Hi! Can someone solve these for me pls?

Y=10/x
y=1/2x+4

solve for x and y

thanks

Hello! Sure, I can help you solve these equations. To find the values of x and y, we need to solve the equations simultaneously. Let's start with the first equation:

Y = 10/x

Since the value of y differs by capitalization, I will assume you meant y in the equation. In that case:

y = 10/x

Now, let's move on to the second equation:

y = (1/2)x + 4

To solve for x and y, we can use substitution or elimination method. Let's use substitution, where we express y in terms of x from one equation and substitute it into the other equation.

First, rearrange the second equation to solve for y:

y = (1/2)x + 4

Now, substitute this value of y into the first equation:

(1/2)x + 4 = 10/x

To get rid of the fraction, we can multiply through by the common denominator, which is 2x:

x * (1/2)x + 4 * 2x = 10

Simplify the equation:

(x^2)/2 + 8x = 10

To make it easier to solve, let's multiply through by 2 to get rid of the denominator:

x^2 + 16x = 20

Rearrange the equation:

x^2 + 16x - 20 = 0

Now, we have a quadratic equation. To solve it, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = 1, b = 16, and c = -20. Substituting these values into the quadratic formula, we get:

x = (-16 ± √(16^2 - 4 * 1 * -20))/(2 * 1)

Simplify further:

x = (-16 ± √(256 + 80))/2

x = (-16 ± √336)/2

Now, we can calculate the two possible values of x:

x = (-16 + √336)/2 or x = (-16 - √336)/2

Using a calculator, we get approximately:

x = (-16 + √336)/2 ≈ 1.47 or x = (-16 - √336)/2 ≈ -17.47

Now we have values for x. To find corresponding values for y, we can substitute these values back into either of the original equations. Let's use the first equation:

y = 10/x

For x = 1.47, we have:

y = 10/1.47 ≈ 6.80

And for x = -17.47, we have:

y = 10/(-17.47) ≈ -0.57

Therefore, the solutions to the system of equations are approximately:

x ≈ 1.47, y ≈ 6.80 and x ≈ -17.47, y ≈ -0.57

I hope this explanation helps! Let me know if you have any further questions.