How do you graph the system #x + 3 2x + 1# and #-4x

Question

How do you graph the system #x + 3 > 2x + 1# and #-4x < -8#?

Eva Adamson · Accepted Answer

Simplify x+3> 2x+1, as 2>x and -4x<-8 as x>2.  Now graph the line x=2. 2>x will represent the region to the left of x=2 , and x>2 will represent the region to the right of x=2, which may be shaded differently for distinguishing the two.

Genesis Allen · Answer

undefined To graph the system of inequalities $x + 3 > 2x + 1$ and $-4x < -8$, you would first graph the boundary lines for each inequality. Then, determine the region that satisfies both inequalities. For $x + 3 > 2x + 1$, you would graph the line $x + 3 = 2x + 1$, which simplifies to $x = 2$. Since the inequality is a strict inequality, the line will be dashed. Then, shade the region to the right of the line since $x$ values greater than 2 satisfy the inequality. For $-4x < -8$, you would graph the line $-4x = -8$, which simplifies to $x = 2$. Again, since the inequality is strict, the line will be dashed. Then, shade the region to the left of the line since $x$ values less than 2 satisfy the inequality. The solution to the system is the overlapping region shaded in between the two boundary lines.

How do you graph the system #x + 3 &gt; 2x + 1# and #-4x &lt; -8#?

How do you graph the system #x + 3 > 2x + 1# and #-4x < -8#?