Given Equation:-
[tex] \sf25 - 12 \div [(8 - 5) \times 1][/tex]
[tex] \\ \\ [/tex]
Step-by-step explanation:-
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - 12 \div [(8 - 5) \times 1][/tex]
first write the equation so that you'll emit minor mistakes of writing wrong digits.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - 12 \div [(3) \times 1][/tex]
subtract 8 and 5 , which will give us result as 3
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - 12 \div [3 \times 1][/tex]
open the brakets that contains 3.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - 12 \div [3][/tex]
multiply 3 with 1 which will result 3.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - 12 \div 3[/tex]
open the brakets of 3.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - \dfrac{12}{3} [/tex]
to make it easy to understand I have converted it into fraction form.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - \dfrac{2 \times 2 \times 3}{3} [/tex]
simplify 12 which is contained in numerator.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - \dfrac{2 \times 2 \times \cancel3}{\cancel3} [/tex]
as we can see after converting into fraction 3 is common both in numerator and denominator. So cancel ir.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 - (2 \times 2)[/tex]
we no longer need the fraction as 3 is canceled.
[tex] \\ [/tex]
[tex] \dashrightarrow \sf25 -4[/tex]
multiply 2 with 2 which will result 4.
[tex] \\ [/tex]
[tex] \dashrightarrow \bf21[/tex]
finally after subtracting 25 and 4 , we got our answer as 21.
[tex] \\ [/tex]
Rule applied:-
B : bracket
O : on/off
D: divide
M : multiply
A : add
S : subtract
