解下列方程组:①{3x+5y+42,x/2=y/3.②{x+y/5=x-y/3=2.③3(x-1)-4(y-4)=0,5(y-1)=3(x

问题描述:

解下列方程组:①{3x+5y+42,x/2=y/3.②{x+y/5=x-y/3=2.③3(x-1)-4(y-4)=0,5(y-1)=3(x
+5).④2m+n/3=3m-2n/8=2

①3x+5y=42,x/2=y/3,则x=2y/3代入第一个方程,得2y+5y=42,所以y=6,x=2*6/3=4②x+y/5=x-y/3=2,则x+y=10,x-y=6两方程相加,得2x=16,则x=8,y=10-8=2③3(x-1)-4(y-4)=0,则3x=4y-135(y-1)=3(x +5),则3x=5y-20所以4y...