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Nostrana [21]
2 years ago
6

2x + 3 = 5 . Solve for x

Mathematics
2 answers:
Lady_Fox [76]2 years ago
7 0

Answer:

\boxed{ \boxed{\huge\bf \; x = 1}}

Step-by-step explanation:

\underline{\bf \: Given  \: equation:-}

2x + 3 = 5

\underline{ \bf \: To   \: Find :-}

\rm Value\:  of \: x

\underline{ \bf \: Solution  :-}

\bf : \sf  \longmapsto2x + 3 = 5

\underline{\rm Subtract  \: 3\: from\: both \; sides:}

\bf : \sf  \longmapsto2x + 3 - 3 = 5 - 3

\underline{\rm On  \: Simplification:}

\bf : \sf  \longmapsto2x + 0 = 2

\bf : \sf  \longmapsto2x = 0

\underline{ \rm \: Divide  \: both \:  sides \:  by \:  2 :}

\bf : \sf  \longmapsto \:  \cfrac{2x}{2}  =  \cfrac{2}{ 2}

\underline{\rm On  \: Simplification:}

\rm Cancel \: 2 \: and \: 2 \: of \: LHS \: and \: then \: Cancel \: 2 \: and \: 2 \: of \: RHS(Leave\: x \: of \: LHS)

:  \sf \longmapsto\cfrac{ \cancel2{x}}{\cancel2}  = \cfrac{ \cancel2{}}{\cancel2}

\underline{\rm \: On \:  Cancelling,}

:  \sf \longmapsto1x = 1

As \rm 1x = xso,

:  \sf \longmapsto \: x =\boxed{ 1}

________________________________

I hope this helps!

Please let me know if you have any questions.I am joyous to help!

:)

STALIN [3.7K]2 years ago
4 0

Answer:

x = 1

Step-by-step explanation:

2x + 3 = 5

=> 2x = 5 - 3

=> 2x = 2

=> x = 2/2

=> x = 1

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Step One
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Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
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Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

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28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


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