x and y

if xy = 0, that means either x or y has to be 0, but because of the second statement, we know that they cannot both be 0 because if that were the case, it would be 1*2 = 0 which is false.

to make the second statement true, one of the brackets must be equal to 0. in the first bracket (x+y+1), we can eliminate the 1 by making another statement, whereas x+y = -1. because one of them must be equal to 0, there are two possibilities for x and y. (0,-1) and (-1,0). do the same with the second side and you'll get 4 possibilities in total.

produto xy sendo igual a 0 temos dois casos:ou x=0 ou y=0 sendo x=0 substituindo na equação fica (y+1)(y+2)=0,novamente temos um produto igual a 0(ou y+1=0 ou y+2=0,logo y=-1,y=-2 respectivamente. Fazendo o mesmo com y sendo igual a zero,vamos ter os mesmos casos: x+1=0 ou x=2=0,logo x=-1 e -2 respectivamente.Temos então: x e y (0,-1),(0-2),(-1,0),(-2,0) 4 pares ordenados.

Since xy = 0, either x or y must be zero. In the case that x = 0, we get (y + 1)(y + 2) = 0, which yields two solutions. In the case that y = 0, we get (x + 1)(x + 2) = 0, which yields an additional two solutions. Therefore, there are four possible solutions.

On the first case, there will be two cases which is either $x$ or $y$ will be 0

Let's take $x=0$

If that's so, the second equation would be:

$(y+1)(y+2)=0$

$y^{2}+3y+2=0$

Therefore there are 2 cases if x is 0.

Therefore, there will also be 2 cases if y is 0

Therefore, 2 cases for x and 2 cases for y would make 4 cases/sets for the equations.

Systems of equations (linear) is solved using substitution:

Solution:

From equation 1: XY = 0 satisfies that X=0 or Y=0.

Take X = 0 and solve Y using equation 2:

            = (x+y+1)(x+y+2) 

            = (0+y+1)(0+y+2) 

            = (y+1)(y+2) = 0

e.i, either (y+1) = 0 or (y+2) = 0, giving us:

for X=0, Y= (-1) or Y=(-2)

or (0,-1) and (0,-2)

Now, take Y=0 from equation 1 and solve for X in equation 2. This gives us (-1,0) and (-2,0)

Therefore, there are 4 possible (x,y) ordered pairs from the given set of equations, which are

(0,-1)

(0,-2)

(-1,0)

(-2,0) ■

From the first equation we find that x = 0

Now, putting value of x as 0 in second equation, we find that there are only two values that satisfy this equation which are (0,-1) and (0,-2).

As we have to find all the possible ordered pairs, from the first equation we can take y=0. Putting y as 0 in second equation we find two more possible ordered pairs which are (-1,0) and (-2,0).

So there are 4 possible ordered set of real numbers that satisfy the two systems of equations which are (0,-1), (0,-2), (-1,0), (-2,0).

If xy = o, that means that x or y = 0.

The second statement clearly shows that they both can't be zero, since you would get 1 * 2 = 2.

Now, either the first or second bracket should be zero. Hence, for the first bracket x is either -1 or 0 and y is 0 and -1 respectively. For the second bracket, x is either equal to -2 and 0 and y is 0 and -2.

There are 4 cases.

• x = -1, y = 0
• x = 0, y = -1
• x = -2, y = 0
• x = 0, y = -2.

We have to know, to get a zero is by multiply something with zero. At first, we know xy = 0, so one of the x or y is 0. If x = 0, then we go to the second condition where (x + y + 1) (x + y + 2) = 0. So the possible value of y = -1 or -2, so one of them can be zero. If y = 0, the possible value for x so (x + y + 1)(x + y + 2) = 0 is -1 or 2. (x,y) possible value are (-2, 0), (-1, 0), (0, -2), (0, -1). So total we have 4 possible values.

the solutions would be (-1,0);(-2,0);(0,-1);(0,-2)

THERE 4 CASE IF X+Y=-1 and if X+Y=-2 so (0,-1)(-1,0)(0,-2)(-2,0)

$xy = 0$ so, $x=0$ or $y=0$ . If $x= 0$ we have $2$ solutions and they are $y=(-1,-2)$ and If $y= 0$ we have $2$ solutions and they are $x = (-1,-2)$ . So total solution $2 \times 2 = \boxed 4$

by hit & trial take the following values a) x=0 y=-1 b)x=-1 y=0 c)x=-2 y =0 c) x=0 y=-2 therefore there are only 4 conditions

xy = 0, so one of them must be 0, first we take x = 0: (x+y+1)(x+y+2) = 0 ==> (y+1)(y+2)=0 ==> Hence, y = -1 or y = -2. We get 2 pairs : (0,-1) and (0,-2). Then, we take y = 0 : (x+1)(x+2) = 0, Hence, x = -1 or x = -2 We also get 2 pairs : (-1,0) and (-2,0). So, the total pairs are 4 pairs.

Since xy = 0,

x = 0 or y = 0. For each of these cases the second equation reduces to a quadratic with two real roots.

