Kopertina e librit Matematika 10 - 11: Pjesa II

Zgjidhja e ushtrimit 9

Zgjidhja e ushtrimit 9 të mësimit 7.5A në librin Matematika 10 - 11: Pjesa II nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Në secilin nga rastet e mëposhtme, gjeni xx dhe yy.

  1. (xy)+(13)=(62)\dbinom{x}{y}+\dbinom{1}{3}=\dbinom{6}{2}
  2. (xy)−(25)=(84)\dbinom{x}{y}-\dbinom{2}{5}=\dbinom{8}{4}
  3. x(32)+y(15)=(113)x\dbinom{3}{2}+y\dbinom{1}{5}=\dbinom{11}{3}
  4. x(43)−y(3−2)=(613)x\dbinom{4}{3}-y\dbinom{3}{-2}=\dbinom{6}{13}

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Zgjidhja

  1. (xy)+(13)=(62)\dbinom{x}{y}+\dbinom{1}{3}=\dbinom{6}{2} ⇒\rArr (x+1y+3)=(62)\dbinom{x+1}{y+3}=\dbinom{6}{2} ⇒\rArr {x+1=6y+3=2\begin{cases}x+1=6\\y+3=2\end{cases} ⇒\rArr {x=6−1y=2−3\begin{cases}x=6-1\\y=2-3\end{cases} ⇒\rArr {x=5y=−1\begin{cases}x=5\\y=-1\end{cases}
  2. (xy)−(25)=(84)\dbinom{x}{y}-\dbinom{2}{5}=\dbinom{8}{4} ⇒\rArr (x−2y−5)=(84)\dbinom{x-2}{y-5}=\dbinom{8}{4} ⇒\rArr {x−2=8y−5=4\begin{cases}x-2=8\\y-5=4\end{cases} ⇒\rArr {x=8+2y=4+5\begin{cases}x=8+2\\y=4+5\end{cases} ⇒\rArr {x=10y=9\begin{cases}x=10\\y=9\end{cases}
  3. x(32)+y(15)=(113)x\dbinom{3}{2}+y\dbinom{1}{5}=\dbinom{11}{3} ⇒\rArr (x×3+y×1x×2+y×5)=(113)\dbinom{x\times3+y\times1}{x\times2+y\times5}=\dbinom{11}{3} ⇒\rArr (3x+y2x+5y)=(113)\dbinom{3x+y}{2x+5y}=\dbinom{11}{3} ⇒\rArr {3x+y=112x+5y=3\begin{cases}3x+y=11\\2x+5y=3\end{cases} ⇒\rArr {y=11−3x2x+5(11−3x)=3\begin{cases}y=11-3x\\2x+5(11-3x)=3\end{cases} ⇒\rArr {y=11−3x2x+55−15x=3\begin{cases}y=11-3x\\2x+55-15x=3\end{cases} ⇒\rArr {y=11−3x−13x=3−55\begin{cases}y=11-3x\\-13x=3-55\end{cases} ⇒\rArr {y=11−3x−13x=−52\begin{cases}y=11-3x\\-13x=-52\end{cases} ⇒\rArr {y=11−3xx=−52−13=4\begin{cases}y=11-3x\\x=\dfrac{-52}{-13}=4\end{cases} ⇒\rArr {y=11−3×4x=4\begin{cases}y=11-3\times4\\x=4\end{cases} ⇒\rArr {y=11−12x=4\begin{cases}y=11-12\\x=4\end{cases} ⇒\rArr {y=−1x=4\begin{cases}y=-1\\x=4\end{cases}
  4. x(43)−y(3−2)=(613)x\dbinom{4}{3}-y\dbinom{3}{-2}=\dbinom{6}{13} ⇒\rArr (x×4−y×3x×3−y×(−2))=(613)\dbinom{x\times4-y\times3}{x\times3-y\times(-2)}=\dbinom{6}{13} ⇒\rArr (4x−3y3x+2y)=(613)\dbinom{4x-3y}{3x+2y}=\dbinom{6}{13} ⇒\rArr {4x−3y=63x+2y=13\begin{cases}4x-3y=6\\3x+2y=13\end{cases} ⇒\rArr {4x=6+3y3x+2y=13\begin{cases}4x=6+3y\\3x+2y=13\end{cases} ⇒\rArr {x=32+34y3(32+34y)+2y=13\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}y\\3(\dfrac{3}{2}+\dfrac{3}{4}y)+2y=13\end{cases} ⇒\rArr {x=32+34y92+94y+2y=13\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}y\\ \dfrac{9}{2}+\dfrac{9}{4}y+2y=13\end{cases} ⇒\rArr {x=32+34y174y=172\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}y\\ \dfrac{17}{4}y=\dfrac{17}{2}\end{cases} ⇒\rArr {x=32+34yy=172174\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}y\\y=\dfrac{\dfrac{17}{2}}{\dfrac{17}{4}}\end{cases} ⇒\rArr {x=32+34yy=172×417=2\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}y\\y=\dfrac{17}{2}\times\dfrac{4}{17}=2\end{cases} ⇒\rArr {x=32+34×2y=2\begin{cases}x=\dfrac{3}{2}+\dfrac{3}{4}\times2\\y=2\end{cases} ⇒\rArr {x=32+64=6+64=3y=2\begin{cases}x=\dfrac{3}{2}+\dfrac{6}{4}=\dfrac{6+6}{4}=3\\y=2\end{cases} ⇒\rArr {x=3y=2\begin{cases}x=3\\y=2\end{cases}