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(x^(2)+3x-a)/(x-2)=x+5
Given the equation for all x!=2, what is the value of a?

x2+3xax2=x+5\frac{x^{2}+3 x-a}{x-2}=x+5\newlineGiven the equation for all x2 x \neq 2 , what is the value of a a ?

Full solution

Q. x2+3xax2=x+5\frac{x^{2}+3 x-a}{x-2}=x+5\newlineGiven the equation for all x2 x \neq 2 , what is the value of a a ?
  1. Given Equation: We are given the equation:\newline(x2+3xa)/(x2)=x+5(x^{2}+3x-a)/(x-2)=x+5\newlineWe need to find the value of aa. To do this, we will multiply both sides of the equation by (x2)(x-2) to eliminate the denominator.\newline(x2+3xa)=(x+5)(x2)(x^{2}+3x-a) = (x+5)(x-2)
  2. Eliminate Denominator: Now we will expand the right side of the equation:\newline(x+5)(x2)=x22x+5x10(x+5)(x-2) = x^2 - 2x + 5x - 10\newlineSimplify the terms:\newlinex2+3x10x^2 + 3x - 10
  3. Expand Right Side: Next, we will compare the coefficients from the left side of the equation to the right side:\newlinex2+3xa=x2+3x10x^2 + 3x - a = x^2 + 3x - 10\newlineSince the x2x^2 and 3x3x terms are identical on both sides, we can equate the constants to find the value of aa:\newlinea=10-a = -10
  4. Compare Coefficients: Finally, we solve for aa:a=10a = 10

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