How do you solve the system of equations #-12d - 7e = 48# and #5d + e = 3#?

There are two common ways you can do this:

• Substitution
• Elimination

SUBSTITUTION

You can solve for one variable and express it in terms of the other:

#-12d - 7e = 48#

#-12d = 48 + 7e#

#color(green)(d) = 48/(-12) + 7/(-12) e#

#= color(green)(-4 - 7/12 e)#

This is only a temporary expression, because we don't know #e# yet. Now, #d# can be used in the second equation to give:

#5d + e = 3#

#5(-4 - 7/12 e) + e = 3#

#-20 - 35/12 e + e = 3#

#-20 - 35/12 e + 12/12e = 3#

#-20 - 23/12 e = 3#

#-23/12 e = 23#

#cancel(12)/cancel(23)*-cancel(23)/cancel(12) e = cancel(23)*12/cancel(23)#

#=> color(blue)(e = -12)#

Therefore, we can plug this back into the easier equation to use (the second one) to get:

#5d + (-12) = 3#

#5d = 15#

#=> color(blue)(d = 3)#

ELIMINATION

The idea here is to take the two equations, add them together, and see what you can do to get rid of one variable early on. Usually, this is faster than substitution.

Try multiplying the second equation by #7#.

#-12d - 7e = 48#

#7(5d + e = 3)#

Notice how the #7e# will cancel out if we add these equations together:

#" "" "-12d - 7e = 48#

#+" "7*5d + 7*e = 7*3#

#"--------------------------------"#

#" "" "-12d - cancel(7e) = 48#

#+" "35d + cancel(7e) = 21#

#"--------------------------------"#

#" "23d" "" " = 69#

#=> d = 69/23 = (cancel(23)*3)/cancel(23)#

#=> color(blue)(d = 3)#

That allows us to solve for #e# by substitution:

#5d + e = 3#

#5(3) + e = 3#

#=> e = 3 - 5(3) = 3 - 15#

#=> color(blue)(e = -12)#

which should match the answer we got from just substitution (and it does!).