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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...

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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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New post Updated on: 11 Jun 2018, 11:31
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If x > 0, y > 0, \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) = 4 and what is the value of \(\frac{{x+y}}{y}\)?

A) \(\frac{3}{4}\)

B) \(\frac{4}{3}\)

C) \(\frac{10}{7}\)

D) \(\frac{7}{4}\)

E) \(\frac{7}{3}\)

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Originally posted by CAMANISHPARMAR on 11 Jun 2018, 10:26.
Last edited by CAMANISHPARMAR on 11 Jun 2018, 11:31, edited 1 time in total.
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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New post 11 Jun 2018, 11:41
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2
CAMANISHPARMAR wrote:
If x > 0, y > 0, \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) = 4 and what is the value of \(\frac{{x+y}}{y}\)?

A) \(\frac{3}{4}\)

B) \(\frac{4}{3}\)

C) \(\frac{10}{7}\)

D) \(\frac{7}{4}\)

E) \(\frac{7}{3}\)


If \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) = 4,

We can simplify this equation as follows

\(7x^2 + 72xy + 4y^2 = 4(4x^2 + 12xy + 5y^2) = 16x^2 + 48xy + 20y^2\)

-> \(16x^2 - 7x^2 + 48xy - 72xy + 20y^2 - 4y^2 = 9x^2 - 24xy + 16y^2 = (3x)^2 - 2(3x)(4y) + (4y)^2 = 0\)

-> \((3x - 4y)^2 = 0\) -> \(3x = 4y\) -> \(\frac{x}{y} = \frac{4}{3}\)

Adding one on both sides,

\(\frac{x}{y} + 1 = \frac{4}{3} + 1\) -> \(\frac{{x+y}}{y} = \frac{7}{3}\) (Option E)

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Re: If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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New post 11 Jun 2018, 11:29
2
CAMANISHPARMAR wrote:
If x > 0, y > 0, \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) and what is the value of \(\frac{{x+y}}{y}\)?

A) \(\frac{3}{4}\)

B) \(\frac{4}{3}\)

C) \(\frac{10}{7}\)

D) \(\frac{7}{4}\)

E) \(\frac{7}{3}\)


CAMANISHPARMAR, Is the question complete??
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Re: If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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New post 11 Jun 2018, 11:34
1
pushpitkc wrote:
CAMANISHPARMAR wrote:
If x > 0, y > 0, \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) and what is the value of \(\frac{{x+y}}{y}\)?

A) \(\frac{3}{4}\)

B) \(\frac{4}{3}\)

C) \(\frac{10}{7}\)

D) \(\frac{7}{4}\)

E) \(\frac{7}{3}\)


CAMANISHPARMAR, Is the question complete??


Thanks pushpitkc - I don't know by mistake I missed typing "= 4", I have done the needful. Thanks for informing :-)
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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New post 06 Feb 2019, 21:37
CAMANISHPARMAR wrote:
If x > 0, y > 0, \(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) = 4 and what is the value of \(\frac{{x+y}}{y}\)?

A) \(\frac{3}{4}\)

B) \(\frac{4}{3}\)

C) \(\frac{10}{7}\)

D) \(\frac{7}{4}\)

E) \(\frac{7}{3}\)


Better to invest time and get this question correct

\(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}\) = 4

\(7x^2 + 72xy + 4y^2\) = 4 \(4x^2 + 12xy + 5y^2\)

just solve this expression to get the inline
9x^2 - 24xy +16y^2 = 0
3x-4y * 3x-4y = 0

x = 4/3 y

\(\frac{{x+y}}{y}\) = 7/3

E
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...   [#permalink] 06 Feb 2019, 21:37
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