How do you solve #(2/5)x + (1/5)y = -1# and #x + (2/5)y = -(8/5)#?

2 Answers
Jun 8, 2018

Answer:

#x = 2, y = -9#

Explanation:

You have a simple system of linear equations and can use the substitution method.
So take one of the equations and solve for x. We are going to use the second equation, because the coefficient of x is already 1.

# x + 2/5y = -8/5#
#x = -8/5 -2/5y #

Now let's substitute the x in our first equation

# 2/5x + 1/5y = -1#
#2/5*(-8/5-2/5y) + 1/5y = -1#
#-16/25 - 4/25y +1/5y = -1#
#1/25y = -9/25#
#y = -9#

To conclude, just solve for x in our substitution equation :
#x = -8/5 -2/5y#
#x = -8/5 -2/5*(-9)#
#x = -8/5 + 18/5 = 10/5 = 2 #
#x = 2#

So the there exists only one solution to this system of linear equations and the solution is :
#x = 2#
#y = -9#

Jun 8, 2018

Answer:

#x=2, y=-9#

Explanation:

If #(2/5)x+(1/5)y=-1#,
1) #2x+y=-5# (i.e. 5 times more)

If #x+(2/5)y=-(8/5)#
2) #5x+2y=-8# (again 5 times more)

(It is best first to get rid of the denominators, so that one multiplies the whole equation with the mcd, here 5.)

From 1) we have
3) #y=-(2x+5)#
Insert this into 2)
#5x-2(2x+5)=-8#
#5x-4x-10=-8#
4) #x=2#

Insert 4) into 3):
#y=-(2x+5)=-(2*2+5)=-9#

Test:
#(2/5)x+(1/5)y=(2/5*2)+(1/5)(-9)=-1# Check
#x+(2/5)y=2+(2/5)(-9)=(10-18)/5=-8/5# Check