Systems of equations help?

#-10x-20y=-20#
#-5x-10y=10#

With work (please!)

1 Answer
maganbhai P.
May 14, 2018

The systems of equns.has no solution.#to phi#

Explanation:

Here,

#-10x-20y=-20#

Dividing each term by #(-10)# ,we get

#color(red)(x+2y=2...to(1)#

Also given that,

#-5x-10y=10#

Dividing each term by #(-5)# ,we get

#color(red)(x+2y=-2...to(2)#

Subtracting equn.#(1)# from #(2)#

#x+2y=2#

#x+2y=-2#

#ul(- -color(white)(.........)+#

#color(white)(..............)0=4 to# which is false statement.

Thus, the pair of equn. has no solution.

Let us draw the graphs of equn. #(1) and(2)#

From the graph ,we can say that the lines are parallel.

i.e.two lines donot intersect anywhere.

So, the systems of equns.has no solution.

enter image source here

Note:
We know that :If for #a_1,b_1,c_1,a_2,b_2,c_2 in RR #

#a_1x+b_1y+c_1=0 , where,a_1^2+b_1^2!=0#

#a_2x+b_2y+c_2=0 , where,a_2^2+b_2^2!=0 and #

#and a_1/a_2=b_1/b_2!=c_1/c_2=>No color(white)(.)Solution.#

In short, #1/1=2/2!=2/(-2) to phi#

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