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Linear Equations Unit

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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.

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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.

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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.

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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

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

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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

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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

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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

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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.

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