What is solution of the set of x^2+y^2=26ANDX-Y=6

Question

What the solution of the problem and how do I slove it and rembe how to Solve it correctly 

Answer 1

Both equations must be true at the same time.  You can do this a couple of ways.

 

1) If you are using a calculator, Graph both equations on the same axes and see where they intersect.  (Note that the graphs are of a circle and a line.)

 

2)  To do the problem analytically, solve the 2nd equation for either x or y and substitute it into the first equation.

 

x=y+6   so (y+6)^2 + y^2 = 26

 

expand the squared term  .. y^2 + 12y + 36 + y^2 = 26  

 

combine like terms  ..  2y^2 + 12 y + 10 = 0

 

Factor and simplify   2[ y^2 + 6y + 5] = 0.

 

Once you find out what y is, plug it back into the equation x = y+6 to get the x value(s).  There may be more than one answer to this kind of problem!

Answer 2

Solve the second equation for x.

 

x - y = 6 so 

x = y + 6

 

Now, plug this into the first equation.

 

(y + 6)² + y² = 26

y² + 12y + 36 + y² = 26

2y² + 12y + 36 = 26

2y² + 12y + 10 = 0

 

Dividing both sides of the equation by 2 gives us:

 

y² + 6y + 5 = 0

 

Factor this to get:

 

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

 

Set each factor equal to 0 and solve for y, which gives us

 

y = -1 and y = -5

 

We have to find x for each value of y. 

 

Using our previous equation x = y + 6, then 

 

When y = -1, x = -1 + 6 = 5

When y = -5, x = -5 + 6 = 1

 

The solutions are (5, -1) and (1, -5)