MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

x \geq y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

x \leq y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

x > y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E

x = y

or relationship cannot be established

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

thumbs-up

2

Similar questions

8 x + 6 y = (6400)^{\frac{1}{2}}

18 x - 15 y = (4356)^{\frac{1}{2}}

I . (x)^{2} = 961

I I . y = \sqrt{961}

x = \sqrt{625}

y = \sqrt{676}

4 x^{2} - 13 x - 12 = 0

y^{2} - 7 y - 60 = 0

\sqrt{289} x + \sqrt{25} = 0

\sqrt{676} y + 10 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Derivative of Standard Functions

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Derivative of Standard Functions

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon