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655-705 (Hard)|   Algebra|      
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CAMANISHPARMAR
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... Updated on: 11 Jun 2018, 12:31
Originally posted by CAMANISHPARMAR on 11 Jun 2018, 11:26.
Last edited by CAMANISHPARMAR on 11 Jun 2018, 12:31, edited 1 time in total.
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65% (hard)

Question Stats:

68% (02:55) correct 32% (02:55) wrong based on 491 sessions

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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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E
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pushpitkc
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... 11 Jun 2018, 12:41
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CAMANISHPARMAR
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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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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pushpitkc
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... 11 Jun 2018, 12:29
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CAMANISHPARMAR
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) \(\frac{7}{3}\)
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CAMANISHPARMAR, Is the question complete??
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CAMANISHPARMAR
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... 11 Jun 2018, 12:34
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pushpitkc
CAMANISHPARMAR
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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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}... 06 Feb 2019, 22:37
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CAMANISHPARMAR
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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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}... 23 Feb 2021, 02:43
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CAMANISHPARMAR
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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\(\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}} = 4\)

Divide numerator and denominator by y^2 to get

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

Put x/y = a

\(\frac{{7a^2 + 72a + 4}}{{4a^2 + 12a + 5}} = 4\)

\(9a^2 - 24a + 16 = 0\)

\((3a - 4)^2 = 0\)

a = 4/3

\(\frac{{x+y}}{y} = 1 + a = 1 + 4/3 = 7/3\)

Answer (E)
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... 30 Jan 2026, 10:47
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7x^2+72xy + 4y^2 / 4x^2 + 12xy + 5y^2 = 4
7x^2 + 72xy + 4y^2 = 16x^2 + 48xy + 20y^2
(3x-4y)^2 = 0
3x-4y = 0
3x = 4y
x= 4y/3

put into que stem.
4y/3 + y / y
7y / 3y
7/3
CAMANISHPARMAR
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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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}... 30 Jan 2026, 10:47
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7x^2+72xy + 4y^2 / 4x^2 + 12xy + 5y^2 = 4
7x^2 + 72xy + 4y^2 = 16x^2 + 48xy + 20y^2
(3x-4y)^2 = 0
3x-4y = 0
3x = 4y
x= 4y/3

put into que stem.
4y/3 + y / y
7y / 3y
7/3
CAMANISHPARMAR
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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