Question

inequality graphs and systems of inequalities

1. solve for x.

$-2x + 3 < 4x - 3$

1. solve for x.

$-2 < -5x + 3$

1. $y < 3$

$y \leq -x + 1$

1. $y \geq x - 3$

$y \geq -x - 1$

1. $x \leq -3$

$y < -|x| + 1$

1. $y > |x + 2|$

$2y < -x + 4$

1. $y < |x - 5|$

$2x - y > 1$

1. $y < |x - 2|$

$y > 2x - 4$

1. $x < 3$

$y > |x| - 4$

Explanation:

Step1: Isolate x terms

$-2x - 4x < -3 - 3$

Step2: Simplify both sides

$-6x < -6$

Step3: Divide by -6, reverse inequality

$x > 1$

Step1: Isolate x terms

$-2 - 3 < -5x$

Step2: Simplify left side

$-5 < -5x$

Step3: Divide by -5, reverse inequality

$1 > x$ or $x < 1$

1. Graph $y=3$ as a dashed horizontal line, shade below it (for $y<3$).
2. Graph $y=-x+1$ as a solid line (slope -1, y-intercept 1), shade below it (for $y\leq -x+1$).
3. The solution is the overlapping shaded region.

1. Graph $y=x-3$ as a solid line (slope 1, y-intercept -3), shade above it (for $y\geq x-3$).
2. Graph $y=-x-1$ as a solid line (slope -1, y-intercept -1), shade above it (for $y\geq -x-1$).
3. The solution is the overlapping shaded region.

Answer:

$x > 1$

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