2Y^5 . Y^3 please help me please
1 answer:
You might be interested in
The solutions for the given system of equations are:
(3, 0), (-4, 7).
<h3>How to solve the given system of equations:</h3>
Here we have the system:
x + y = 3
y = x² - 9
To solve this, we can replace the second equation into the first one, so we get:
x + (x² - 9) = 3
Now we can solve this quadratic equation for x, we need to solve:
x² + x - 12 = 0.
The solutions are given by Bhaskara's formula:
Then the two solutions are:
x = (-1 + 7)/2 = 3
x = (-1 - 7)/2 = -4
To get the y-values correspondent, we can evaluate the linear equation in these two values:
y = 3 - x.
For x = 3:
y = 3 - 3 = 0
For x = -4:
y = 3 + 4 = 7
Then the two solutions are: (3, 0), (-4, 7).
If you want to learn more about systems of equations:
#SPJ1