Solve the inequality |x-3| > 5
(1) by algebra
(2) by sketching a graph
(1) If we remove the modulus (absolute value) signs, x-3 may be positive or it may be negative.
We need to solve for each possibility.
If x-3 is negative: x-3 < -5
x < -5 + 3
x < -2
If x-3 is positive: x-3 > 5
x > 5 + 3
x > 8
Combining both gives Solution: -2 > x and x > 8
(2) Graphs of the modulus of linear functions are “V” shaped.
The vertex of the V will be on the x-axis where the value of x makes the function equal to
zero. In this case at x = 3.
Draw a line given by y = the number the function is compared to. In this case y = 5.
Pick a value for x to the right of the vertex and compute the value of the function so that you
have another point that allows you draw the blue line going up to the right. The blue line
going up to the left will be a reflection of this in a vertical line through the vertex.
You can now read off the graph what values of x cause the graph to be under, on or above the