Question

svlc algebra 1a - standard (15260)
solving systems of linear inequalities
describing a system of two - variable inequalities
which statements are true about the graph of ( yleq3x + 1 ) and ( ygeq - x + 2 )? choose three correct answers.

• both inequalities are shaded below the boundary lines.
• both boundary lines are solid.
• a solution to the system is ( (1,3) ).
• the boundary lines intersect.
• the slope of one boundary line is 2.

Explanation:

1. Both inequalities are shaded below the boundary lines: For \( y \leq 3x + 1 \), we shade below the line. For \( y \geq -x + 2 \), we shade above the line. So this statement is false.
2. Both boundary lines are solid: The inequalities are \( y \leq 3x + 1 \) and \( y \geq -x + 2 \). In both cases, the inequality is "less than or equal to" or "greater than or equal to", so the boundary lines are solid (since equality is included). This statement is true.
3. A solution to the system is \((1, 3)\): Check if \((1, 3)\) satisfies both inequalities. For \( y \leq 3x + 1 \): \( 3 \leq 3(1) + 1 = 4 \), which is true. For \( y \geq -x + 2 \): \( 3 \geq -1 + 2 = 1 \), which is true. So \((1, 3)\) is a solution. This statement is true.
4. The boundary lines intersect: The first line is \( y = 3x + 1 \) (slope 3, y - intercept 1) and the second is \( y = -x + 2 \) (slope - 1, y - intercept 2). Since the slopes are different, the lines are not parallel and will intersect. This statement is true.
5. The slope of one boundary line is 2: The slopes are 3 (for \( y = 3x + 1 \)) and - 1 (for \( y=-x + 2\)). So this statement is false.

Answer:

B. Both boundary lines are solid
C. A solution to the system is \((1, 3)\)
D. The boundary lines intersect