How do you solve #4x+5y=-23# and #x= -2-5y# using substitution?

Question

Ethan Adkins · Accepted Answer

#x=-7#
#y=1# Given: #" "4x+5y=-23# ...........................(1) #" "x=-2-5y#..................................(2) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Substitute for #x# in equation (1) using equation (2) giving: #" "4(-2-5y)+5y=-23" "...................(1_a)# Multiply out the bracket #" "-8-20y+5y=-23# #color(brown)(" "-8-15y=-23)# Add #color(blue)(15y)# to both sides #color(brown)(" -8-15ycolor(blue)(+15y)=-23color(blue)(+15y))# But #color(brown)(-15y)color(blue)(+15y)color(green)(=0)# #-8color(green)(+0)=-23+15y# Add 23 to both sides #-8+23=15y# #15=15y# Divide both sides by 15 #y=1# '~~~~~~~~~~~~~~~~~~~~~~~~~~~ Substitute #y=1# into equation (2) #x=-2-5(1) = -7#

Jameson Allen · Answer

undefined To solve the system of equations $ 4x + 5y = -23 $ and $ x = -2 - 5y $ using substitution, you can substitute the expression for $ x $ from the second equation into the first equation and solve for $ y $. Then, substitute the value of $ y $ back into either equation to find the value of $ x $.

First, substitute $ x = -2 - 5y $ into $ 4x + 5y = -23 $:

$$ 4(-2 - 5y) + 5y = -23 $$
$$ -8 - 20y + 5y = -23 $$
$$ -15y = -15 $$
$$ y = 1 $$

Now, substitute $ y = 1 $ back into $ x = -2 - 5y $:

$$ x = -2 - 5(1) $$
$$ x = -2 - 5 $$
$$ x = -7 $$

So, the solution to the system of equations is $ x = -7 $ and $ y = 1 $.