The equations 4x+2y=4 and -x^2+y=-6 intersect at the coordinates (a, b) and (c, d).
2 x + y = 2 ... y = 2 - 2 x
substituting ... -x^2 - 2 x + 2 = -6 ... x^2 + 2x - 8 = 0
factoring ... (x + 4) (x - 2) = 0 ... x = -4 , x = 2 ... these are the "a" values
substitute back to find the corresponding y values ("b" values)
Well, the coordinates (a, b) and (c, d) seem to be keeping themselves a secret. Maybe those equations are just shy and don't want to reveal their intersection points. Let's try to coax them out with a little humor!
Hey, equations! Why so serious? We just want to know where you intersect. Don't be so negative, -x^2+y=-6! And 4x+2y=4, don't be too linear with us. Come on, show us your coordinates (a, b) and (c, d)! Let's find some common ground and solve this math mystery together!
On a serious note, to find the intersection points, you can solve the system of equations simultaneously. But remember, laughter is always the best solution in life!
To find the coordinates (a, b) and (c, d) where the equations 4x + 2y = 4 and -x^2 + y = -6 intersect, we can solve these two equations simultaneously.
Step 1: Solve one equation for one variable in terms of the other variable.
Let's solve the second equation -x^2 + y = -6 for y:
y = x^2 - 6
Step 2: Substitute this value of y into the first equation and solve for x.
4x + 2(x^2 - 6) = 4
4x + 2x^2 - 12 = 4
2x^2 + 4x - 16 = 0
Step 3: Simplify the quadratic equation and solve for x by factoring, using the quadratic formula, or completing the square.
Dividing the equation by 2, we get:
x^2 + 2x - 8 = 0
Factoring the quadratic equation gives:
(x + 4)(x - 2) = 0
Setting each factor equal to 0 and solving for x, we have:
x + 4 = 0 --> x = -4
x - 2 = 0 --> x = 2
So, we have two possible x-values: x = -4 and x = 2.
Step 4: Substitute these x-values into either of the original equations to find the corresponding y-values.
Using the first equation 4x + 2y = 4, let's substitute x = -4:
4(-4) + 2y = 4
-16 + 2y = 4
2y = 20
y = 10
Therefore, one point of intersection is (-4, 10).
Now, let's substitute x = 2 into the first equation:
4(2) + 2y = 4
8 + 2y = 4
2y = -4
y = -2
Therefore, the other point of intersection is (2, -2).
To summarize, the equations 4x + 2y = 4 and -x^2 + y = -6 intersect at the coordinates (-4, 10) and (2, -2).
To find the points of intersection between the equations 4x+2y=4 and -x^2+y=-6, we need to solve the system of equations simultaneously.
1. Start by rearranging the equations:
4x + 2y = 4 ---> Equation 1
-x^2 + y = -6 ---> Equation 2
2. Solve Equation 1 for x:
4x = 4 - 2y
x = (4 - 2y)/4
x = (2 - y)/2
3. Substitute the expression for x in Equation 2:
-(2 - y)^2 + y = -6
4. Expand and simplify Equation 2:
-(4 - 4y + y^2) + y = -6
-4 + 4y - y^2 + y = -6
-y^2 + 5y - 10 = 0
5. Rearrange Equation 2:
y^2 - 5y + 10 = 0
6. This equation is a quadratic equation in the form of ay^2 + by + c = 0, where a = 1, b = -5, and c = 10.
7. Use the quadratic formula to solve for y:
y = (-b ± √(b^2 - 4ac)) / (2a)
y = (-(-5) ± √((-5)^2 - 4(1)(10))) / (2(1))
y = (5 ± √(25 - 40)) / 2
y = (5 ± √(-15)) / 2
8. Since the discriminant (b^2 - 4ac) is negative (-15), there are no real solutions for y. This means the two equations do not intersect in the real number plane.
Therefore, there are no points (a, b) and (c, d) where the equations 4x+2y=4 and -x^2+y=-6 intersect.