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How do you solve the following system: #-2x + 5y = 20, 2x – 5y = 5 #?

Answer 1

No solution

Some( in fact many) system of equations doesn't have answers... as you'll solve this you'll always get #20=5# which is impossible

it is useful in #11^(th)# standard... so wait till then

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Answer 2

Eva Abramson

No solutions

#-2x + 5y =20# #2x - 5y = 5#

We need to solve #-2x+5y=20# for #x#

#-2x + 5y = 20#

#-2x = 20 - 5y#

#x=(20-5y)/(-2)#

#x=5/2 y -10#

Now we can substitute #5/2y-10# for #x# in #2x-5y=5#

#2x - 5y=5#

#2(5/2y-10)-5y=5#

Distribute

#(2)(5/2y)+(2)(-10)-5y=5#

#(10/2y)-20-5y=5#

#5y - 20 - 5y = 5#

Combine like terms

#cancel(5y) cancel(-5y) - 20 = 5#

#-20 = 5#

Add #20# to both sides

#-20+20=5+20#

#0=25#

Thus,

There are no solutions!

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Answer 3

Hazel Anglin

To solve the system of equations:

-2x + 5y = 20 2x - 5y = 5

Add the two equations together to eliminate the variable x:

(-2x + 5y) + (2x - 5y) = 20 + 5 -2x + 2x + 5y - 5y = 25 0 = 25

Since 0 ≠ 25, the system is inconsistent, meaning there is no solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.