How do you solve the simultaneous equations #7a - 3b = 17 # and #2a + b = 16#?

2 Answers
Jul 23, 2015

Answer:

#(a,b) = (5,6)#

Explanation:

[1]#color(white)("XXXX")##7a-3b=17#
[2]#color(white)("XXXX")##2a+b =16#

Multiply both sides of [2] by #3# (to get the same coefficient for #b# as in [1])
[3]#color(white)("XXXX")##6a+3b = 48#

Add [1] and [3]
[4]#color(white)("XXXX")##13a = 65#

Divide both sides by #13#
[5]#color(white)("XXXX")##a = 5#

Substitute #5# for #a# in [2]
[6]#color(white)("XXXX")##2(5) + b =16#

Simplify
[7]#color(white)("XXXX")##b = 6#

Jul 23, 2015

Answer:

#a=5#

#b=6#

Explanation:

We are given #7a-3b=17# and #2a+b=16#

Let's get the #b#'s equal to each other but opposite in sign by multiplying #2a+b=16# by #3#:

#2a+b=16#

#3*2a+3*b=3*16#

#6a+3b=48#

Now, let's add #6a+3b=48# to #7a-3b=17#

#6a+3b=48#
#7a-3b=17#

#13a=65#

#a=65/13#

#a=5#

Now, substitute #a# into #7a-3b=17#

#7a-3b=17#

#7*5-3b=17#

#35-3b=17#

#-3b=17-35#

#-3b=-18#

#b=6#

Let's check to see if our answers are correct by plugging in the values we found for both #a# and #b# into #2a+b=16#

#2a+b=16#

#2*5+6=16#

#10+6=16#

#16=16#