z + 6 \\ 4z - 3z = 6 + 4 \\ \large{ \boxed{z = \underline{ \underline{10}}}} \\ \\ \large{\underline{Let's \: \: verify} :} \\ \frac{z - 1}{3} = 1 + \frac{z - 2}{4} \\ \frac{10 - 1}{3} = 1 + \frac{10 - 2}{4} \\ \frac{9}{3} = 1 + \frac{8}{4} \\ 3 = 1 + 2 \\ \large{ \boxed{ \underline{ \underline{3 = 3}}}}$