Question

Given that y is a solution to the simultaneous equations: 4x2 + y2 + 10y = 472x - y = 5What is the value of y when x > 0?​

Answer 1

$\displaystyle \left \{ {{4x^2+y^2+10y=47} \atop {2x-y=5}} \right.~~~\Rightarrow~~~\left \{ {{4x^2-4xy+y^2+10y+4xy=47} \atop {2x-y=5}} \right.\\ \\ \\ \left \{ {{(2x-y)^2+10y+4xy=47} \atop {2x-y=5}} \right.~~~\Rightarrow~~~\left \{ {{5^2+10y+4xy=47} \atop {y=2x-5}} \right.\\ \\ 25+10\cdot(2x-5)+4x\cdot (2x-5)=47\\ \\ 25+20x-50+8x^2-20x=47\\ \\ 8x^2=72\\ \\ x^2=9\\ \\ x=\pm3$

$if~~x=-3~~~ than~~~y=2\cdot (-3)-5=-11\\ \\ if ~~ x=3~~~ than~~~ y=2\cdot 3-5=1$

$y=1,~~~ when~~~ x>0$

Answer 2

To find the value of y when x > 0, we need to solve the system of simultaneous equations:

1. 4x^2 + y^2 + 10y = 47
2. 2x - y = 5

Let's proceed with the solution:

From equation (2), we can express y in terms of x: y = 2x - 5

Now, we can substitute this expression for y into equation (1):

4x^2 + (2x - 5)^2 + 10(2x - 5) = 47

Simplify the equation:

4x^2 + (4x^2 - 20x + 25) + 20x - 50 = 47

Combine like terms:

8x^2 - 30x - 22 = 47

Now, move all terms to one side of the equation:

8x^2 - 30x - 69 = 0

Now we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 8, b = -30, and c = -69.

We can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Substitute the values:

x = (30 ± √((-30)^2 - 4 * 8 * -69)) / 2 * 8

x = (30 ± √(900 + 2208)) / 16

x = (30 ± √3108) / 16

Now, we'll calculate the two possible values of x:

1. x = (30 + √3108) / 16 ≈ 4.61
2. x = (30 - √3108) / 16 ≈ -1.11

Since we're interested in the case where x > 0, we can discard the negative value of x.

Now, to find the corresponding value of y for x ≈ 4.61, we'll use the expression we obtained earlier for y in terms of x:

y = 2x - 5

y ≈ 2 * 4.61 - 5 ≈ 4.22

So, the value of y when x > 0 is approximately 4.22.