Question

solve the pair of linear equation
3/root x + 4/root y = 2, 5/root x + 7/root y = 41/12​

Answer 1

Answer:

x+√y=7

√x+y=11

from eq 1...

√y=7-x

y=(7-x)^2

from eq 2...

√x=11-y

substitute (7-x)^2 for y

√x=11-(7-x)^2

square both sides

x=121-22(7-x)^2+(7-x)^4

(7-x)^4=(x-7)^2*(x-7)^2=(x^2-14x+49)(x^2-14x-49)=

x^4-28x^3+294x^2-1372x+2401

x=121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

0=-x+121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

x^4-28x^3+272x^2-1065x+1444=0

from the rational root theorem, possible roots are factors of 1444

4 is a factor of 1444 and a root of the equation...

256-1792+4352-4260+1444=0

-1536+4352-4260+1444=0

2816-4260+1444=0

-1444+1444=0

0=0

x=4

x+√y=7

4+√y=7

√y=7-4

√y=3

square both sides

y=9

x=4 and y=9

Answer 2

Answer:

Step-by-step explanation:

3/√x   +   4/√y   =   2

5/√x   +   7/√y   =   41/12

15√x   +   21√y    =   41/4

-15√x   -   20√y   =   -10

-------------------------------------

                1/√y       =   1/4

y = 16

3/√x + 4/(4) = 2

3/√x + 1 = 2

3/√x = 1

√x = 3

x = 9