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Derivative of Standard Functions

I .4 x +3 y =...

Question

$I . 4 x + 3 y = (1600)^{\frac{1}{2}}$
$I I . 6 x - 5 y = (484)^{\frac{1}{2}}$

A

x < y

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B

x \geq y

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C

x \leq y

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D

x > y

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E

x = y

or relationship cannot be established

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Solution

The correct option is D $x > y$
Equation I
4x + 3y = 40 ...(i)
Equation II
6x - 5y = 22 ...(ii)
On solving Eqs. (i) and (ii), we get
$4 x + 3 y = 40... \times 6$
$6 x - 5 y = 22... \times 4$
$i . e 24 x + 18 y = 240$
$24 x - 20 y = 88 - + - - --------------- - 38 y = 152 y = 4$
On putting the value of y in Eq. (i). we get
$4 x + 3 \times 4 = 40 \Rightarrow 4 x = 40 - 12 \Rightarrow 4 x = 28 ∴ x = 7$
Hence, x >y
Alternate Method
$4 x + 3 y = 40 x = \frac{40 - 3 y}{4}$
On putting the value of x in Eq. (ii), we get
$6 (x) - 5 y = 22 6 (\frac{40 - 3 y}{4}) - 5 y = 22 \frac{120 - 9}{2} - 5 y = 22 120 - 9 y - 10 y = 44 - 19 y - 76 \Rightarrow y = 4$
On putting the value of y in Eq. (i). we get
$9 x + 3 (4) = 40 4 x = 40 - 12 x = \frac{28}{4} ∴ x = 7$
Hence, x > y

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