How do you solve #y-6x=22# and #4y+17=-11x# using substitution?

2 Answers
May 17, 2018

Answer:

#x = -3#
#y = 4#

Explanation:

Solving through Substitution

#y - 6x = 22#
#4y + 17 = -11x#

First, we want to solve for a variable that can be easily found. Then, we'll plug the variable in to solve for the variables. Let's start with finding what y is since it looks easy to solve for.

#y - 6x = 22#

Add 6x to both sides to negate #-6x#. This'll give us #y#'s value. You should now have:

#y = 6x + 22#

Substitute in this value for y in the second equation to solve for the variables:

#4y + 17 = -11x#
#4(6x + 22) + 17 = -11x#

Distribute.

#24x + 88 + 17 = -11x#

Combine like terms. #88 + 17# is #105#, so:

#24x + 105 = -11x#

Subtract #24x# from both sides to negate #24x# in order to solve for x. The reason why we aren't adding #11x# to both sides is because it'll make the equation unbalanced. #-11x - 24x# is basically the same as #11 + 24 #because they're the same sign, so they combine to become #-35x#. So:

#105 = -35x#

Divide by -35 to solve for x.

#105/-35# = #x#

#x = -3#

Now, plug this variable back into this equation to solve for y:

#y = 6x + 22#
#y = 6(-3) + 22#

Distribute.

#y = -18 + 22#

Add.
#y = 4#

Confirm this by plugging it back into the two equations:

#y - 6x = 22#
#4 - 6(-3) = 22#

Distribute.

#4 + 18 = 22#
#22 = 22#

#4y + 17 = -11x#
#4(4) + 17 = -11(-3) #16 + 17 = 33# #33 = 33#

Confirmed: these are the correct variables.

May 17, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#y - 6x + color(red)(6x) = 22 + color(red)(6x)#

#y - 0 = 22 + 6x#

#y = 22 + 6x#

Step 2) Substitute #(22 + 6x)# for #y# in the second equation and solve for #x#:

#4y + 17 = -11x# becomes:

#4(22 + 6x) + 17 = -11x#

#(4 * 22) + (4 * 6x) + 17 = -11x#

#88 + 24x + 17 = -11x#

#88 + 17 + 24x = -11x#

#105 + 24x = -11x#

#105 - color(red)(105) + 24x + color(blue)(11x) = -11x + color(blue)(11x) - color(red)(105)#

#0 + (24 + color(blue)(11))x = 0 - 105#

#35x = -105#

#(35x)/color(red)(35) = -105/color(red)(35)#

#(color(red)(cancel(color(black)(35)))x)/cancel(color(red)(35)) = -3#

#x = -3#

Step 3) Substitute #-3# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#;

#y = 22 + 6x# becomes:

#y = 22 + (6 xx -3)#

#y = 22 - 18#

#y = 4#

The Solution Is:

#x = -3# and #y = 4#

Or

#(-3, 4)#