Answer:
[tex]x=2[/tex]
Step-by-step explanation:
When given the following equation;
[tex]\frac{x+3}{3x-2}-\frac{x-3}{3x+2}=\frac{44}{9x^2-4}[/tex]
One has to solve for the variable (x). Remember, when working with fractions, one must have a common denominator in order to perform operations. Since the denominators on the left side of the equation are unlike, one must change them so that they are like denominators. Multiply each fraction by the other fraction's denominator on the respective side. Remember to multiply both the numerator and denominator by the value to ensure that the equation remains true.
[tex]=\frac{x+3}{3x-2}*(\frac{3x+2}{3x+2})-\frac{x-3}{3x+2}*(\frac{3x-2}{3x-2})=\frac{44}{9x^2-4}[/tex]
Simplify,
[tex]=\frac{(x+3)(3x+2)}{(3x-2)(3x+2)}-\frac{(x-3)(3x-2)}{(3x+2)(3x-2)}=\frac{44}{9x^2-4}\\\\=\frac{3x^2+11x+6}{9x^2-4}-\frac{3x^2-11x+6}{9x^2-4}=\frac{44}{9x^2-4}[/tex]
Distribute the negative sign to simplify the left side of the equation;
[tex]=\frac{3x^2+11x+6}{9x^2-4}-\frac{3x^2-11x+6}{9x^2-4}=\frac{44}{9x^2-4}\\\\=\frac{3x^2+11x+6-(3x^2-11x+6)}{9x^2-4}=\frac{44}{9x^2-4}\\\\=\frac{3x^2+11x+6-3x^2+11x-6}{9x^2-4}=\frac{44}{9x^2-4}\\\\=\frac{22x}{9x^2-4}{=\frac{44}{9x^2-4}[/tex]
Since the denominators on opposite sides of the equation are like, one can now ignore the denominators,
[tex]=22x=44[/tex]
Inverse operations,
[tex]=22x=44[/tex]
÷[tex]2[/tex] ÷[tex]2[/tex]
[tex]x=2[/tex]
