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3 years ago

Simplify the expression

Mathematics

1 answer:

ollegr [7]3 years ago

4 0

Answer:

Step-by-step explanation:

$\dfrac{x^2-8x+12}{x^2-7x+10}\\\\x^2-7x+10\ne0\ \iff\ x=\frac{7\pm\sqrt{49-40}}{2}\ne0\ \iff\ x\ne5\ \wedge\ x\ne2\\\\\\\dfrac{x^2-8x+12}{x^2-7x+10}=\dfrac{x^2-2x-6x+12}{x^2-2x-5x+10}=\dfrac{x(x-2)-6(x-2)}{x(x-2)-5(x-2)}=\\\\\\ =\dfrac{(x-2)(x-6)}{(x-2)(x-5)}=\dfrac{x-6}{x-5}\\\\\\f(x)=x-6\\\\g(x)=x-5$

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Your friend says that the absolute value equation |3x+8|−9=−5 has no solution because the constant on the right side of the equa

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$Solution, solve\:for\:x,\:\left|3x+8\right|-9=-5\quad :\quad x=-4\quad \mathrm{or}\quad \:x=-\frac{4}{3}$

$Steps:$

$\left|3x+8\right|-9=-5$

$\mathrm{Add\:}9\mathrm{\:to\:both\:sides},\\\left|3x+8\right|-9+9=-5+9$

$\mathrm{Simplify},\\\left|3x+8\right|=4$

$|f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a,\\3x+8=-4\quad \quad \mathrm{or}\quad \:\quad \:3x+8=4$

$3x+8=-4,\\\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides},\\3x+8-8=-4-8,\\\mathrm{Simplify},\\3x=-12,\\\mathrm{Divide\:both\:sides\:by\:}3,\\\frac{3x}{3}=\frac{-12}{3},\\\mathrm{Simplify},\\x=-4$

$3x+8=4,\\\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides},\\3x+8-8=4-8,\\\mathrm{Simplify},\\3x=-4,\\\mathrm{Divide\:both\:sides\:by\:}3,\\\frac{3x}{3}=\frac{-4}{3},\\\mathrm{Simplify},\\x=-\frac{4}{3}$