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

Question 1 (iv)
Solve the following pair of linear equations by the elimination method and the substitution method:
x2+2y3=1 and xy3=3


Open in App
Solution

x2+2y3=1 and xy3=3

By elimination method:

x2+2y3=1 ... (i)
xy3=3 ... (ii)

Multiplying equation (i) by 2, we get
x+4y3=2 ... (iii)
xy3=3 ... (ii)
Subtracting equation (ii) from equation (iii), we get

5y3=5

Dividing by 5 and multiplying by 3, we get

y=155
y = - 3

Putting this value in equation (ii), we get

xy3=3 ... (ii)
x(3)3=3

x + 1 = 3
x = 2

Hence our answer is x = 2 and y = -3.

By substitution method

xy3=3 ... (ii)

Add y/3 both side, we get

x=3+y3... (iv)

Putting this value in equation (i), we get

x2+2y3=1 ... (i)

(3+y3)2+2y3=1

32+y6+2y3=1

Multiplying by 6, we get

9 + y + 4y = - 6

5y = -15

y = - 3

Hence, our answer is x = 2 and y = -3.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon