write a system of linear equations in which (3,-5) is a solution of equation 1 but not a solution of equation 2 and (-1,7) is a
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Answer:
0
Step-by-step explanation:
We are given the following:
(equ. 1)
(equ. 2)
We are asked:
(equ. 3) , what is
?
The 's represent the quotients you get.
is the remainder of dividing
by 5.
We know that is a number in {0,1,2,3,4}.
(I got this by multiplying both sides of equ 1. by 5.)
(I got this by multiplying both sides of equ 2. by 5.)
Let's add these equations together:
Factoring the 5 out for the 's part and simplify 4+1 gives:
So can't be the remainder of dividing something by 5 but see that we can factor this right hand expression more as:
So while there is no remainder (the remainder is 0).
Let's do an example if you are not convinced at this point that the remainder will be 0.
So choose x=9 since 9/5 gives a remainder of 4.
And choose y=16 since 16/5 gives a remainder of 1.
x+y=9+16=25 and 25/5 gives a remainder of 0.