<span>1. X = -2 or 3
2. X = -5 or 3
3. X = -2.5 or 3
4. X = -4 or 2
5. X = 3 or -3
6. X = -4 or 2
I am assuming that you're looking for the intersections between the two equations for each problem. The general approach to each of the given problems is to solve both equations for y (only need to do this with problems 4 through 6 since you've already been given the equations solved for y with problems 1 through 3). After you have two equations solved for y, simply set them equal to each other and then manipulate until you have a quadratic equation of the form:
Ax^2 + Bx + C = 0
After you've gotten your quadratic equation, just find the roots to the equation and you'll know both X values that will result in the same Y value as the equations you've been given for each problem. I'm personally using the quadratic formula for getting the desired roots, but you can also factor manually. So let's do it.
1. y = x+2, y = x^2 - 4
Set the equations equal to each other
x + 2 = x^2 - 4
2 = x^2 - x - 4
0 = x^2 - x - 6
Using the quadratic formula with A=1, B=-1, C=-6, you get the solutions -2 and 3.
2. y = x^2 + 3x - 1, y = x+14
Same thing, set the equations equal to each other.
x^2 + 3x - 1 = x + 14
x^2 + 2x - 1 = 14
x^2 + 2x - 15 = 0
Use the quadratic formula with A=1, B=2, C=-15. Roots are -5 and 3.
3. y = 2x^2 + x - 7, y = 2x + 8
Set the equations equal to each other again.
2x^2 + x - 7 = 2x + 8
2x^2 - x - 7 = 8
2x^2 - x - 15 = 0
Quadratic formula with A=2, B=-1, C=-15, gives you the roots of -2.5 and 3
4. y = x(x + 3), y - x = 8
A little more complicated. Solve the second equation for y
y - x = 8
y = x + 8
Multiply out the 1st equation
y = x(x + 3)
y = x^2 + 3x
Now set the equations equal to each other
x + 8 = x^2 + 3x
8 = x^2 + 2x
0 = x^2 + 2x - 8
And use the quadratic formula with A=1, B=2, C=-8. Roots are -4, 2
5. y = -3x^2 - 2x + 5, y + 2x + 22 = 0
Solve the 2nd equation for y
y + 2x + 22 = 0
y + 22 = -2x
y = -2x - 22
Set equal to 1st equation
-2x - 22 = -3x^2 - 2x + 5
-22 = -3x^2 + 5
0 = -3x^2 + 27
Use the quadratic formula with A=-3, B=0, C=27, giving roots of 3 and -3
6. y + 6 = 2x^2 + x, y + 3x = 10
Solve the 1st equation for y
y + 6 = 2x^2 + x
y = 2x^2 + x - 6
Solve the 2nd equation for y
y + 3x = 10
y = -3x + 10
Set the solved equations equal to each other
2x^2 + x - 6 = -3x + 10
2x^2 + 4x - 6 = + 10
2x^2 + 4x - 16 = 0
Use the quadratic formula with A=2, B=4, C=-16, getting roots of -4 and 2.</span>
red marbles=3/10,green marbles=3/10,blue marbles=4/10
Step-by-step explanation:
The first thing I did is find the outcome which is the total number of marbles and finding the outcome to each marble knowing how much marbles are in one specific color like there are 3 marbles which are red and 10 marbles total so the outcome would be 3/10 chance in picking a red marble,the same goes to the rest of the colored marbles...sorry if I wrote too much
