Equations with absolute values in practice

Hello, and welcome back to a step in practice. We are dealing today with equations that involve absolute values. In exercise 1, the equation is absolute value of 3x minus 7 equals to minus 8. Well, it is clear at the first glance that this equation has no solutions. Why? Because the absolute value of something is greater or equal than 0, whereas minus 8 is strictly negative. And so the set of solutions is the empty set.

Alternatively, one could have used the method that Francis showed us in the last step. More precisely, the method is that to consider two different cases, depending on the sign of 3x minus 7.

So alternatively, if 3x minus 7 is greater or equal than 0, that is x greater or equal than 7/3, the equation is equivalent to 3x minus 7 equal to minus 8, that is x equal to minus 1/3.

However, minus 1/3– well, does not belong to the interval 7/3 plus infinity. So the first set of solutions is the empty set. The second case is 3x minus 7, strictly less than 0, that is, x strictly less than 7/3. Well in that case the equation is equivalent to minus 3x minus 7 equals 2 minus 8, which is minus 3x equals to minus 15, which gives x equal 5. However, 5– well, it’s bigger than 7/3, and so the second set of solutions is the empty set. And the conclusion, again is that the set of solutions which is the union of the two sets of solutions, considering the two cases, is again the empty set.

So two different methods to find the same conclusion.

In exercise 2, we are dealing with two absolute values.

Well, the equation is absolute value of 2x minus 4 minus absolute value of x minus 1 equal 3.

We suggest to consider different cases, depending on the sign of 2x minus 4 and x minus 1. Now, x minus 1 vanishes when x equals 1, whereas 2x minus 4 vanishes when x is equal to 2. So let us consider a table with the values 1 and 2. And here, let us put the sign of x minus 1 and the sign of 2x minus 4.

Now, 2x minus 4 vanishes in 2, whereas x minus 1 vanishes in 1. Here we have the positive sign, negative sign, here we have a negative sign, and then a positive sign, Now we consider these three different cases. If x is less than 1, then both terms are negative. So the equation is equivalent to minus 2x minus 4, minus and then again minus x minus 1 equal to 3. And this is minus 2x plus 4 plus x minus 1 equal to 3, and this is again is equivalent to minus x plus 3 equal to 3, which gives x equal to 0. 0 less than 1. So we get a set of solution 0.

Now, the second case is x greater than 1 and less than 2.

Well, in that case, x minus 1 is positive, 2x minus 4 is negative, so the equation is equivalent to minus 2x minus 4, which is the absolute value of 2x minus 4, and x minus 1 is positive. So minus x minus 1 equal to 3, and this is equivalent to minus 3x plus 4 plus 1 to this plus 5 equal to 3, which gives 3x equals to 2, and then x equal to 2/3. Now, 2/3 is less than 1, and thus we get a second set of solutions which is the empty set.

Finally, the third case, x greater or equal than 2. Well, every term is positive. So the equation is equivalent to 2x minus 4 minus x minus 1 equal to 3. But this is also x minus 4 plus 1, so minus 3 equal to 3, and this is equivalent to x equal to 6. Well, let us look at the compatibility– 6 is greater than 2, and thus we get 6 as a third set of solutions. Now, the set of solutions to our equality is the union of the three sets that we found in each of the different cases.

And so we get 0 and 6, and this is the solution of exercise 2.

In exercise 3, we consider the equation absolute value of x minus 1 equals to minus 4 plus 2x. So we can split this equality into two equations by following what Francis showed us. So in case where x minus 1 is greater or equal than 0, that is x greater or equal than 1, we get x minus 1 equal to minus 4 plus 2x, that is, x equals minus 1 plus 4, 3. So 3 is greater or equal than 1. And so we have a first solution, which is 3. Let us check– 3 minus 1 is 2. The absolute value of 2 is 2, and minus 4 plus 6 is 2. So we get the equality.

If x minus 1 is less than 0, that is, x is less than 1, and we take care of this condition, the equality is equivalent to minus x minus 1 equals to minus 4 plus 2x, which is 3x equal to 1 plus 4, 5, that is x equal to 5/3. However, 5/3 is greater than 1, so we have to exclude the solution. And the second set of solutions is the empty set. So finally, the set of solutions is just S1 that is the singleton 3. And this ends this step in practice. And in the next step in practice, you meet again Alberto. Be careful. He is a little bit too a fan of algebra. See you next week. Bye.