Absolute Value Equations – Example

Senin, 17 Februari 2020

Example 1 Solve each of the following.

(a)     | 2x – 5 | = 9  

(b)    | 1 – 3t | = 20

(c)     | 5y – 8 | = 1  

Solution

Now, remember that absolute value does not just make all minus signs into plus signs! To solve these we’ve got to use the formula above since in all cases the number on the right side of the equal sign is positive.

(a)   | 2x – 5 | = 9  

There really isn’t much to do here other than using the formula from above as noted above. All we need to note is that in the formula above p represents whatever is on the inside of the absolute value bars and so in this case we have,

2x – 5 = – 9                 or                     2x – 5 = 9

At this point we’ve got two linear equations that are easy to solve.

2x = – 4                         or                   2x = 14

x = – 2                         or                     x = 7

So, we’ve got two solutions to the equation x = – 2 and x = 7.

(b)    | 1 – 3t | = 20

This one is pretty much the same as the previous part so we won’t put as much detail into this one.

1 – 3t = – 20                  or                   1 – 3t = 20  

   – 3t = – 21                  or                      – 3t = 19  

       t = 7                        or                      

The two solutions to this equation areand t = 7.

(c)     | 5y – 8 | = 1  

Again, not much more to this one.

                                    5y – 8 = – 1                             or                                 5y – 8 = 1

                                         5y  = 7                                or                                       5y = 9

                                                                  or                                    

In this case the two solutions areand.

Example 2 Solve each of the following.

(a)      | 10x – 3 | = 0

(b)      | 5x + 9 | = – 3

Solution

(a)  Let’s approach this one from a geometric standpoint. This is saying that the quantity in the absolute value bars has a distance of zero from the origin. There is only one number that has the property and that is zero itself. So, we must have,

                                    10x – 3 = 0                              ⇒                         

In this case we get a single solution.

(b)  Now, in this case let’s recall that we noted at the start of this section that |p| ≥ 0. In other words, we can’t get a negative value out of the absolute value. That is exactly what this equation is saying however. Since this isn’t possible that means there is no solution to this equation.

Example 3 Solve each of the following.

(a)    | x – 2 | = 3x + 1

(b)    | 4x +  3 | = 3 – x

(c)     | 2x – 1 | = | x + 3 |

Solution

At first glance the formula we used above will do us no good here. It requires the right side of the equation to be a positive number. It turns out that we can still use it here, but we’re going to have to be careful with the answers as using this formula will, on occasion introduce an incorrect answer. So, while we can use the formula we’ll need to make sure we check our solutions to see if they really work.

(a)  | x – 2 | = 3x + 1

So, we’ll start off using the formula above as we have in the previous problems and solving the two linear equations.

                        x – 2 = – (3x + 1) = – 3x – 1               or                     x – 2 = 3x + 1

                            4x = 1                                              or                     – 2x = 3

                                                                  or                      

Okay, we’ve got two potential answers here. There is a problem with the second one however. If we plug this one into the equation we get,

We get the same number on each side but with opposite signs. This will happen on occasion when we solve this kind of equation with absolute values. Note that we really didn’t need to plug the solution into the whole equation here. All we needed to do was check the portion without the absolute value and if it was negative then the potential solution will NOT in fact be a solution and if it’s positive or zero it will be solution.

We’ll leave it to you to verify that the first potential solution does in fact work and so there is a single solution to this equation : x = ¼  and notice that this is less than 2 (as our assumption required) and so is a solution to the equation with the absolute value in it.

So, all together there is a single solution to this equation: x = ¼ .

(b)   | 4x + 3 | = 3 – x

This one will work in pretty much the same way so we won’t put in quite as much explanation.

                        4x + 3 = – (3 – x) = – 3 + x                 or                     4x + 3 = 3 – x

                              3x = – 6                                         or                           5x = 0

                                x = – 2                                         or                             x = 0

Now, before we check each of these we should give a quick warning. Do not make the assumption that because the first potential solution is negative it won’t be a solution. We only exclude a potential solution if it makes the portion without absolute value bars negative. In this case both potential solutions will make the portion without absolute value bars positive and so both are in fact solutions. So in this case, unlike the first example, we get two solutions: x = – 2 and x = 0.

(c)   | 2x – 1 | = | 4x + 9 |

This case looks very different from any of the previous problems we’ve worked to this point and in this case the formula we’ve been using doesn’t really work at all. However, if we think about this a little we can see that we’ll still do something similar here to get a solution.

Both sides of the equation have contain absolute values and so the only way the two sides are equal will be if the two quantities inside the absolute value bars are equal or equal but with opposite signs. Or in other words, we must have,

                        2x – 1 = – (4x + 9) = – 4x – 9             or                     2x – 1 = 4x + 9

                              6x = – 8                                         or                        – 2x = 10

                                                      or                            x = – 5

Now, we won’t need to verify our solutions here as we did in the previous two parts of this problem. Both with be solutions provided we solved the two equations correctly. However, it will probably be a good idea to verify them anyway just to show that the solution technique we used here really did work properly.

Let’s first check.

In the case the quantities inside the absolute value were the same number but opposite signs. However, upon taking the absolute value we got the same number and so is a solution. Now, let’s check x = – 5.

In the case we got the same value inside the absolute value bars.

So, as suggested above both answers did in fact work and both are solutions to the equation.



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Labels: Mathematician

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