if (x,y) is a solution to the system of equations above and x > 0 what is the value of x?

y-x^2 +2 = 9x
y = 4x + 4

What is the value of y in the solutions of the system of equations above?

8x-4y =7
5y - 4x = 10

Well, well, well, we have ourselves a system of equations, don't we? Let's solve it with a touch of humor!

In the first system of equations, we have:
1) y - x^2 + 2 = 9x
2) y = 4x + 4

To find the value of x, let's substitute equation 2 into equation 1. We get:
4x + 4 - x^2 + 2 = 9x

By moving things around a bit, we get:
8x - x^2 - 2 = 0

Now, let's solve this quadratic equation. I'm not good at finding the roots, but I'm great at cracking jokes! So, while you do the math, here's a joke for you:

Why don't scientists trust atoms?

Because they make up everything!

Once you find the value of x, I'm sure you'll chuckle at that joke.

To find the value of x in the first system of equations:

y - x^2 + 2 = 9x
y = 4x + 4

We can start by substituting the value of y from the second equation into the first equation:
4x + 4 - x^2 + 2 = 9x

Combining like terms, we get:
-x^2 + 3x - 6 = 0

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -1, b = 3, and c = -6. Plugging these values into the quadratic formula, we get:
x = (-3 ± √(3^2 - 4(-1)(-6))) / (2(-1))

Simplifying further:
x = (-3 ± √(9 - 24)) / (-2)

Since x > 0, we can discard the negative solution, and we are left with:
x = (-3 + √15) / (-2)

Now, to find the value of y in the second system of equations:
8x - 4y = 7
5y - 4x = 10

We can solve this system of equations by using substitution or elimination method. Let's use elimination method:

Multiply the first equation by 5 and the second equation by 8 to cancel out the x term:
40x - 20y = 35
40x - 32y = 80

Subtracting the first equation from the second equation, we get:
40x - 32y - (40x - 20y) = 80 - 35
40x - 32y - 40x + 20y = 45
-12y = 45
y = -45/12
y = -15/4

Thus, the value of y in the solutions of the second system of equations is -15/4.

To find the value of x in the first system of equations, we need to solve the following:

1. y - x^2 + 2 = 9x
2. y = 4x + 4

First, substitute equation 2 into equation 1:

4x + 4 - x^2 + 2 = 9x

Rearrange the equation to isolate the x terms:

-x^2 + 5x + 6 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula. Factoring it, we have:

-(x - 2)(x - 3) = 0

Setting each factor to zero, we get:

x - 2 = 0 -> x = 2
x - 3 = 0 -> x = 3

Given that we're told x > 0, the only valid solution is x = 2.

To find the value of y in the solutions of the first system of equations, substitute the value of x = 2 into equation 2:

y = 4(2) + 4
y = 8 + 4
y = 12

So the value of y in the first system of equations is 12.

Now, let's move on to the second system of equations:

1. 8x - 4y = 7
2. 5y - 4x = 10

To find the value of y in the second system of equations, we can use the method of substitution. Rearrange equation 2 to solve for y:

5y = 4x + 10
y = (4/5)x + 2

Now, substitute this expression for y into equation 1:

8x - 4((4/5)x + 2) = 7

Simplify the equation by distributing the -4:

8x - (16/5)x - 8 = 7

Combine like terms:

(40/5)x - (16/5)x - 8 = 7
(24/5)x - 8 = 7

Add 8 to both sides:

(24/5)x = 15

To solve for x, multiply both sides by the reciprocal of (24/5), which is (5/24):

x = (15)(5/24)

Simplify:

x = 75/24

So the value of x in the second system of equations is 75/24.

Now, substitute this value of x back into equation 2 to find the value of y:

5y - 4(75/24) = 10

Simplify:

5y - (300/24) = 10
5y - 25/2 = 10

Add (25/2) to both sides:

5y = 10 + 25/2

Combine the fraction and the whole number:

5y = 20/2 + 25/2
5y = 45/2

To solve for y, multiply both sides by the reciprocal of 5, which is 1/5:

y = (45/2)(1/5)

Simplify:

y = 45/10

Reduce the fraction:

y = 9/2

So the value of y in the second system of equations is 9/2 (or 4.5).

for the first, use substitution , that is, replace the y in the first with the value of y of the 2nd

y-x^2 +2 = 9x
4x+4 - x^2 + 2 = 9x
-x^2 - 5x + 6 = 0
x^2 + 5x - 6 = 0
take over, you have a nice quadratic which factors.
remember x > 0 , so reject the negative x value

In the second pair, I would multiply the 2nd equation by 2 and then add them.
easy after that