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# Math121, Chapter 2.1 Notes Math 121

OleMiss

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This 4 page Class Notes was uploaded by Mallory McClurg on Monday September 5, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 108 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/05/16

MATH121 Chapter 2 Notes LESSON 2.1 – Linear Equations in One Variable Example 1. 6(5x - 5) = -31(3 - x) (Multiply 6 and 5x, and 6 and -5) (Multiply -31 and 3, and -31 and –x) 30x - 30 = -93 + 31x (Get similar values on same sides) -x = -63 (Divide by -1 to get x by itself) x = 63 (One Solution!) Example 2. 3.4x + 5 = 4.4x (Subtract 3.4x from 4.4x to get the similar values on the same side) 5 = x (One solution!) Example 3. (2z - 9)/5 – (1/10) = (-13z + 12)/10 (Since each is a fraction, you can easily see the similar denominators. To eliminate the denominators, multiply all the fractions by 10, which will double all values in the fraction with 5 as the denominator.) 4z – 38 – 1 = -13z + 12 (Combine like values.) 4z – 39 = -13z + 12 (Add 13z to 4z, and add 39 to 12.) 17z = 51 (Divide by 17 to isolate z.) z = 3 (One solution!) Example 4. (3/5)*(y – 2) – (23/5) = -5 (Since there are fractions present, we can again multiply all separate values by the largest denominator.) 3(y – 2) – 23 = -25 (Multiply y and -2 by 3.) 3y – 6 – 23 = -25 (Combine like values.) 3y – 29 = -25 (Add 29 to -25 to get y on its own side.) 3y = 4 (Divide 4 by 3 to isolate y.) y = (4/3) (One Solution!) Example 5. |-7y – 2| = 0 (Remove the absolute value signs, because we know the value in the signs would equal to a positive or negative real number.) -7y – 2 = 0 (Add 2 to both sides.) -7y = 2 (Divide both sides by -7.) y = (-2/7) (One Solution!) Example 6. |8x – 4 | = 7 (Since we know that whatever is in the absolute value signs can either be a positive number or a negative number, we must set this equal to 7 AND - 7 when we remove the signs.) 8x – 4 = 7 8x – 4 = -7 (Add 4 to both sides.) 8x = 11 8x = -3 (Divide 11 and -3 by 8.) x = (11/8) x = (-3/8) (Two solutions!) Example 7. |6y + 9| - 16 = 0 (Add 16 to both sides to put the equation in the same format as Example 6.) |6y + 9| = 16 (Eliminate the absolute value signs and set the equation equal to 16 AND -16.) 6y + 9 = 16 6y + 9 = -16 (Subtract 9 to both sides.) 6y = 7 6y = -25 (Divide 7 and -25 by 6.) y = (7/6) y = (-25/6) (Two solutions!) Example 8. |14x + 4| + 4 = 2 (Subtract 4 from both sides to put in format like previous 2 examples.) |14x +4| = -2 (Since we know absolute values mean that the value inside the bars is positive, it obviously cannot equal to a negative number, so there’s NO SOLUTION.) Example 9. |3x – 5| = |2x + 10| (Rewrite the equation twice, removing the absolute value signs from both sides, and then choosing ONE side of ONE equation and reverse the signs of the numbers.) 3x – 5 = 2x + 10 -3x + 5 = 2x + 10 (Get similar values on the same sides.) x = 15 -5x = 5 (Divide the 5 in second equation by -5, to isolate x.) x = 15 x = -1 (Two solutions!) Example 10. |y – 7| - |y - 41| = 0 (Rewrite the equation by adding the second absolute value to the other side of the equation, and remove the absolute value signs. Rewrite the equation again and reverse the signs of ONE side of ONE equation.) y – 7 = y – 41 -y + 7 = y – 41 (Get like values on the same side and solve each for y.) 0 = 34 -2y = -48 (Obviously the first equation isn’t true, so no solution for that one. On the second equation, divide both sides by -2.) NO SOLUTION y = 24 (One solution!)

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