How do you solve the system of equations #{(y =7 - x), (x^2 + y^2 = 65):}#?

2 Answers
Mar 15, 2017

Answer:

There are two points that are solutions, #(8,-1) and (-1,8)#

Explanation:

Given:

#y^2+x^2=65 and y + x = 7#

Write the line as y as a function of x

#y^2+x^2=65 and y = 7-x#

Substitute #(7 -x)# for y into the circle:

#(7-x)^2+x^2=65#

Square first term:

#49-14x+x^2+x^2=65#

#2x^2-14x-16=0#

#x^2-7x-8=0#

#(x-8)(x+1)=0#

#x = 8 and x = -1#

Find the corresponding y values, using the equation, #y = 7-x#

y = -1 and y = 8

Mar 15, 2017

Answer:

#(8, -1);(-1,8)#

Explanation:

From the second equation, we have #y = 7 - x#. Substituting:

#(7- x)^2 + x^2 = 65#

#49 - 14x + x^2 + x^2 = 65#

#2x^2 - 14x - 16 = 0#

#2(x^2 - 7x - 8) = 0#

#x^2 - 7x - 8 = 0#

#(x - 8)(x + 1) = 0#

#x = 8 and -1#

Since this system involves the equation of a circle (e.g #x^2 + y^2 = 65#), the solution set will be of the form #(a, b); (b, a)#.

The solution set therefore is #(8, -1);(-1, 8)#.

Hopefully this helps!