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Presenting Solving Absolute Value Equations

Solving absolute value equations

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PresentingSolving

Absolute Value Equations

Solving Absolute Value Equations

Howdy! I’m hoping you’ll help me

escape these bars!

An absolute value equation is one with a

variable trapped between the bars!

What does it mean to be

between bars?

It represents distance from a given value!

|x - 3| = 2The distance between a number and 3 is 2!

Check out these 2 examples! What is happening in each?

Example 1a|-4 + 5x| = 16

-4 + 5x = 16 or -4 + 5x = -16

5x = 16+ 4 5x = -16+45x = 20 5x = -12 x = 20/5 x = -12/5 x = 4 Solutions: {-12/5, 4}

Example 1b|1 - 5a| = 29

1 - 5a = 29 or 1 - 5a = -29

-5a = 29 - 1 -5a = -29 - 1-5a = 28 -5a = -30 a = -28/5 a = -30/-5 a = 6Solutions: {-28/5, 6}

Example 1a|-4 + 5x| = 16

-4 + 5x = 16 or -4 + 5x = -16

5x = 16+ 4 5x = -16+45x = 20 5x = -12 x = 20/5 x = -12/5 x = 4 Solutions: {-12/5, 4}

Example 1b|1 - 5a| = 29

1 - 5a = 29 or 1 - 5a = -29

-5a = 29 - 1 -5a = -29 - 1-5a = 28 -5a = -30 a = -28/5 a = -30/-5 a = 6Solutions: {-28/5, 6}

What do you notice? What steps occur in both examples?

Now it is your turn!Solve for x: |-2x - 1| = 11

Let’s look 2 more examples? What do you

notice?

Example 2a-5|3 + 4k| = -115-5(3 + 4k) = -115 or -5(3 + 4k) = 1153 + 4k = -115/-5 3 + 4k = 115/-53 + 4k = 23 3 + 4k = -23 4k = 23 - 3 4k = -23 - 3 4k = 20 4k = - 26 k = 5 k = -26/4 k = -13/2Solutions: {-13/2, 5}

Example 2b-8|3 - 8k| = 40-8|3 - 8k| = 40 or -8|3 - 8k| = -40 |3 - 8k| = -5 3 - 8k = 5 not possible -8k = 2 k = -2/8 k = -1/4 doesn’t workNo solution because one solution is extraneous and the other not possible.

What do you notice? What are key ideas you need in your notes?

Example 2a-5|3 + 4k| = -115-5(3 + 4k) = -115 or -5(3 + 4k) = 1153 + 4k = -115/-5 3 + 4k = 115/-53 + 4k = 23 3 + 4k = -23 4k = 23 - 3 4k = -23 - 3 4k = 20 4k = - 26 k = 5 k = -26/4 k = -13/2Solutions: {-13/2, 5}

Example 2b-8|3 - 8k| = 40-8|3 - 8k| = 40 or -8|3 - 8k| = -40 |3 - 8k| = -5 3 - 8k = 5 not possible -8k = 2 k = -2/8 k = -1/4 doesn’t workNo solution because one solution is extraneous and the other not possible.

Now it is your turn!Solve for x: 10|7x + 3| = 0

Last 2 examples! Again - what do you notice?

Example 3a3|3 - 5x| - 3 = 183(3 - 5x) = 21 or 3(3 - 5x) = -21 3 - 5x = 7 3 - 5x = -7 -5x = 4 -5x = -10 x = -4/5 x = 2

Solutions both check {-⅘, 2}

Example 3b5|9 - 5x| - 7 = 385(9 - 5x) = 45 or 5(9 - 5x) = -45 9 - 5x = 9 9 - 5x = -9 -5x = 0 -5x = -18 x = 0 x = 18/5

Solutions both check {0, 18/5}

What do you notice? What are key ideas you need to add to your notes?

Example 3a3|3 - 5x| - 3 = 183(3 - 5x) = 21 or 3(3 - 5x) = -21 3 - 5x = 7 3 - 5x = -7 -5x = 4 -5x = -10 x = -4/5 x = 2

Solutions both check {-⅘, 2}

Example 3b5|9 - 5x| - 7 = 385(9 - 5x) = 45 or 5(9 - 5x) = -45 9 - 5x = 9 9 - 5x = -9 -5x = 0 -5x = -18 x = 0 x = 18/5

Solutions both check {0, 18/5}

Now it is your turn! Three to move on!

1) Solve for x: |-5x| + 4 = -11

2) Solve for x: 3 - |8x - 6| = 3

3) Solve for x: 6|1 - 5x| - 9 = 57

{No solution}

{¾}

{-2, 12/5}

That’s a Wrap for Solving

Absolute Value Equations