Question

svlc algebra 1a - standard (15260)
graphing two - variable linear inequalities
which points are solutions to the linear inequality (y < 0.5x + 2)? choose three correct answers.
((-1,-2)) ((-3,-2))
((1,-2)) ((-2,1))
((-1,2))

Explanation:

To determine which points satisfy the inequality \( y < 0.5x + 2 \), we substitute the \( x \)- and \( y \)-values of each point into the inequality and check if the statement is true.

Step 1: Test Point \((-1, -2)\)

Substitute \( x = -1 \) and \( y = -2 \) into \( y < 0.5x + 2 \):
\( -2 < 0.5(-1) + 2 \)
\( -2 < -0.5 + 2 \)
\( -2 < 1.5 \)
This is true.

Step 2: Test Point \((-3, -2)\)

Substitute \( x = -3 \) and \( y = -2 \) into \( y < 0.5x + 2 \):
\( -2 < 0.5(-3) + 2 \)
\( -2 < -1.5 + 2 \)
\( -2 < 0.5 \)
This is true.

Step 3: Test Point \((1, -2)\)

Substitute \( x = 1 \) and \( y = -2 \) into \( y < 0.5x + 2 \):
\( -2 < 0.5(1) + 2 \)
\( -2 < 0.5 + 2 \)
\( -2 < 2.5 \)
This is true.

Step 4: Test Point \((-2, 1)\)

Substitute \( x = -2 \) and \( y = 1 \) into \( y < 0.5x + 2 \):
\( 1 < 0.5(-2) + 2 \)
\( 1 < -1 + 2 \)
\( 1 < 1 \)
This is false (since \( 1 \) is not less than \( 1 \)).

Step 5: Test Point \((-1, 2)\)

Substitute \( x = -1 \) and \( y = 2 \) into \( y < 0.5x + 2 \):
\( 2 < 0.5(-1) + 2 \)
\( 2 < -0.5 + 2 \)
\( 2 < 1.5 \)
This is false.

Answer:

The points that satisfy \( y < 0.5x + 2 \) are:
\((-1, -2)\), \((-3, -2)\), and \((1, -2)\).