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

Question 1 (vii)
Solve the following pairs of equations by reducing them to a pair of linear equations:
10x+y+2xy=4
15x+y5xy=2

Open in App
Solution

10x+y+2xy=4
15x+y5xy=2
10x+y+2xy=4
15x+y5xy=2
Putting 1x+y=p and 1xy=q in the given equations, we get;
10p+2q=4
10p+2q4=0...(i)
15p5q=2
15p5q+2=0...(ii)
Using cross multiplication, we get
p420=q60(20)=15030
p16=q80=180
p16=180 and q80=180
p=15 and q=1
p=1x+y=15 and q=1xy=1
x+y=5...(iii) and xy=1...(iv)
Adding equation (iii) and (iv), we get
2x=6x=3
Putting x = 3 in (iii), we get
3+y=5y=2
Hence, x = 3 and y = 2


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