algebra ii\n5.5 practice solving systems of inequalities\nsketch the solution to each system of…

algebra ii\n5.5 practice solving systems of inequalities\nsketch the solution to each system of inequalities.\n1) ( y geq-\frac{1}{2} x-2 )\n( y leq-\frac{1}{2} x+2 )\n2) ( y<-\frac{4}{3} x-1 )\n( y<-\frac{1}{3} x+2 )\n3) ( y<-2 x-3 )\n( y leq 2 x+1 )\n4) ( y leq-x-2 )\n( y leq-x+3 )

algebra ii\n5.5 practice solving systems of inequalities\nsketch the solution to each system of inequalities.\n1) ( y geq-\frac{1}{2} x-2 )\n( y leq-\frac{1}{2} x+2 )\n2) ( y<-\frac{4}{3} x-1 )\n( y<-\frac{1}{3} x+2 )\n3) ( y<-2 x-3 )\n( y leq 2 x+1 )\n4) ( y leq-x-2 )\n( y leq-x+3 )

Answer

Explanation:

Step1: Analyze the first inequality (y\geq-\frac{1}{2}x - 2)

The boundary line is (y =-\frac{1}{2}x - 2), which has a slope of (-\frac{1}{2}) and a (y) - intercept of (-2). Since the inequality is (y\geq-\frac{1}{2}x - 2), we shade above the line (including the line because of the "(\geq)" sign).

Step2: Analyze the second inequality (y\leq-\frac{1}{2}x + 2)

The boundary line is (y =-\frac{1}{2}x + 2), which has a slope of (-\frac{1}{2}) and a (y) - intercept of (2). Since the inequality is (y\leq-\frac{1}{2}x + 2), we shade below the line (including the line because of the "(\leq)" sign).

Step3: Find the intersection region

The solution of the system of inequalities is the region that is shaded for both inequalities. Since the two lines (y =-\frac{1}{2}x - 2) and (y =-\frac{1}{2}x + 2) are parallel (same slope (m =-\frac{1}{2})), the solution region is the region between the two parallel lines (including the lines).

Answer:

The solution of the system (y\geq-\frac{1}{2}x - 2) and (y\leq-\frac{1}{2}x + 2) is the region between the two parallel lines (y =-\frac{1}{2}x - 2) and (y =-\frac{1}{2}x + 2) (including the lines).