Mathematics: Did this several times but still didn t get the answer .pls do try this :)

28 testStep-by-step:you do keep change flip and you get 4/1 and 3/1 then times to get 12/1 then roungd to get 12 then add the 16 to get 28...

Did this several times but still didn t get the

Ответ:

taniyahbenyamin2

30.01.2022

Mathematics

x=-15/2 and x=1

x=6 and x=3/4

Step-by-step explanation:

You have the following equations:

$\frac{4x-3}{x+2}=\frac{2x}{x+5}$

$\frac{2}{x-2}+\frac{3}{x}-\frac{9}{x+3}$

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:

$[\frac{4x-3}{x+2}=\frac{2x}{x+5}](x+2)(x+5)\\\\(4x-3)(x+5)=2x(x+2)\\\\4x(x)+4x(5)-3(x)-3(5)=2x(x)+2x(2)\\\\4x^2+20x-3x-15=2x^2+4x\\\\4x^2-2x^2+20x-3x-4x-15=0\\\\2x^2+13x-15=0$

In this case you use the quadratic formula:

$x_{1,2}=\frac{-13\pm \sqrt{(13)^2-4(2)(-15)}}{2(2)}\\\\x_{1,2}=\frac{-13 \pm 17}{4}\\\\x_1=-\frac{15}{2}\\\\x_2=1$

Then, for the first equation the solutions are x=-15/2 and x=1

second equation:

$[\frac{2}{x-2}+\frac{3}{x}=\frac{9}{x+3}]x(x-2)(x+3)\\\\2x(x+3)+3(x-2)(x+3)=9x(x-2)\\\\2x^2+6x+3(x^2+3x-2x-6)=9x^2-18x\\\\2x^2+6x+3(x^2+x-6)=9x^2-18x\\\\2x^2+6x+3x^2+3x-18-9x^2+18x=0\\\\-4x^2+27x-18=0$

Again, you use the quadratic formula:

$x_{1,2}=\frac{-27\pm \sqrt{(27)^2-4(-4)(-18)}}{2(-4)}\\\\x_{1,2}=\frac{-27\pm 21}{-8}\\\\x_1=6\\\\x_2=\frac{3}{4}$

Then, the solutions for the second equation are x=6 and x=3/4

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Did this several times but still didn't get the answer .pls do try this :)

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