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1/3(n+1)=1/6(3n-5)
I really need help on how to solve this i know the answer but im confused on how we got it

Sagot :

SAM9
1/3(n+1)=1/6(3n-5)
1/3n+1/3 = 1/2n-5/6
1/3 = 1/6n - 5/6
7/6 = 1/6n
so n = (7/6) / (1/6)
     n = 7
frika

Consider the equation [tex] \dfrac{1}{3}(n+1)=\dfrac{1}{6}(3n-5). [/tex]

1 step. Multiply this equation by 6:

[tex] \dfrac{6}{3}(n+1)=\dfrac{6}{6}(3n-5),\\ 2(n+1)=3n-5,\\2n+2=3n-5. [/tex]

2 step. Separate expressions in such way, that variables are in the left side and the numbers are in the right side:

[tex] 2n-3n=-5-2. [/tex]

3 step. Add expressions in both sides:

[tex] n\cdot (2-3)=-(5+2),\\-1\cdot n=-7,\\-n=-7. [/tex]

4 step. Multiply this equaiton by -1:

n=7.

Answer: n=7.

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