1. Linear Equations Unit

2.  Lines contain an infinite number of points.  These points are solutions to the equations that represent the lines.  To find a point on the line, choose any x-value since lines are infinite and find the corresponding y-value.

3.  The equations of lines contain both x and y as variables to represent the coordinates of the lines’ points.  A line may be represented by an equation.  The slope-intercept form of a line is y = mx+b where m represents the slope and b represents the y-intercept.

4.  The y-intercept is a point where the line crosses the y-axis. The b in the equation y=mx+b is the value of the y-coordinate in the ordered pair (x,y) of the y-intercept.  The x-intercept is a point where the line crosses the x-axis.  The slope is the rise divided by the run or the difference in the y-values divided by the difference in the x-values.

5.  Solve for b using y=mx+b where the constant rate of change is 4 and with P(-2,3) on the line.  Substitute.  3 = 4(-2)+b  Simplify.  3 = -8+b  Solve.  3+8=-8+8+b  11=b  Check

6.  Solve for x.  2x + 1 = 5.  Solve.  2x+1-1=5-1  2x=4  x=2

7.  Solve for m.  2m + 9 =.5(2m + 2)  Apply Distributive Property  2m+9=m+1  Take Variables to One Side  2m-m+9=m-m+1  m+9=1  Solve  m+9-9=1-9  m=-8  Check

8.  Solve for h.  2h + 2 = 2(h+4)  Apply Distributive Property  2h+2=2h+8  Take Variables to One Side  2h-2h+2=2h-2h+8  2=8  No Solution

9.  Solve for c.  2(2c + 2) = 4c + 4  Apply Distributive Property  4c+4=4c+4  Take Variables to One Side  4c-4c+4=4c-4c+4  4=4  Infinitely-Many Solutions

10.  Solve for a.  2a – 4 + 3a +16 – 12 = 3(2a+4) + 2  Apply the Distributive Property  2a – 4 + 3a +16 – 12 = 6a +12 + 2  Combine Like Terms  2a+3a-4+16-12=6a+12+2  5a=6a+14  Take Variables to One Side  5a-6a=6a-6a+14  -a=14  a=-14  Check.