Respuesta :
Let us simplify the equation using cross multiplication.
Using cross multiplication: [a/b = c/d = a × d = b × c]
[tex]\dfrac{2x + 3}{3x - 1} = \dfrac{4x - 1}{x + 3}[/tex]
[tex]\dfrac{(2x + 3)(x + 3)}{1} = \dfrac{(4x - 1)(3x - 1)}{1}[/tex]
[tex]\dfrac{(2x^{2} + 6x + 3x + 9)}{1} = \dfrac{(12x^{2} - 4x - 3x + 1)}{1}[/tex]
[tex]2x^{2} + 6x + 3x + 9} = \dfac{12x^{2} - 4x - 3x + 1[/tex]
Now, combine like terms to simplify.
Combining like terms on both sides:
[tex]2x^{2} + 6x + 3x + 9} = \dfac{12x^{2} - 4x - 3x + 1[/tex]
[tex]2x^{2} + 9x + 9} = \dfac{12x^{2} - 7x + 1[/tex]
Now, let's set the value of the R.H.S to 0.
Setting the value of the R.H.S, to 0:
[tex]2x^{2} + 9x + 9} = \dfac{12x^{2} - 7x + 1[/tex]
[tex]2x^{2} + 9x + 9} - \dfac{12x^{2} + 7x - 1= 0[/tex]
Now, combine like terms and simplify.
Combining like terms on the L.H.S:
[tex]2x^{2} + 9x + 9} - \dfac{12x^{2} + 7x - 1= 0[/tex]
[tex]-10x^{2} + 16x + 8} = 0[/tex]
Rewrite 16x in such a way such that the product of the coefficients of -10x² and 8 is equal to the product of the coefficients of the addends that sum up to 16x. Clearly, the product of the coefficients of -10x² and 8 is 80. Thus, the only suitable addends could be 20x and -4x, as the product of the coefficients of 20x and -4x is 80 and their sum is 16x. Therefore, we get:
[tex]-10x^{2} + 20x - 4x + 8} = 0[/tex]
Factorize the expressions as needed:
[tex]-10x^{2} + 20x - 4x + 8} = 0[/tex]
[tex](-10x^{2} - 4x) + (20x + 8)} = 0[/tex]
[tex]-2x(5x + 2) + 4(5x + 2)} = 0[/tex]
Combine the factorized expressions:
[tex]-2x(5x + 2) + 4(5x + 2)} = 0[/tex]
[tex](-2x + 4)(5x + 2) = 0[/tex]
Therefore, we obtain the following equations:
[tex](-2x + 4)(5x + 2) = 0[/tex]
[tex](-2x + 4) = 0\ ; (5x + 2) = 0[/tex]
Open the parentheses and solve for x:
[tex](-2x + 4) = 0\ ; (5x + 2) = 0[/tex]
[tex]-2x + 4 = 0\ ; 5x + 2 = 0[/tex]
[tex]\boxed{x = 2\ ; \ x = \dfrac{-2}{5}}[/tex]
Learn more about solving for the variable in quadric equations:
⇒ https://brainly.com/question/19048987
[tex]\\ \rm\Rrightarrow \dfrac{2x+3}{3x-1}=\dfrac{4x-1}{x+3}[/tex]
[tex]\\ \rm\Rrightarrow (2x+3)(x+3)=(4x-1)(3x-1)[/tex]
[tex]\\ \rm\Rrightarrow 2x(x+3)+3(x+3)=4x(3x-1)-1(3x-1)[/tex]
[tex]\\ \rm\Rrightarrow 2x^2+6x+3x+9=12x^2-4x-3x+1[/tex]
[tex]\\ \rm\Rrightarrow 2x^2+9x+9=12x^2-7x+1[/tex]
[tex]\\ \rm\Rrightarrow 9x+7x+9-1=10x^2[/tex]
[tex]\\ \rm\Rrightarrow 16x+8=10x^2[/tex]
[tex]\\ \rm\Rrightarrow 8x+4=5x^2[/tex]
[tex]\\ \rm\Rrightarrow 5x^2-8x-4=0[/tex]
[tex]\\ \rm\Rrightarrow 5x^2-10x+2x-4=0[/tex]
[tex]\\ \rm\Rrightarrow 5x(x-2)+2(x-2)=0[/tex]
[tex]\\ \rm\Rrightarrow (5x+2)(x-2)=0[/tex]
[tex]\\ \rm\Rrightarrow x=\dfrac{-2}{5}\:or\:2[/tex]
