Answer:
[tex]x=\frac{1}{2}[/tex] or 0.5 in decimal form
Step-by-step explanation:
The expression is [tex]\frac{x+1}{x-2}-\frac{x-3}{x+2}=0[/tex]
Find Least Common Multiplier (LCM) of the denominators [tex]x-2,\:x+2[/tex]
This is [tex](x-2)(x+2)[/tex] the product of the two denominators
Multiply entire equation by this LCM: [tex](x-2)(x+2)[/tex]
We get
[tex]\frac{x+1}{x-2}\left(x-2\right)\left(x+2\right)-\frac{x-3}{x+2}\left(x-2\right)\left(x+2\right)=0\cdot \left(x-2\right)\left(x+2\right)[/tex]
Simplify the first term
[tex]\frac{x+1}{x-2}\left(x-2\right)\left(x+2\right)[/tex] [tex]=\left(x+1\right)\left(x+2\right)[/tex]
Simplify the second term
[tex]-\frac{x-3}{x+2}\left(x-2\right)\left(x+2\right) =\quad -\left(x-3\right)\left(x-2\right)[/tex]
RHS is [tex]0[/tex]
Therefore we get
[tex]\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x-2\right)=0[/tex]
Expand the first term using the FOIL method [tex](x+a) (x + b) = x^2 + ax + bx + ab[/tex]
Here a = 1, b = 2. So [tex](x+1)(x+2) = x^2 + 2x + 1x + (1)(2) = x^2 + 3x + 2[/tex]
[tex]\left(x+1\right)\left(x+2\right)[/tex][tex]=x^2+3x+2[/tex]
Expanding [tex]-\left(x-3\right)\left(x-2\right)[/tex] gives us [tex]-x^2 + 5x - 6[/tex]
(Use the FOIL method: [tex](x+b) (x + a) = x^2 + ax + bx + ab[/tex] Here a = -2, b = -3. Note there is a negative sign before the entire expression)
So the original expression is
[tex]x^2+3x+2-x^2+5x-6 = 0[/tex]
Collecting like terms
[tex]x^2-x^2 + 3x +5x + 2 - 6[/tex] [tex]= 8x -4[/tex] = [tex]0[/tex]
Add 4 to both sides
[tex]8x-4+4=0+4[/tex]
[tex]8x=4[/tex]
Divide both sides by 8:
[tex]\frac{8x}{8}=\frac{4}{8}[/tex]
[tex]x=\frac{1}{2}[/tex] or [tex]0.5[/tex] in decimal
