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Please help: x/3x+8 - x+4/x+3 =1
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\[\frac{ x }{ 3x+8 }-\frac{ x+4 }{ x+3 }=1\]
find the solutions
i multiplied (3x+8)(x+4) -x(x+3)
and multiplied 1 by the denominator
i got 2x^2+23x+32=3x^2+17x+24
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hello???
hi
almost, but no, here amplify by (x+3)(3x+8) x(x+3) - (x+4)(3x+8) = (x+3)(3x+8) \[x^2+3x-(3x^2+8x+12x+32)=3x^2+8x+9x+24\] by ordering \[5x^2+34x+56=0\] Now you solve the second grade equation (I'm gonna use this method, feel free to use factorization or wathever) \[\dfrac {-b \pm \sqrt{b^2-4ac}} {2a}=x1,x2\] for ax^2+bx+c=0 And you'll get -4 and - (14/5)
thank u kind person u deserve a medal
hello
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I revised it, it gives -4 and -14/5, try using WolframAlpha to check it
i got it thank u !!
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