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2x-3y=15

y=-(1)/(3)(x^(2)-16 x+63)
If 
(x,y) is a solution to the system of equations shown and 
y > 0, what is the value of 
x ?

2x3y=15 2 x-3 y=15 \newliney=13(x216x+63) y=-\frac{1}{3}\left(x^{2}-16 x+63\right) \newlineIf (x,y) (x, y) is a solution to the system of equations shown and y>0 , what is the value of x x ?

Full solution

Q. 2x3y=15 2 x-3 y=15 \newliney=13(x216x+63) y=-\frac{1}{3}\left(x^{2}-16 x+63\right) \newlineIf (x,y) (x, y) is a solution to the system of equations shown and y>0 y>0 , what is the value of x x ?
  1. Given Equations: Given the system of equations:\newline11) 2x3y=152x - 3y = 15\newline22) y=13(x216x+63)y = -\frac{1}{3}(x^2 - 16x + 63)\newlineWe need to find the value of xx for which y > 0.\newlineLet's solve the first equation for yy:\newliney=23x5y = \frac{2}{3}x - 5
  2. Solving for y: Now we will substitute the expression for yy from the first equation into the second equation: -\left(\frac{\(1\)}{\(3\)}\right)\left(x^\(2 - 1616x + 6363\right) = \left(\frac{22}{33}\right)x - 55
  3. Substitution and Simplification: Multiply both sides of the equation by 3-3 to get rid of the fractions:\newlinex216x+63=2x+15x^2 - 16x + 63 = -2x + 15
  4. Setting up Quadratic Equation: Now, let's move all terms to one side to set the equation to zero:\newlinex216x+2x+6315=0x^2 - 16x + 2x + 63 - 15 = 0\newlinex214x+48=0x^2 - 14x + 48 = 0
  5. Factoring the Quadratic Equation: We can factor the quadratic equation: x - \(6)(x - 88) = 00\
  6. Finding Possible Solutions: Setting each factor equal to zero gives us two possible solutions for xx:x6=0x - 6 = 0 or x8=0x - 8 = 0So, x=6x = 6 or x=8x = 8
  7. Checking Valid Solutions: We need to determine which of these solutions will give us a positive value for yy. Let's substitute x=6x = 6 into the first equation to find yy:y=(23)(6)5y = \left(\frac{2}{3}\right)(6) - 5y=45y = 4 - 5y=1y = -1Since yy is not greater than 00, x=6x = 6 is not a valid solution.
  8. Checking Valid Solutions: We need to determine which of these solutions will give us a positive value for yy. Let's substitute x=6x = 6 into the first equation to find yy:y=(23)(6)5y = \left(\frac{2}{3}\right)(6) - 5y=45y = 4 - 5y=1y = -1Since yy is not greater than 00, x=6x = 6 is not a valid solution.Now let's substitute x=8x = 8 into the first equation to find yy:y=(23)(8)5y = \left(\frac{2}{3}\right)(8) - 5y=(163)5y = \left(\frac{16}{3}\right) - 5y=(163)(153)y = \left(\frac{16}{3}\right) - \left(\frac{15}{3}\right)y=13y = \frac{1}{3}Since yy is greater than 00, x=8x = 8 is the valid solution.

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