Hello there!
Question:-
[tex]\sf \longmapsto2(3x+1)=3[/tex]
We need to find the value of x.
Solution:-
[tex]\sf \longmapsto2(3x+1)=3[/tex]
Distribute:-
[tex]\sf \longmapsto \: (2)(3x)+(2)(1)=3[/tex]
On Simplification:-
[tex]\sf \longmapsto \: 6x + 2 = 3[/tex]
Subtract 2 from both sides :-
[tex]\sf \longmapsto 6x+2 - 2=3 - 2[/tex]
On Simplification:-
[tex]\sf \longmapsto6x + 0 = 3 - 2[/tex]
[tex]\sf \longmapsto6x = 1[/tex]
Divide the two sides by 6:-
[tex]\sf \longmapsto \: \dfrac{6x}{6} = \dfrac{1}{6} [/tex]
Cancel 6/6, leave x, answer is 1x. 1/6 can't be cancelled.
[tex]\sf \longmapsto \: \dfrac{ \cancel6x}{ \cancel6} = \dfrac{1}{6} [/tex]
[tex]\sf \longmapsto \: x = \dfrac{1}{6} [/tex]
In Decimal:-
[tex]\sf \longmapsto \: x = 0.16[/tex]
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Henceforth,the value of x is :-
[tex]\boxed{\huge\tt x = \dfrac{1}{6}} [/tex]
[tex] \bold{OR,}[/tex]
[tex]\boxed{\huge\tt x = 0.16}[/tex]
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Please let me know if you have any questions.
~MisterBrian
