Which ordered pair makes both inequalities true? y < –x + 1 y > x which ordered pair makes both inequalities true? y < –x + 1 y > x (–3, 5) (–2, 2) (–1, –3) (0, –1) (–3, 5) (–2, 2) (–1, –3) (0, –1) mark this and return

  (-2, 2)

Step-by-step explanation:

You can graph the equations and points and see which points fall into the doubly-shaded area. The only one that does is (-2, 2).

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You can also evaluate the inequalities at the given points to see which might work. The inequality y > x is the simplest to evaluate, and it immediately eliminates the last two choices:

  -3 > -1 . . . not true

  -1 > 0 . . . not true

So, we can check the first two choices in the first inequality:

  -(-3) +1 > 5 . . . not true

  -(-2) +1 > 2 . . . TRUE

The ordered pair (-2, 2) makes both inequalities true.

2nd Option is correct.

Step-by-step explanation:

Given Inequalities are,

y < x + 1  (1)

y > x (2)

To find: Ordered pair which makes he inequalities true.

Pair 1:

x = -3 , y = 5

Inequality (1),

LHS = 5

RHS = -(-3) + 1 = 4

⇒ LHS > RHS

Thus, this pair does not satisfy the pair of inequalities.

Pair 2:

x = -2 , y = 2

Inequality (1),

LHS = 2

RHS = -(-2) + 1 = 3

⇒ LHS < RHS

Inequality (2),

LHS = 2

RHS = -2

⇒ LHS > RHS

Thus, this pair satisfies the pair of inequalities.

Pair 3:

x = -1 , y = -3

Inequality (1),

LHS = -3

RHS = -(-1) + 1 = 0

⇒ LHS < RHS

Inequality (2),

LHS = -3

RHS = -1

⇒ LHS < RHS

Thus, this pair does not satisfy the pair of inequalities.

Pair 4:

x = 0 , y = -1

Inequality (1),

LHS = -1

RHS = -(0) + 1 = 1

⇒ LHS < RHS

Inequality (2),

LHS = -1

RHS = 0

⇒ LHS < RHS

Thus, this pair does not satisfy the pair of inequalities.

Therefore, 2nd Option is correct.

b.my answer is b.

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other answer is correct. i had used the radius of the radius somehow. a website: