Question

Which ordered pairs are in the solution set of the system of linear inequalities?

y > Negative one-halfx

y < One-halfx + 1

On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (0, 0) and (4, negative 2). Everything above the line is shaded. The second dashed line has a positive slope and goes through (negative 2, 0) and (2, 2). Everything below the line is shaded.
(5, –2), (3, 1), (–4, 2)
(5, –2), (3, –1), (4, –3)
(5, –2), (3, 1), (4, 2)
(5, –2), (–3, 1), (4, 2)

Answer 1

After evaluating each ordered pair, (5, –2) and (3, 1) satisfy both inequalities and are part of the solution set, but (–4, 2) does not and is excluded.

Step-by-step explanation:

We need to determine which ordered pairs satisfy both inequalities
`y > -\((1)/(2)\)x and y < \((1)/(2)\)x + 1.` This means we are looking for points that are above the line with a negative slope (a) and below the line with a positive slope (b).

To check each point, substitute the x and y values into both inequalities:

• (5, –2):
  `y > -\((1)/(2)\)x` becomes
  `–2 > -\((1)/(2)\)(5)`which simplifies to
  `–2 > -\((5)/(2)\`), which is true. y < \(\frac{1}{2}\)x + 1 becomes –2 < \(\frac{1}{2}\)(5) + 1, which simplifies to –2 < 3.5, which is also true. So, (5, –2) is a solution.
• 
  `(3, 1): y > -\((1)/(2)\)x becomes 1 > -\((1)/(2)\)(3) which simplifies to 1 > -\((3)/(2)\), which is true. y < \((1)/(2)\)x + 1 becomes 1 < \((1)/(2)\)(3) + 1, which simplifies to 1 < 2.5, which is also true. So, (3, 1) is a solution.`
• 
  `(–4, 2): y > -\((1)/(2)\)x`becomes
  `2 > -\((1)/(2)\)(–4)` which simplifies to 2 > 2, which is not true. So, (–4, 2) is not a solution.

Therefore, the ordered pair (5, –2) and (3, 1) are in the solution set of the system, but (–4, 2) is not.