To find the values of [tex]\( x \)[/tex] that satisfy the equation, we need to set the two given expressions equal to each other and solve for [tex]\( x \)[/tex].
We are given two expressions:
[tex]\[ \text{Expression a: } \frac{6x}{x - 2} \][/tex]
[tex]\[ \text{Expression b: } \frac{x^2 + 3x}{x^2 + 2x - 3} \][/tex]
First, we'll set these two expressions equal to each other:
[tex]\[ \frac{6x}{x - 2} = \frac{x^2 + 3x}{x^2 + 2x - 3} \][/tex]
Next, we solve this equation by cross-multiplying to eliminate the denominators:
[tex]\[ 6x (x^2 + 2x - 3) = (x - 2) (x^2 + 3x) \][/tex]
Expanding both sides, we get:
[tex]\[ 6x^3 + 12x^2 - 18x = x^3 + 3x^2 - 2x^2 - 6x \][/tex]
Simplifying the right-hand side:
[tex]\[ 6x^3 + 12x^2 - 18x = x^3 + x^2 - 6x \][/tex]
Now, bringing all terms to one side of the equation:
[tex]\[ 6x^3 + 12x^2 - 18x - x^3 - x^2 + 6x = 0 \][/tex]
Combining like terms:
[tex]\[ 5x^3 + 11x^2 - 12x = 0 \][/tex]
Factor out the common term [tex]\( x \)[/tex]:
[tex]\[ x (5x^2 + 11x - 12) = 0 \][/tex]
We set each factor equal to zero:
[tex]\[ x = 0 \][/tex]
[tex]\[ 5x^2 + 11x - 12 = 0 \][/tex]
Next, we solve the quadratic equation [tex]\( 5x^2 + 11x - 12 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -12 \)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 11^2 - 4(5)(-12) = 121 + 240 = 361 \][/tex]
[tex]\[ x = \frac{-11 \pm \sqrt{361}}{2 \cdot 5} = \frac{-11 \pm 19}{10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-11 + 19}{10} = \frac{8}{10} = \frac{4}{5} \][/tex]
[tex]\[ x = \frac{-11 - 19}{10} = \frac{-30}{10} = -3 \][/tex]
However, we must check for any potential restrictions arising from the original denominators not being zero:
[tex]\[ x - 2 \neq 0 \Rightarrow x \neq 2 \][/tex]
[tex]\[ x^2 + 2x - 3 \neq 0 \Rightarrow (x - 1)(x + 3) \neq 0 \Rightarrow x \neq 1 \text{ and } x \neq -3 \][/tex]
Clearly, [tex]\( x = 4/5 \)[/tex] and [tex]\( x = 0 \)[/tex] satisfy these conditions, while [tex]\( x = -3 \)[/tex] does not.
Thus, the solutions to the problem are:
[tex]\[ \boxed{0 \text{ and } \frac{4}{5}} \][/tex]
