Question

Ex 5 va rog urgent !!! Dau coroana!!!

Answer 1

 

Ex. 5)

Fiecare paranteza contine coordonatele carteziene ale unui punct.

x ∈ R;   y ∈ R

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$\displaystyle\bf\\a)\\(2x+1,~5)=(y,~2x+y)\\\\2x+1=y\\2x+y=5\\-----\\2x-y=-1\\2x+y=5\\------- Adunam~ecuatiile.\\4x~~~/~=4\\\\x=\frac{4}{4}\\\boxed{\bf~x=1}\\2x+1=y\\y=2x+1 = 2\cdot1+1=2+1\\\boxed{\bf~y=3}\\\\\\b)\\(2x-5,~-5)=(y,~x-3y)\\\\2x-5=y~~~\implies~substitutia:~~\boxed{\bf~y=2x-5}\\x-3y=-5\\-----\\x-3(2x-5)=-5\\x-6x+15=-5\\-5x=-5-15\\-5x=-20\\\\x=\frac{-20}{-5}\\\\\boxed{\bf~x=4}\\Ne~intoarcem~la~substitutie.\\y=2x-5=2\cdot4-5=8-5\\\\\boxed{\bf~y=3}$

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$\displaystyle\bf\\c)\\(3x+y,~24)=(8,~9x+3y)\\\\3x+y=8\\9x+3y=24~\Big|:3\\-----\\3x+y=8\\3x+y=8\\\textbf{Ecuatiile sunt identice (dependente).}\\\textbf{Sistemul este echivalent cu o singura ecuatie cu 2 necunoscute.}\\\implies~\textbf{Sistemul are o infinitate de solutii.}$

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$\displaystyle\bf\\d)\\(x^2-4,~2)=(5,~y^2-2)\\\\x^2-4=5\\y^2-2=2\\-----\\x^2-4-5=0\\y^2-2-2=0\\-----\\x^2-9=0\\y^2-4=0\\-----\\x^2-3^2=0\\y^2-2^2=0\\-----\\(x-3)(x+3)=0~~\implies~~(x-3)=0~si~(x+3)=0\\(y-2)(y+2)=0~~\implies~~(y-2)=0~si~(y+2)=0\\\\\boxed{\bf~x_1=3~si~x_2=-3}\\\boxed{\bf~y_1=2~si~y_2=-2}$