Answer:
x=-15/2 and x=1
x=6 and x=3/4
Step-by-step explanation:
You have the following equations:
[tex]\frac{4x-3}{x+2}=\frac{2x}{x+5}[/tex]
[tex]\frac{2}{x-2}+\frac{3}{x}-\frac{9}{x+3}[/tex]
To solve both equations you can first multiply by the m.c.m of the denominators, and then solve for x, just as follow:
first equation:
[tex][\frac{4x-3}{x+2}=\frac{2x}{x+5}](x+2)(x+5)\\\\(4x-3)(x+5)=2x(x+2)\\\\4x(x)+4x(5)-3(x)-3(5)=2x(x)+2x(2)\\\\4x^2+20x-3x-15=2x^2+4x\\\\4x^2-2x^2+20x-3x-4x-15=0\\\\2x^2+13x-15=0[/tex]
In this case you use the quadratic formula:
[tex]x_{1,2}=\frac{-13\pm \sqrt{(13)^2-4(2)(-15)}}{2(2)}\\\\x_{1,2}=\frac{-13 \pm 17}{4}\\\\x_1=-\frac{15}{2}\\\\x_2=1[/tex]
Then, for the first equation the solutions are x=-15/2 and x=1
second equation:
[tex][\frac{2}{x-2}+\frac{3}{x}=\frac{9}{x+3}]x(x-2)(x+3)\\\\2x(x+3)+3(x-2)(x+3)=9x(x-2)\\\\2x^2+6x+3(x^2+3x-2x-6)=9x^2-18x\\\\2x^2+6x+3(x^2+x-6)=9x^2-18x\\\\2x^2+6x+3x^2+3x-18-9x^2+18x=0\\\\-4x^2+27x-18=0[/tex]
Again, you use the quadratic formula:
[tex]x_{1,2}=\frac{-27\pm \sqrt{(27)^2-4(-4)(-18)}}{2(-4)}\\\\x_{1,2}=\frac{-27\pm 21}{-8}\\\\x_1=6\\\\x_2=\frac{3}{4}[/tex]
Then, the solutions for the second equation are x=6 and x=3/4
