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# Linear equations and inequalities

1.Give 3 real-world examples where something is increasing at a constant rate or decreasing at a constant rate. The starting amount would be the y-intercept, "b", and the rate at which it increases would be the slope, "m". Give the equation of the lines for your examples using y = mx + b with your "m" and "b".

2. Why do we have to shade the region above the line for y > mx +b or y > mx + b, and shade below the line for y < mx + b or y < mx + b? Why can't the answer for these linear inequalities be simply the line y = mx + b? Why do we have to do all the shading?

#### Solution Preview

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1. Real-world example:
a. You have a job that pays \$8, and plus 1\$ for each hour worked. Then, the equation is y=x+8
Graph: see 1.jpg
m=1, b=8
The intercept is 8, so draw a point at y-axis.
The slope is 1, so draw a slope triangle with run = 1 and rise = 1. This bring you to Keep going, the next points are (2,10), (3,11), (4,12),...

b. You save your money on a box. You start with \$5. And save one more dollar per hour. Then, the equation is y=x+5
Graph:see 2.jpg
m=5, b=1
The intercept is 5, so draw a point at y-axis.
The slope is 1, so draw a slope triangle with run = 1 and rise = 1. This bring you to the second point (1, 6).
Keep going, the next ...

#### Solution Summary

This provides several examples of real-world examples modeled by linear equations, and explains the shading method for linear inequalities.

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