Answered

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hi i need some help with this problem can you help me?find the value of x in this equation...1/3(6x-5)-x=1/3-2(x+1)

Sagot :

Answer:

x = 0

Explanation:

We want to solve

[tex]\frac{1}{3}\mleft(6x-5\mright)-x=\frac{1}{3}-2(x+1)[/tex]

for x.

Now expanding both sides gives

[tex]\frac{6x}{3}-\frac{5}{3}-x=\frac{1}{3}-2x-2[/tex]

now further simplification gives

[tex](\frac{6}{3}x-x)-\frac{5}{3}=\frac{1}{3}-2x-2[/tex][tex]\Rightarrow x-\frac{5}{3}=\frac{1}{3}-2x-2[/tex]

Now adding 5/3 to both sides gives

[tex]x=\frac{1}{3}-2x-2+\frac{5}{3}[/tex]

then adding 2x to both sides gives

[tex]x+2x=\frac{1}{3}+\frac{5}{3}-2[/tex][tex]\Rightarrow3x=\frac{6}{3}-2[/tex][tex]3x=2-2[/tex][tex]3x=0[/tex][tex]\Rightarrow\boxed{x=0}[/tex]

which is our answer!

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