Therefore there are four solutions to the system of equations.

xy=0

either x=0 or y=0

Suppose x=0 then (x+y+1)(x+y+2)=0 will be (y+1)(y+2)=0

hence y={-1,-2}

and we have the following order sets { (0,-1),(0,-2) }

Suppose y=0 then (x+y+1)(x+y+2)=0 will be (x+1)(x+2)=0

hence x={-1,-2}

and we have the following order sets { (-1,0),(-2,0) }

and hence there are 4 possible order sets of real numbers that satisfy the system

xy= 0 so x or y must be zero. now for (x+y+1)(x+y+2)=0 after putting 0 into x or y. it becomes. (x+1)(x+2)=0 and (y+1)(y+2)=0 so count how many x's and y' in these new equations. 4.

One of the parentheses below should equal to zero therefore possible sets are (0,-1)(-1,0)(0,-2)(-2,0)

4

xy = 0, then x = 0 or y = 0

(x + y +1)(x + y +2) = (x + y)^2 + 3(x + y) + 2

Solve equation for both x = 0 and y = 0.

x = 0, y = -1, -2 or y = 0, x = -1, -2

xy=0 ==> x=0 or y=0

if x=0 ==> (y+1)(y+2) = 0 == > either y=-1 or y=-2

if y=0 ==> (x+1)(x+2) = 0 == > either x=-1 or x=-2

4 pairs (0,-1),(0,-2),(-1,0),(-2,0)

xy=0; therefore, x or y must be equal to zero. In the bottom equation there are two numbers that are known, 1 and 2. Therefore, x or y must also equal to -1 or -2 to make one side equal to zero. According to this there are only four possible choices.(0,-1); (-1,0); (0,-2);(-2;0)

Na segunda equação chame x+y=p Multiplicando chega-se na equação do segundo grau: p^2+3p+2=0, que tem raízes -2 e -1. Como da primeira equação ao menos um dos números tem que ser zero, e lembrando que as raízes da equação são os possíveis resultados para x + y, os possíveis pares são: (0,-20);(-2,0);(0,-1);(-1,0). São 4 pares

if xy=0 it either x=0 , or y=0 substitute x= 0 to the 2nd equation, we get y= -1, or -2 substitute y= 0 to " " " , we get x=-1, or -2 thus we got 4 solutions

1. xy=0 , this means that either x or y should be equals to 0
2. (x+y+1)(x+y+2)=0 this means that either x+y+1= or x+y+2=0 or x+y=-1 or x+y=-2 but from 1 we know that either x or y must be 0 so the ordered pairs are (0,-1),(0,-2), (-1,0),(-2,0).

xy = 0 . So , x = 0 or y = 0 . If x = 0 , then ( y + 1 ) ( y + 2 ) = 0 . So , y = -1 or -2 . These give 2 sets ( 0,-1 ; 0,-2 ) . If y = 0 , then ( x + 1 ) ( x + 2 ) = 0 . So , x = -1 or -2 . These give 2 more sets ( -1,0 ; -2,0 ) . So , there are total of four sets ( -1,0 ; -2,0 ; 0,-1 ; 0,-2 )

1.xy = 0 implies that x or y is 0.
2.Suppose x = 0, this implies (y+1)(y+2)=0 which is a quadratic equation. 3.Suppose y = 0, this implies (x+1)(x+2)=0 which also is a quadratic equation. We know that a quadratic equation has 2 roots. Here two quadratic equations are present and hence 2+2=4 set of ordered pairs are possible. Following are four sets of possible pairs : 1. (-2,0) 2.(-1,0) 3.(0,-2) 4.(0,-1)

We have to solve equation as following one of the xy pair need to be 0 at any given moment due to the constraint xy=0 second of the pair needs to be -1 or -2 for second equation to be true. therefore (0;-1), (0-2) (-1;0) and (-2;0) are the only solutions

xy=0 then, x should be zero or y should be zero now,in second case, (x+1)(x+2)=0 as (y=0) therefore x=-1 or x=-2 therefore there are two solutions for x if y=0 similarly two solutions if x=0 total number of ways=2+2=4

From the second equation, x+y must equal -1 or -2 for (x+y+1)(x+y+2) to equal 0. From the first equation, we get that one of (x, y) must be a zero(They can't both be zeros, or the second equation would be false) For x+y=-1, we get (-1, 0) and (0, -1). Likewise, for x+y=-2, we get (0, -2) and (-2, 0)

If xy=0,either x=0 or y=0.If x=0,(x+y+1)(x+y+2)=(y+1)(y+2)=0.Thus y=-1 or y=-2.Similarly,if y=0,(x+y+1)(x+y+2)=(x+1)(x+2)=0.Thus,x=-1 or x=-2.Thus the possible ordered pairs are (0,-1),(0,-2),(-1,0),(-2,0).i.e,4 ordered pairs.

one of x and y must be 0.if x = 0 y =-1 or -2 , if y - 0 x = -1 or -2