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stepladder [879]
3 years ago
15

id="TexFormula1" title="{ \color{darkred}{(2+√3)+(4-√3)}} = ?" alt="{ \color{darkred}{(2+√3)+(4-√3)}} = ?" align="absmiddle" class="latex-formula">
︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎︎​
Mathematics
2 answers:
BabaBlast [244]3 years ago
8 0

(2 +  \sqrt{3} ) + (4 -  \sqrt{3} ) \\  = 2 +  \sqrt{3}  + 4 -  \sqrt{3}  \\  = 2 + 4 \\  = 6

Answer:

6

Hope you could get an idea from here.

Doubt clarification - use comment section.

barxatty [35]3 years ago
3 0

\huge \bf༆ Answer ༄

Here's the solution ~

  • \sf(2 +  \sqrt{3} ) + (4 -  \sqrt{3})

  • \sf2 +  \cancel {\sqrt{3} } + 4 -   \cancel{\sqrt{3} }

  • \sf6

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<h3>Given :-</h3>

\\

  • x=y-1
  • x+2y=8

\\

<h3>To find:</h3>

\\

  • Value of x
  • Value of y

\\

<h3>Solution:-</h3>

Let say it is first equation:-

x=y-1. . . . (1)

and this is second equation:-

x+2y=8 . . . . (2)

\\  \\

Simplifying 1 equation:-

  • x = y - 1
  • x +1 = y
  • y = x + 1

Put this value of y in second equation.

\\

\dashrightarrow \sf \: x+2y=8

\\

\dashrightarrow \sf \: x+2(x + 1)=8

\\

\dashrightarrow \sf \: x+2x +2=8

\\  \\

\dashrightarrow \sf \: 3x +2=8

\\  \\

\dashrightarrow \sf \: 3x=8 - 2

\\  \\

\dashrightarrow \sf \: 3x=6

\\  \\

\dashrightarrow \sf \: x=6 \div 3

\\  \\

\dashrightarrow \sf \: x= \bf2

\\

to find value of y :-

y = x + 1

y = 2 + 1

y = 3

\\

verification:-

\\

1 equation:-

x = y - 1

put value of x and y

2 = 3 - 1

2 = 2

LHS = RHS

Hence verified!

\\

2 equation:-

\\

  • x+2y=8

put value of x and y

  • 2 + 2 × 3 = 8
  • 2 + 6 = 8
  • 8 = 8

LHS = RHS

Hence verified!

\\

Both equation verified!

.°. value of x and y is 2 and 3 respectively

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