Q. 2x−3y=15y=−31(x2−16x+63)If (x,y) is a solution to the system of equations shown and y>0, what is the value of x ?
Given Equations: Given the system of equations:1) 2x−3y=152) y=−31(x2−16x+63)We need to find the value of x for which y > 0.Let's solve the first equation for y:y=32x−5
Solving for y: Now we will substitute the expression for y from the first equation into the second equation: -\left(\frac{\(1\)}{\(3\)}\right)\left(x^\(2 - 16x + 63\right) = \left(\frac{2}{3}\right)x - 5
Substitution and Simplification: Multiply both sides of the equation by −3 to get rid of the fractions:x2−16x+63=−2x+15
Setting up Quadratic Equation: Now, let's move all terms to one side to set the equation to zero:x2−16x+2x+63−15=0x2−14x+48=0
Factoring the Quadratic Equation: We can factor the quadratic equation: x - \(6)(x - 8) = 0\
Finding Possible Solutions: Setting each factor equal to zero gives us two possible solutions for x:x−6=0 or x−8=0So, x=6 or x=8
Checking Valid Solutions: We need to determine which of these solutions will give us a positive value for y. Let's substitute x=6 into the first equation to find y:y=(32)(6)−5y=4−5y=−1Since y is not greater than 0, x=6 is not a valid solution.
Checking Valid Solutions: We need to determine which of these solutions will give us a positive value for y. Let's substitute x=6 into the first equation to find y:y=(32)(6)−5y=4−5y=−1Since y is not greater than 0, x=6 is not a valid solution.Now let's substitute x=8 into the first equation to find y:y=(32)(8)−5y=(316)−5y=(316)−(315)y=31Since y is greater than 0, x=8 is the valid solution.