Question

what can be the possible values of X and y?? ax+by=2​

Answer 1

Answer:

put x=1 and y=2 in

ax+by=2

we get

a+2b=2

or a=2 - 2b (1)

again put x=1, y=2 in

bx+a^2y=10

we get

b+2a^2=10 (2)

put value of a from (1) to (2), we have

b+2(2-2b)^2=10

b+2(4+4b^2-8b)-10=0

b+8+8b^2-16b-10=0

8b^2-15b-2=0

factorize the above equation

8b^2-(16 - 1 )b - 2 = 0

8b^2-16b+b-2=0

8b(b-2) + 1(b-2)=0

(b-2)(8b+1)=0

hence b-2=0 or 8b+1=0

so b = 2 or b= -1/8

from (1)

a=2-2b=2-2*2=2-4=-2

or a = 2 - 2*(-1/8)=2+1/4=(8+1)/2=9/2

Therefore a= -2 for b=2 and a = 9/2 for b = -1/8

Step-by-step explanation:

i hope this helps you a lot