Question

Help me please!!! It’s quadratic equations.

Answer 1

The other solution of the system is
`(x,y) = ((4)/(5) ,(3)/(5) )`

Explanation:

Here, the given set of equations are: y = 2x -1 and
`x^(2)  + y^(2)  =1`

Now, to solve them put y = 2x -1 in the second equation, so we get

`x^(2)  + y^(2)  =1` ⇒
`x^(2)  + (2x-1)^(2)  =1`

Also, by ALGEBRAIC IDENTITY :
`(a-b)^(2)  = a^(2) +  b^(2) -2ab`

So, implying above , we get

`x^(2)  + (2x-1)^(2)  =1` ⇒
`x^(2)  + 4x^(2)  + 1 -4x  =1  or,  5x^(2)  -4x  =0`

Now,
`5x^(2)  -4x  =0` ⇒x(5x-4) =0

So, either x =0, or x = 4/5

As (0, -1) is already a given solution:

So, when x = 4/5 ,put in equation (1)

y = 2(5/2) -1 = (3/5)

Hence, the other solution of the system is
`(x,y) = ((4)/(5) ,(3)/(5) )`