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{:[-7x^(2)=(y+5)(y-5)],[5y=15 x]:} If (a,b) is a solution to the system of equations shown and a > 0, what is the value of a ?

−7x2=(y+5)(y−5)5y=15x \begin{aligned} -7 x^{2} & =(y+5)(y-5) \\ 5 y & =15 x \end{aligned} \newlineIf (a,b) (a, b) is a solution to the system of equations shown and a>0 a>0 , what is the value of a a ?

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Q. −7x2=(y+5)(y−5)5y=15x \begin{aligned} -7 x^{2} & =(y+5)(y-5) \\ 5 y & =15 x \end{aligned} \newlineIf (a,b) (a, b) is a solution to the system of equations shown and a>0 a>0 , what is the value of a a ?
  1. Simplify second equation: We have the system of equations:\newline11) −7x2=(y+5)(y−5)-7x^2 = (y + 5)(y - 5)\newline22) 5y=15x5y = 15x\newlineFirst, let's simplify the second equation to find a relationship between xx and yy.\newline5y=15x5y = 15x\newlineDivide both sides by 55 to isolate yy:\newliney=3xy = 3x
  2. Substitute y=3xy = 3x: Now, let's substitute y=3xy = 3x from the second equation into the first equation to solve for xx.\newline−7x2=((3x)+5)((3x)−5)-7x^2 = ((3x) + 5)((3x) - 5)
  3. Expand right side of equation: Next, we will expand the right side of the equation using the difference of squares formula:\newline−7x2=(3x+5)(3x−5)-7x^2 = (3x + 5)(3x - 5)\newline−7x2=9x2−25-7x^2 = 9x^2 - 25
  4. Move terms to one side: Now, let's move all terms to one side of the equation to set it equal to zero:\newline−7x2−9x2+25=0-7x^2 - 9x^2 + 25 = 0\newlineCombine like terms:\newline−16x2+25=0-16x^2 + 25 = 0
  5. Solve for x2x^2: To find the value of xx, we need to solve for x2x^2:
    -16x2=−2516x^2 = -25
    Divide both sides by -1616:
    x2=2516x^2 = \frac{25}{16}
  6. Find value of x: Taking the square root of both sides gives us two possible solutions for xx:x=±2516x = \pm\sqrt{\frac{25}{16}}x=±54x = \pm\frac{5}{4}Since we are looking for the positive value of aa (and aa corresponds to xx), we choose the positive solution:a=54a = \frac{5}{4}

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