Pleasee helpp 20 pointss!
1 answer:
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Answer:
The system has no real solution.
Step-by-step explanation:
Hi there!
We have the following system of equations:
x² + y² = 1
5x + 6y = 12
The solutions of the system are those (x,y) pairs that satisfy both equations.
Let´s take the second equation and solve it for y:
5x + 6y =12
subtract 5x to both sides of the equation
6y = 12 - 5x
divide by 6 both sides
y = (12- 5x)/6
Now, let´s replace the y in the first equation:
x² + y² = 1
x² + [(12- 5x)/6]² = 1
Apply distributive property
x² + (12 - 5x)²/36 = 1
x² + (12 - 5x)(12 - 5x)/36 = 1
Apply distributive property
x² + (144 - 120x + 25x²)/36 = 1
Apply distributive property
x² + 4 - 10/3 · x + 25/36 · x² = 1
subtract 1 to both sides
61/36 · x² - 10/3 · x + 3 = 0
Using the quadratic formula, let´s find the solutions to this quadratic equation:
a = 61/36
b = -10/3
c = 3
[-b ± √(b² - 4ac)]/2a
[10/3 ± √(100/9 - 4(61/36)(3))]/2(61/36)
(10/3 ± √-83/9)/61/18
This expression has no real solution because √-83/9 is not defined in the set of real numbers. This means that both curves don´t intersect (see attached figure).
Have a nice day!
