October 2018 1 38 Report

Prosze o rozwiązanie tego układu rownań: podpowiedź to3x-y jest nad kreską ułamkową :D a)3x-y/x+2y=2 4(x-1)-3(y+5)=-6


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\left \{ {{\frac{3x - y}{x + 2y} = 2} \atop {4(x - 1) - 3(y + 5) = -6}} \right

 

\left \{ {{3x - y = 2(x + 2y)} \atop {4x - 4 - 3y - 15 = -6}} \right

 

\left \{ {{3x - y = 2x + 4y} \atop {4x - 3y = -6  + 4 + 15}} \right

 

\left \{ {{3x - 2x - y - 4y = 0} \atop {4x - 3y = 13}} \right

 

\left \{ {{x - 5y = 0} \atop {4x - 3y = 13}} \right

 

\left \{ {{x = 5y} \atop {4*5y - 3y = 13}} \right

 

\left \{ {{x = 5y} \atop {17y = 13/:17}} \right

 

\left \{ {{x = 5*\frac{13}{17}} \atop {y = \frac{13}{17}}} \right

 

\left \{ {{x = \frac{65}{17}} \atop {y = \frac{13}{17}}} \right

 

\left \{ {{x = 3\frac{14}{17}} \atop {y = \frac{13}{17}}} \right

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