Question #59cd3

2 Answers
Feb 7, 2018

The solution to the system of equations is #(7.5,1.5)#

Explanation:

Given: #3x+3y=27# and #x-3y=3#

We have a system of equations in which we have two unknowns, #x# and #y#. We can solve for the system by solving for one variable and substitute that value into either equation to find the other variable.

So we have:

#3x+3y=27#

#x-3y=3#

We have to decide which variable we want to solve for first. This means that we will have to get rid of or "eliminate' a variable also known as applying the elimination method. It does not matter which variable you choose to solve for first BUT in terms of mathematical computations, it's best to go with the one with less math.

Thus we can add the two equations to eliminate the #y# variable:

#" "3x+color(purple)(3y)=27#
#+#
#" "ul(x-color(purple)(3y)=3)#
#" "4x=30#

Now all we have is a simple equation that we can easily for #x#:

#4x=30#

#x=30/4#

#color(blue)(x=7.5#

Now that we have found the #x# value, we need to find the #y# value. Again, we can do this by substituting #x=7.5# into either equations given and then solving for #y#. I'll use #x-3y=3#

So

#color(blue)(7.5)-3y=3#

#-3y=3-color(blue)(7.5#

#-3y=-4.5#

#y=-4.5/-3#

#color(red)(y=1.5#

So we have arrived at the solution #(color(blue)(7.5),color(red)(1.5))#

We shall verify that this indeed is the solution by substituting the #x# and #y# values into both equations:

Check 1:

#3x+3y=27#

#3(color(blue)(7.5))+3(color(red)(1.5))=27#

#22.5+4.5=27#

#27=27#

Check 2:

#x-3y=3#

#color(blue)7.5-3(color(red)(1.5))=3#

#7.5-4.5=3#

#3=3#

So yes, the answers check and therefore the solution is indeed #(7.5,1.5)#

Here is the solution graphically:

https://www.desmos.com/calculator/43ishtkmac

Feb 7, 2018

#x = (15) / (2)#

#y = (3)/(2)#

Explanation:

Because both of the #y# terms in the two equations have the same coefficient ( namely, #3# ), you can add the equations to make the #y# terms go to #0#.

That will leave you with an equation with just one unknown (namely, the #x# terms) -- which you can easily solve.

Add the equations to let the #y# terms drop out

#color(white)(....)# #3x+ 3y=27#
#+# #color(white.mm.........)#  #   x −3y=   3#
.......................................
#color(white)(....)# #4x# #color(white)(.......)# #= 30#

1) Divide both sides by #4# to isolate #x#

#x = (30)/(4)#

2) Reduce to lowest terms

#x = (15)/ (2)# #larr# answer for #x#
#color(white)(................)# _____

Use the value of #x# to find the value of #y#

1) Sub in #(15)/(2)# in the place of #x#, then solve for #y#

#color(white)(.)# #x −3y= 3#

#(15)/ (2) - 3y = 3#

2) Clear the fraction by multiplying all the terms on both sides by #2# and letting the denominator cancel out
#15 - 6y = 6#

3) Subtract #15# from both sides to isolate the #-6y# term
#- 6y = - 9#

4) Divide both sides by #-6# to isolate #y#
#y = (9)/(6)#

5) Reduce to lowest terms
#y = (3)/(2)# #larr# answer for #y#

#color(white)(................)# _____

Answer:

#x = (15) / (2)#

#y = (3)/(2)#

#color(white)(................)# _____

Check

Sub in the values for #x# and #y# and see if the equation still equals its given value.

#3x+ 3y=27#

#3( (15)/ (2)) + 3((3)/(2))# should equal #27#

Clear the parentheses by distributing the #3#s

#(45)/(2) + (9)/(2)# should equal #27#

Add the fractions, which already have a common denominator

#(54)/(2)# should equal #27#

Write the fraction in lowest terms

#27# does equal #27#

#Check#