How do you solve the following system: #5x+8y= -29 , x+3y=3 #?

#color(green)(5x)+8y =-29#.............equation #(1)#

#x+3y=3#, multiplying by #5#
#color(green)(5x)+15y=15#.................equation #(2)#

Solving by elimination:
Subtracting equation #2# from #1# would eliminate #color(green)(5x)#

#cancelcolor(green)(5x)+8y =-29#
#-cancelcolor(green)(5x)-15y=-15#

#-7y =-34#

#y =(-44)/(-7)#

#color(green)(y =44/7#

Finding #x# from equation #2#:

#color(green)(5x)+15y=15#

#5x=15 -15y#

#5x=15 -15 xx(44/7)#

#5x=15 -(660/7)#

#5x=105/7 -(660/7)#

#5x=(-555/7)#

#x=(-555/(7xx5))#

#color(green)(x=-111/7#

From the given equations
#5x+8y=-29" "#first equation
#x+3y=3" "#second equation

start with the second equation
Let #x=3-3y# substitute in the first equation

#5x+8y=-29" "#first equation
#5(3-3y)+8y=-29" "#
#15-15y+8y=-29#
#-7y=-29-15#
#-7y=-44#
#y=44/7#

Use #y=44/7# in the second equation

#x=3-3y#
#x=3-3(44/7)#
#x=(21-132)/7#
#x=-111/7#

checking
#5x+8y=-29" "#first equation
#5(-111/7)+8(44/7)=-29" "#

#-555/7+352/7=-29#
#-203/7=-29#
#-29=-29#

checking
#x+3y=3" "#second equation
#(-111/7)+3(44/7)=3" "#
#-111/7+132/7=3#
#21/7=3#
#3=3#

God bless....I hope the explanation is useful..