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telo118 [61]
3 years ago
8

Hello:) anyone able to help with this question ? Thank you!

Mathematics
2 answers:
svetlana [45]3 years ago
6 0

Answer:

look at the photo below for the Answer.

:)

OverLord2011 [107]3 years ago
5 0
<h3>Answer: a = 4, b = 2</h3>

===================================================

Explanation:

Multiply both sides by the denominator a+b*sqrt(7). Then expand out the RHS and get that RHS into the form p+q*sqrt(7), where p and q are algebraic expressions.

2+4*sqrt(7) = (4-sqrt(7))(a+b*sqrt(7))

2+4*sqrt(7) = 4(a+b*sqrt(7))-sqrt(7)*(a+b*sqrt(7))

2+4*sqrt(7) = 4a+4b*sqrt(7)-a*sqrt(7)-7b

2+4*sqrt(7) = (4a-7b)+(4b-a)*sqrt(7)

The (4a-7b) part on the RHS matches up with 2 on the LHS, so

2 = 4a-7b or 4a-7b = 2

Similarly, the (4b-a)*sqrt(7) on the RHS matches up with 4*sqrt(7)

So,

4*sqrt(7) = (4b-a)*sqrt(7)

4 = 4b-a

-a+4b = 4

--------------

We have this system of equations

4a-7b = 2 and -a+4b = 4

Let's solve the second equation for a

-a+4b = 4

4b = 4+a

4b-4 = a

a = 4b-4

Plug this into the first equation; solve for b.

4a - 7b = 2

4( a ) - 7b = 2

4(4b-4)-7b = 2

16b-16-7b = 2

9b-16 = 2

9b = 18

b = 2

From here, we can compute the value of 'a'

a = 4b-4

a = 4(2)-4

a = 4

Your method of rationalizing the denominator would work out, though it seems like it would involve more steps.

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{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In the above question, it is given that :

\qquad \sf  \boxed{ \sf f(x) =  \frac{x + 3}{2} }

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\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{2 + 3}{2}

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\qquad \sf  \dashrightarrow \: let \: y = f (x)

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\qquad \sf  \dashrightarrow \: y =  \dfrac{x + 3}{2}

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