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# MATH121, 3.3 Notes Math 121

OleMiss

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This 2 page Class Notes was uploaded by Mallory McClurg on Tuesday September 20, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 9 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/20/16

MATH121 CHAPTER 3 Lesson 3.3 – Forms of Linear Equations EXAMPLE 1. (-6, 8) and (-1, -4) (We’re asked to find the slope of a line that passes through the two points. We can use the formula (y – 2 /x1– 2 ) 1o find the slope.) (-4 – 8)/(-1 + 6) = -12/5 Slope = -12/5 EXAMPLE 2. (1, 6) and (1, -8) (We’re ask to find the slope of the a line that passes through the two points. Use the same formula as EXAMPLE 1.) (-8 – 6)/(1 – 1) = -14/0 (We know that a fraction with the denominator of zero is undefined, so our slope here is undefined!) Slope = undefined EXAMPLE 3. (-8, -8) and (-9, -8) (Again, find the slope. Use the same formula as EXAMPLE 1.) (-8 + 8)/(-9 + 8) = 0/1 (A fraction with zero as the numerator means the slope is zero!) Slope = 0 EXAMPLE 4. 3x + 3y = 9 (We’re asked to find the slope of this line, but first we need to put it in it’s proper linear “y = mx + b” form, so get y on a side by itself.) 3y = 3x + 9 (Divide both sides by 3.) y = x + 3 (Now, this is in proper linear form, so we can look at “m”, the coefficient in front of x… and know that is our slope. Note: the coefficient of “x” is 1.) Slope = 1 EXAMPLE 5. -x – 6y = 3 (Find the slope of this line by putting it in proper linear form.) -6y = x + 3 (Divide both sides by -6.) y = (-1/6)x – 2 (Remember, slope is the coefficient of x.) Slope = -1/6 EXAMPLE 6. 3y + 2 = (x/-2) (Find the slope of this line by putting it in proper linear form. First, multiply both sides by -2 to get ride of the fraction.) -6y – 4 = x (Now, get y on a side by itself.) -6y = x + 4 MATH121 CHAPTER 3 y = (-1/6)x – (2/3) Slope = -1/6 EXAMPLE 7. 7y = 1 (Find the slope of the line by first putting it in linear form.) 7y = 0x + 1 Slope = 0 EXAMPLE 8. -6x = 14 (Find the slope of the line by first putting it in linear form. We should realize that there is no y…so this is undefined.) Slope = undefined EXAMPLE 9. (-1, -10), slope = 4/3 (We’re asked to come up with the linear equation for a line that passes through this point and has a slope of 4/3. So first, we need to substitute our x and y values and our slope from the question into the y = mx + b equation.) -10 = (4/3)(-1) + b (Simplify.) -10 = (-4/3) + b (Add “-4/3” to both sides, to isolate the variable.) (-26/3) = b (Now, substitute our “b” and the given slope into the y = mx + b equation.) y = (4/3)x – (26/3) (This is our linear equation!) EXAMPLE 10. (-2, -11) and (6, 5) (We’re asked to write the linear equation of the line that passes through these two points. So first, we’ll use the “y 2 y 1 x 2 – x1” formula to find the slope, first.) (5 + 11)/(6 + 2) = (16/8) = 2 = slope (Now that we’ve found the slope, we can substitute the x and y values from one of our points to find the value of b in the “y = mx + b” equation.) 5 = 2 (6) + b (Simplify to get b on a side by itself.) b = -7 (So, now we have the slope and the value for b, so write it out as a linear equation!) y = 2x – 7 (This is our linear equation!) EXAMPLE 11. -5x + 4y = 34 (First, we need to get y on a side by itself, so we can get closer to the y = mx + b form.) 4y = 5x + 34 (Divide both sides by 4.) y = (5/4)x + (17/2) (The question asks us to find the value of y when x = 2, so substitute a 2 in for x and to solve for y.) (5/4)(2) + (17/2) = y = 11 (11 is the value for y in this equation when x = 2.)

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