wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve: x+y2+3x−5y4=2,x14+y18=1

A
7,8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
7,9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6,10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 7,9
Given,
x+y2+3x5y4=2.............Eq(1)
x14+y18=1....................Eq(2)
On solving eq (1) we get
x+y2+3x5y4=2
2(x+y)+(3x5y)4=2
2x+2y+3x5y=8
5x3y=8.............................Eq(3)..multiplied by 7
on solving Eq(2) we get
x14+y18=1
9x+7y126=1
9x+7y=126..........................Eq(4)...multipied by3
on adding both Eq 3 and Eq 4 we get
35x21y=56
27x+21y=378
_________________________
62x=434
x=43462=7
putting the value of x=7 in eq 4 we get
9×7+7y=126
7y=12663
7y=63
y=637=9
Hence x=7 and y=9

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Multiplication
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon