How do you solve #4x – y = 5# and #x + y = 10# and which method do you use?

3 Answers
May 29, 2018

#x=3# and #y = 7#

Explanation:

Here is the answer...

First take the second equation...
#x+y=10#
#x=10-y# ...#(eq. 1)#

Now substitute #(eq. 1)# in the given equation #4x - y = 5#.

#4 (10-y)-y = 5#

#40 - 4y - y = 5#

#40 - 5y = 5#

#40-5 = 5y#

#35 = 5y#

#y = 7#

Substituting this value of #y# in #(eq. 1)#, we get #x=3#.

Therefore, #x=3# and #y = 7#.

May 29, 2018

#(3,7)#

Explanation:

#4x-y=5--(1)#

#x+y=10--(2)#

solve by elimination

#(1)+(2)#

#5x=15#

#=>x=3#

substitute into #(2)#

#3+y=10#

#=>y=7#

check in #(1)#

#4xx3-7=12-7=5 #

#=RHS#(1)#

#:. #consistent

#(3,7)#

May 29, 2018

See explanation.

Explanation:

The system is:

#{(4x-y=5),(x+y=10):}#

It can be solved using any of 3 methods:

  • Using substitution:

From the second equation we can calculate that: #y=10-x#

If we put this in the first equation we get:

#4x-(10-x)=5#

#4x-10+x=5#

#5x-10=5#

#5x=15=>x=3#

Now we can calculate that #y=10-3=7#
So the solution is

#{(x=3),(y=7):}#

  • By adding both equations:

In the initial system the coefficients of #y# are opposite numbers #-1# and #1#, so if we add both equations we get an equation with #x# variable only:

#5x=15#
#x=3#

Now we can calculate the remaining variable #y# by substitution:

#3+y=10=>y=7#

  • Graphically

Both equations represent linear functions, so we can solve the system by graphing the lines and seeing if they intersect:

graph{(y-4x+5)(x+y-10)((x-3)^2+(y-7)^2-0.05)=0 [-10, 10, -8, 8]}

As we can see the lines intersect at #(3,7)#, so the solution is:

#{(x=3),(y=7):}#

The choice of method depends on the system of equations. Here the easiest (for me) is the second method but others may prefer different ones.