@rise_too 2-2×3+3
Now BODMAS
Multiply 2×3 first
2-2×3+3 = 2-6+3
Left to right calculus BODMAS rules 3
First subtract 6 from 2
2-6+3= -4+3
Then add 3 to -4
-4+3 = -1
Answer is -1
Tell me is not correct make I sue you
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@_Rezelvy Let us use an expansion method we must still get the same answer
Let check
7(x-5)+10= 31
Open the brack and multiply it by 7
7×x-7×5+10=31
7x-35+10=31
7x-25= 31
Combine the like terms
7x=31+25
7x=56
Divide through by 7
7x/7= 56/7
X= 8
Is the same answer 8
Both method same
@_Rezelvy Solution by Equadratic Equation
7(x-5)+10= 31
Subrating all through by coefficient of 10
7(x-5)+10-10= 31-10
7(x-5)= 21
Divide through by the coefficient of 7
7(x-5)/7= 21/7
X-5=3
Collect the like terms
X= 3+5
X= 8 tell me is not correct and we meet int world center
@AlexaBliz_ Final answer is 12 and that is Hiding ARITHMETICS unused to get the answer tell me is not correct and let us meet in the world cabinet and see
@AlexaBliz_ HIDE ARITHMETICS IS EQUAL TO 6 FRONTS AND X BACK
ACCORDING TO THE RULES OF HIDING ARETHEMETIS FRONT MUST BE EQUAL TO THE BACK.
XF + XB =HF+HB
2(F) + 2(B) = 2HFB
THE VALUE OF FRONT (F) IS 6 EQUIVALENT TO THE VALUE OF BACK WHICH SAME 6
2(6)+2(6) = 2HFB
12+12 = 2HFB
HFB = 24/2
12