Linear Quadratic Function

x² + y²  =  26

x - y   =  6

Solve the second equation for x:  x - y  =  6   --->   x  =  y + 6

Replace this value into the first equation:  x² + y²  =  26   --->   (y + 6)² + y²  =  26

Multiply out:  y² + 12y + 36 + y²  =  26

Simplify:       2y² + 12y + 36  =  26

Get all terms to one side:  2y² + 12y + 10  =  0

Divide by 2:                              y² + 6y + 5  =  0

Factor:                                 (y + 5)(y + 1)  =  0

Solve:                       y  =  -5     or     y  -1

Place these values into the linear equation:  x  =  -5 + 6   --->   x  =  1   --->   (1, -5)

                                                           x  =  -1 + 6   --->   x  =  5   --->   (5, -1)

(If you replace back into the quadratic equation, you will get extraneous answers.)

Let's do this...using x - y = 6   ..... we can just rearrange it and  say that y = x - 6

Now...in  x^2 + y^2 = 26, let's substitute what we just got for y...so this gives us

x^2 + (x -6)^2  = 26  simplify

x^2 + x^2 - 12x + 36 = 26

2x^2 - 12x + 10 = 0    divide through by 2

x^2 - 6x + 5 = 0     factor

(x -5 ) ( x -1)  = 0    so setting each factor to 0 , we find that x = 5  or x =1

And using y = x -6    when x =5 , y = -1  and when x = 1, y = -5

So..our solutions are (5, -1) and (1, -5)

This is just the intersection of a line with a circle...here's the graph....https://www.desmos.com/calculator/gv9ias9ptz