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

Given Equation: We are given the equation: (x^(2)+3x-a)/(x-2)=x+5 To find the value of a, we need to solve this equation for a. Since the equation is valid for all x not equal to 2, we can multiply both sides by (x-2) to eliminate the denominator. (x^(2)+3x-a) = (x+5)(x-2)Eliminate Denominator: Now we will expand the right side of the equation: (x+5)(x-2) = x^2 - 2x + 5x - 10 Simplify the terms: x^2 + 3x - 10Expand Right Side: Next, we will set the left side of the equation equal to the simplified right side: x^2 + 3x - a = x^2 + 3x - 10Set Equal: Since the x^2 and 3x terms are on both sides of the equation, they cancel each other out. We are left with: -a = -10Cancel Terms: To find the value of a, we divide both sides by -1: a = 10

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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 ?

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

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Algebra 2

Composition of linear and quadratic functions: find a value

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\frac{x^{2}+3 x-a}{x-2}=x+5

\newline

Given the equation for all

x \neq 2

, what is the value of

a

Given Equation: We are given the equation: $\newline$ $(x^{2}+3x-a)/(x-2)=x+5$ $\newline$ To find the value of $a$ , we need to solve this equation for $a$ . Since the equation is valid for all $x$ not equal to $2$ , we can multiply both sides by $(x-2)$ to eliminate the denominator. $\newline$ $(x^{2}+3x-a) = (x+5)(x-2)$
Eliminate Denominator: Now we will expand the right side of the equation: $\newline$ $(x+5)(x-2) = x^2 - 2x + 5x - 10$ $\newline$ Simplify the terms: $\newline$ $x^2 + 3x - 10$
Expand Right Side: Next, we will set the left side of the equation equal to the simplified right side: $x^2 + 3x - a = x^2 + 3x - 10$
Set Equal: Since the $x^2$ and $3x$ terms are on both sides of the equation, they cancel each other out. We are left with: $\newline$ $-a = -10$
Cancel Terms: To find the value of $a$ , we divide both sides by $-1$ : $a = 10$