Graph the line given by 5x+6y=12 and the circle given by x²+y²=1.find all solutions to the system of equations.

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!

ur mom gae no homo

answer: i'm great at algebra

~coco

step-by-step explanation: