SOLUTION: Find the points of intersection of the line y = 7x-42 and the circle with equation x^2+y^2-4x+6y = 12

Algebra ->  Functions -> SOLUTION: Find the points of intersection of the line y = 7x-42 and the circle with equation x^2+y^2-4x+6y = 12      Log On


   

Question 1180172: Find the points of intersection of the line y = 7x-42 and the circle with equation x^2+y^2-4x+6y = 12
Found 2 solutions by ikleyn, mananth:
Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.

You substitute y = 7x-42 from the first equation into the second equation.


Doing this way, you obtain a quadratic equation for x.


You solve it and obtain the solutions for x (if they do exist).


Then for each found value of x you find the corresponding value of y, using the first equation.


Having my instructions, the rest is standard mechanical work, which I leave to you.

---------------

To see many other similar solved problems (yours TEMPLATES), look into the lesson
    - Solving systems of algebraic equations of degree 2 and degree 1
in this site.



Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
y = 7x-42 and the circle with equation x^2+y^2-4x+6y = 12
substitute y in equation of circle
x%5E2+%2B%287x-42%29%5E2-4x%2B6%287x-42%29=12
x%5E2+%2B49x%5E2-588x+%2B1764%2B42x-252=12
50x%5E2+-550x+%2B1500=0
+x%5E2-11x%2B30=0
(x-6)(x-5)=0
x=6, x=5
plug value of x to get values of y
(6,0), (5,-7)
.