Question

Ejercicio 179 del Algebra de Baldor - Número 1 Resolver los siguientes sistemas

Answer 1

$\left \{ {{8x-5=7y-9} \atop {6x=3y+6}} \right. \\ \\ M\'etodo \ de \ reducci\'on \\ \\ 1) ordenamos \to \left \{ {{8x-7y=-9+5} \atop {6x-3y=6}} \right. \to \left \{ {{8x-7y=-4} \atop {6x-3y=6}} \right. \\ \\ 2) anulamos \ "y" \ multiplicando \to \left \{ {{-3(8x-7y=-4)} \atop {7*(6x-3y=6)}} \right. \\ \\ \left \{ {{-24x+21y=+12} \atop {42x-21y=+42}} \right. \quad Sumamos \\ \\ (-24x+42x)= +18\quad (21y +(-21y)=0\quad (12+42)= 54 \\ \\ 18x=54 \to x= \frac{54}{18}\to \boxed{x=3 }$


$3) anulamos \ "x" \ multiplicando \to \left \{ {{-6(8x-7y=-4)} \atop {8*(6x-3y=6)}} \right. \\ \\ \left \{ {{-48x+42y=+24} \atop {48x-24y=+48}} \right. \quad Sumamos \\ \\ (-48x+48x)=0\quad (42y +(-24y)=18\quad (24+48)= 72 \\ \\ 18y=72 \to y= \frac{72}{18}\to \boxed{y=4 }$

El resultado del sistema es (3;4)

espero que te sirva, salu2!!!!