Question

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

Answer 1

Answer:

The other solution of the system is  $(x,y) = (\frac{4}{5} ,\frac{3}{5} )$

Step-by-step 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  $(x,y) = (\frac{4}{5} ,\frac{3}{5} )$