The solutions of the given equations are given by equating the
variables to gives;
- Second part; (0, -6) and (1, -5)
- Third part; (4, 11) and (0, 7)
- Fourth part: y = x² + 3·x + 2; x + y = 2
<h3>How can the (common) solution be found?</h3>
First part;
The given functions are;
y = -3·x² + 24·x + 3, and y = -4·x² + 25·x + 15
When the balloons collided, we have;
The height are equal, which gives;
y = y
-3·x² + 24·x + 3 = -4·x² + 25·x + 15
-3·x² + 24·x + 3 - (-4·x² + 25·x + 15) = x² - x - 12 = 0
Which gives;
x² - x - 12 = (x - 4) × (x + 3) = 0
x = 4, and x = -3
The time at which the balloon collide at the highest point is therefore;
- At the time, x =<u> 4.00 seconds</u>
Second part;
The given system of equations are;
x - y = 6
y = x² - 6
y = x - 6
Which gives;
x - 6 = x² - 6
x = x²
Which gives;
x = 0 or x = 1
Similarly, we have;
x = y + 6
Which gives;
y = (y + 6)² - 6
y² + 11·y + 30 = 0
(y + 5)·(y + 6) = 0
Which gives;
y = -5, or y = -6
When y = -5, we have; x = -5 + 6 = 1
The solution are therefore;
Third part;
The equations are; y = x + 7 and y = x² - 3·x + 7
At the solution, the u-values are equal, which gives;
y = y
x + 7 = x² - 3·x + 7
x² - 3·x + 7 - (x + 7) = 0
x² - 4·x = 0
x·(x - 4) = 0
x = 4, or x = 0
When x = 4, y = 4 + 7 = 11
When x = 0, y = 0 + 7 = 7
The solutions are located at points;
Fourth part;
Given that the point of intersection of the line and the quadratic function
is the point (2, 0), we have;
On the linear function, when y = 0, x = 2
The possible linear function functions is therefore;
x + y = 2
From the quadratic function that accompanies the above linear function, we have;
y = x² + 3·x + 2
When x = 0, y = 0² + 3×0 + 2 = 2
Therefore;
- The correct option is; y = x² + 3·x + 2, and x + y = 2
Learn more about the solutions of equations here:
brainly.com/question/24400554
