Решите систему уравнений, методом подстановки. А ) х=5у-6 3х-4у=4 Б) 4х-5у=-2 3х+2у=-13

A.
$\left \{ {{x=5y-6} \atop {3x-4y=4}} \right. \\ \\ \left \{ {{x=5y-6} \atop {3\cdot (5y-6)-4y=4}} \right. \\ \\ \left \{ {{x=5y-6} \atop {15y-18-4y=4}} \right. \\ \\ \left \{ {{x=5y-6} \atop {11y=22}} \right. \\ \\ \left \{ {{x=10-6} \atop {y=2}} \right. \\ \\ \left \{ {{x=4} \atop {y=2}} \right.$
Б.
$\left \{ {{4x-5y=-2} \atop {3x+2y=-13}} \right. \\ \\ \left \{ {{4x-5y=-2} \atop {y= \frac{ -13-3x}{2}}} \right. \\ \\ \left \{ {{4x-5\cdot( \frac{ -13-3x}{2}) =-2} \atop {y= \frac{ -13-3x}{2}}} \right. \\ \\ \left \{ {{8x-5\cdot( -13-3x) =-4} \atop {y= \frac{ -13-3x}{2}}} \right. \\ \\ \left \{ {{8x+65+15x) =-4} \atop {y= \frac{ -13-3x}{2}}} \right. \\ \\ \left \{ {{23x=69} \atop {y= \frac{ -13-3x}{2}}} \right. \\ \\ \left \{ {{x=3} \atop {y= \frac{ -13-3\cdot 3}{2}}} \right. \\ \\$
$\left \{ {{x=3} \atop {y= \frac{ -24}{2}}} \right. \\ \\ \left \{ {{x=3} \atop {y= -12}} \right. \\ \\$