Question 550289
{5x-y=4
1+y2=2x

y=5x-4

substitute the value of y in the equation (2)

1+(5x-4)^2=2x
1+25x^2-40x+16=2x
25x^2-42x+17=0

Find the roots of the equation by quadratic formula						
						
a=	25	b=	-42	c=	17	
						
b^2-4ac=	1764	-	-1700			
b^2-4ac=	64			{{{sqrt(	64	)}}}=
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}						
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}						
x1=(	42	+	8	)/	50	
x1=	1		1    			
x2=(	42	-8	) /	50		
x2= 	0.68		 2/3		

x ( 1, 0.68)

plug the value of x in any equation to get the value of y
5x-y=4
5-y=4
y=1
(1,1)
5*0.68-y=4
3.4-y=4	
-0.6
(0.68,0.6)

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