Question

Solve (2x+3y-5)^2+(5x-y-4)^2=0
Use the hint as a^2+b^2=0 & a=b=0

Answer 1

a=b=0
so
a=0
b=0
here let a=2x+3y-5
            b=5x-y-4
so
2x+3y=5...................1
5x-y=4.....................2
multiplying eq. 2 with 3 we will get
15x-3y=12
now by elimination method
17x=17
so x=1.............................3
similarly y=1.......................4
putting the valve of x and y
lhs=0
rhs=0
so lhs=rhs

Answer 2

let a = 2x+3y-5=0
    b=5x-y-4=0
⇒2x+3y=5 and 5x-y=4
10x+15y=25
10x-2y=8
by solving
17y=17
y=1
and x=1
then [2.1+3.1-5]²+[5.1-1-4]²=0²+0²=0