16-(7m+3)=8(1-m)/8

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Solution for 16-(7m+3)=8(1-m)/8 equation: 



16-(7m+3)=8(1-m)/8

We move all terms to the left:

16-(7m+3)-(8(1-m)/8)=0

We add all the numbers together, and all the variables

-(7m+3)-(8(-1m+1)/8)+16=0

We get rid of parentheses

-7m-(8(-1m+1)/8)-3+16=0

We multiply all the terms by the denominator


-7m*8)-(8(-1m+1)-3*8)+16*8)=0

We add all the numbers together, and all the variables

-7m*8)-(8(-1m+1)=0

Wy multiply elements

-56m^2=0

a = -56; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-56)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{-112}=0$

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