Question

svlc algebra 1a - standard (15260)
solving systems of linear inequalities
mr. hernandez plotted the point (1, 1) on han’s graph of ( y leq \frac{1}{2}x + 2 ). he instructed han to add a second inequality to the graph that would include the solution (1, 1).
which inequality could miguel write?
( y geq 2x + 1 )
( y leq 2x - 1 )
( y < 2x - 1 )
( y > 2x + 1 )

Explanation:

Step1: Substitute (1,1) into each inequality

For \( y \geq 2x + 1 \): Substitute \( x = 1 \), \( y = 1 \). We get \( 1 \geq 2(1)+1 = 3 \). \( 1 \geq 3 \) is false.

Step2: Substitute into \( y \leq 2x - 1 \)

Substitute \( x = 1 \), \( y = 1 \). \( 1 \leq 2(1)-1 = 1 \). \( 1 \leq 1 \) is true.

Step3: Substitute into \( y < 2x - 1 \)

Substitute \( x = 1 \), \( y = 1 \). \( 1 < 2(1)-1 = 1 \). \( 1 < 1 \) is false.

Step4: Substitute into \( y > 2x + 1 \)

Substitute \( x = 1 \), \( y = 1 \). \( 1 > 2(1)+1 = 3 \). \( 1 > 3 \) is false.

Answer:

B. \( y \leq 2x - 1 \)