How do you solve the system of equations #3x - 7y = 56# and #4x + 7y = - 7#?

#3x-7y=56#
#4x+7y=-7#

There are two ways to solve problem, though, the first way may not apply to every system of equations.

1st:
add the upper equation to the bottom equation like this:

#3x-7y=56#
#ul(4x+7y=-7)#
this cancels the # -7y and 7y# because they add to #0#
#3x+4x =7x#

#-7y +7y =0#

#56-7=49#

#7x=49#

Divide by #7# on both sides, you will get x=7

Replace #x " in " 4x+7y = -7# with #7#

#4(7)+7y=-7#
#28+7y=-7" "#minus 28 on both sides
#7y=-35#

#7y/7=-35/7#

y=-5

2nd:
#3x-7y=56#
#3x=56+7y#

Divide by #3# on both sides, and you will get #x=(56+7y)/3#
Replace #x " in " 4x+7y=-7# with #x=(56+7y)/3#

#4((56+7y)/3)+7y=-7#
#=(224+28y)/3+7y=-7#

#7y=(7y)/1#, so make the denominator similar to #(224+28y)/3# by multiply 3 which result of #(21y)/3#
#(224+28y)/3+(21y)/3=-7#
#(224+49y)/3=-7#

multiplying by # 3# on both sides

#224+49y=-21" "# minus #224# on both sides

#49y=-245" "#divide 49 on both sides
y=-5

replace #y " in " 3x-7y=56 " with" -5#
#3x-7(-5)=56#

#3x+35=56# minus #35# on both sides

#-3x=21 #

divide by # 3#

x=7

sorry if it's a bit too long, but always use the second method as it's universal to the system equation. good luck!