Answer:
We conclude that:
[tex]\frac{x+2}{x+4}-\frac{x-1}{x+6}=\frac{5x+16}{\left(x+4\right)\left(x+6\right)}[/tex]
Step-by-step explanation:
Given the expression
[tex]\frac{x+2}{x+4}-\frac{x-1}{x+6}[/tex]
Least Common Multiple of x+4, x+6: (x+4) (x+6)
Adjusting fractions based on the LCM
[tex]=\frac{\left(x+2\right)\left(x+6\right)}{\left(x+4\right)\left(x+6\right)}-\frac{\left(x-1\right)\left(x+4\right)}{\left(x+6\right)\left(x+4\right)}[/tex]
since the denominators are equal, combine the fractions:
[tex]\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}[/tex]
so the expression becomes
[tex]=\frac{\left(x+2\right)\left(x+6\right)-\left(x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x+6\right)}[/tex]
[tex]=\frac{x^2+8x+12-\left(x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x+6\right)}[/tex]
[tex]=\frac{x^2+8x+12-x^2-3x+4}{\left(x+4\right)\left(x+6\right)}[/tex]
simplify
[tex]=\frac{5x+16}{\left(x+4\right)\left(x+6\right)}[/tex]
Therefore, we conclude that:
[tex]\frac{x+2}{x+4}-\frac{x-1}{x+6}=\frac{5x+16}{\left(x+4\right)\left(x+6\right)}[/tex]
