Question

The number of integral order pair (x,y) that satisfy the system of equation
|x+y-4|=5 and |x-3|+|y-1|=5 is /are

Answer 1

@hartnn

Answer 2

why don't you use elimination or substitution method

Answer 3

How many times??

Answer 4

For that i have to repeatedly put up the values and see whether they satisfy or not...

Answer 5

yea thats a bit tedious

Answer 6

Actually!! It is...that's why posted it..

Answer 7

ok i would give you the answer if am wrong tell me

Answer 8

(x+y−4)=5

Answer 9

x-intercept: (9,0)
y-intercept: (0,9)

Answer 10

It is mod here...

Answer 11

| | symbol is modulus

Answer 12

that means every answer is positive

Answer 13

No!

Answer 14

X ,y can be negative too!

Answer 15

@ganeshie8

Answer 16

@baru

Answer 17

@Michele_laino

Answer 18

the only way I can think of to do this without looking at a calculator is to...
think of all the possible cases:
\[\text{
Case 1: } x+y-4 \ge0 \text{ ; } x-3\ge0 \text{ ; }y-1 \ge 0\]
There are 7 more cases:
The first is assume +,+,+ (which I have above)
Second is +,+,-
Third is +,-,+
Fourth is +,-,-
Fifth is -,+,-
Sixth is -,+,+
Seventh is -,-,+
Eighth is -,-,-

I will show you the rest of the first case...


So assuming the first case we have the following equations:
\[x+y-4=5 \\ x-3+y-1=5 \\ \text{ simplifying } \\ x+y=9 \\ x+y=9 \\ \text{ so we only have to look at } x+y=9\]

\[\text{ now remember in this case we have } x \ge 3 \text{ and }y \ge 1\]

So x+y can be the following:
3+6
4+5
5+4
6+3
7+2
8+1
I got no further because y would be less than 1 and notice I started with x is 3... since that is the least number x can be


anyways... if you were to continue
you would go to Case 2:
\[\text{ Case 2 : }=x+y-4 \ge0 \text{ ; } x-3 \ge 0 \text{ ; } y-1 \le 0 \\ \text{ so this would make your equations look like } \\ x+y-4=5 \text{ and } x-3-(y-1)=5 \\ x+y=9 \text{ and } x-y-2=5 \\ x+y=9 \text{ and } x-y=7 \\ \\ \text{ solve the system } \\ 2x=16 \\ x=8 \\ y=1 \\ \text{ make sure this fits in with our assumptions; it does }\]


anyways continue with the other cases...
I'm trying to think if there is a way to rule out some of these cases...
Like that last case gave us a x and y we already had...

Answer 19

Well what i think is we can see the similarity between the two mod terms .
Both are 5 ...so we can write the first term as |x-3+y-1|=5

Answer 20

It was |x-3+y-1|=5
And second term is |x-3|+|y-1|=5
So that means the first term is simply
|x-3+y-1|=|x-3|+|y-1|

Answer 21

|x+y-4|=5 and |x-3|+|y-1|=5

|x+y-4|=5
|x-3+y-1|=5
assume x>3,y>1
|x-3|+|y-1|=5 has
(4,5),(5,4),(6,3),(7,2),(8,1),(3,6)

Answer 22

be careful ... this is only the first set of solutions i think..
i like what @freckles did but a bit of too much..

Answer 23

We can say that this cN be true only when either x-3>=0 and y-1>=0
Or x-3<=0 and y-1<=0

Answer 24

I m talking about my method ...

Answer 25

well I think @samigupta8 is right on her equation
if you use triangle inequality
\[5=|x-3+y-1| \le |x-3|+|y-1|=5 \\ \text{ but this inequality only makes sense if } \\ |x-3+y-1|=|x-3|+|y-1|\]

Answer 26

so everything she has I think can be reduce to solving the equation
\[|x+y-4|=|x-3|+|y-1|\]
but also make sure after solving for x and y that |x+y-4|=5

Answer 27

Sure!!

Answer 28

@ikram002p is your way more reduced than my way?

Answer 29

like less cases?

Answer 30

8 cases is so much :p

Answer 31

same boat they are :P...

Answer 32

even after applying that triangle inequality I see no way out of getting out of those 8 cases

Answer 33

We have simplified our equation like this
X-3>=0 and y-1>=0
Which gives 6 ordered pairs ...

Answer 34

And a case when both are less than 0

Answer 35

I think I'm going to look at just this for now
solving for integer u and v such that the following are true:
\[|u+v|=5 \\ |u|+|v|=5\]
I think this will be easier to look at

Answer 36

I got 12 cases...

Answer 37

Squaring first equation:
\[(u+v)^2=25 \\ u^2+v^2+2uv=25 \]
Squaring second equation:
\[(|u|+|v|)^2=25 \\ u^2+v^2+2|u||v|=25\]

\[2uv=2|u||v| \\ uv=|u||v| \\ uv=|uv| \text{ is true only if } uv \ge 0 \\ u=x-3 \text{ while } v=y-1 \\ \text{ so we need \to solve } (x-3)(y-1) \ge 0\]
so x-3 is positive and y-1 is positive
or
x-3 is negative and y-1 is negative
but they can't have different sign

So I think we can reduce this down to 2 cases maybe if I haven't made some error

Answer 38

Correct!! Hereby you will get 6 cases for u and v both positive
And 6 cases for u and v both negative

Answer 39

oh answers is what you meant
yes 12 answers looks right

Answer 40

Ryt ..