Question Video: Solving Absolute Value Linear Equations | Nagwa Question Video: Solving Absolute Value Linear Equations | Nagwa

Question Video: Solving Absolute Value Linear Equations Mathematics • Second Year of Secondary School

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Find algebraically the solution set of the equation |π‘₯ βˆ’ 3| βˆ’ |π‘₯ + 1| = 4.

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Video Transcript

Find algebraically the solution set of the equation the absolute value of π‘₯ minus three minus the absolute value of π‘₯ plus one is equal to four.

In this question, we’re given an equation involving the absolute value of some linear functions. We need to find the solution set to this equation algebraically. We can start by recalling the solution set of an equation is just the set of all values of π‘₯ which make both sides of the equation equal. And there are a few different ways we can solve this. For example, we could sketch the graph of the left-hand side of this equation and then check when the output values of this function are four. However, this is a graphical solution, and the question wants us to do this algebraically. So instead, we’ll try and solve this algebraically.

We recall the absolute value of a number is the magnitude of the number. In other words, it’s the value of the number where we ignore its sign. Therefore, when our inner function changes sign, the definition of the absolute value will change. Therefore, when π‘₯ is greater than or equal to three, taking the absolute value of π‘₯ minus three will just give us π‘₯ minus three because this is a positive number. However, if π‘₯ is less than three, the absolute value of π‘₯ minus three is taking the absolute value of a negative number. So we would need to multiply this by negative one. And we get a very similar story for the absolute value of π‘₯ plus one. If π‘₯ is greater than or equal to negative one, then π‘₯ plus one is nonnegative. So its absolute value is just equal to itself, π‘₯ plus one.

However, if π‘₯ is less than negative one, π‘₯ plus one is negative. So we need to multiply this value by negative one. And this gives us useful information. Depending on which of the inequalities our input value of π‘₯ solves, the equation will change. We can solve this equation by just considering all of the possible values of π‘₯ in turn. Let’s start with π‘₯ being greater than or equal to three. We note that when π‘₯ is greater than or equal to three, of course, it can’t be less than three and it’s always greater than or equal to negative one. So π‘₯ can’t be less than negative one. And this is going to allow us to rewrite our equation in a way we can solve it.

First, when π‘₯ is greater than or equal to three, π‘₯ minus three is nonnegative. So the absolute value of π‘₯ minus three is just equal to π‘₯ minus three. The same is true for π‘₯ plus one. When π‘₯ is greater than or equal to three, π‘₯ plus one is positive. So the absolute value of π‘₯ plus one is just equal to π‘₯ plus one. Therefore, our equation simplifies to give us π‘₯ minus three minus π‘₯ plus one is equal to four.

We can then simplify the left-hand side of this equation. We’ll distribute the negative over our parentheses, giving us π‘₯ minus three minus π‘₯ minus one. And of course, π‘₯ minus π‘₯ is equal to zero. So we just have negative four on the left-hand side of the equation. However, the right-hand side of this equation is four. Therefore, no value of π‘₯ greater than or equal to three solves the equation.

But we’re not done yet. We also need to check all of the values of π‘₯ less than three. And when we do this, we can see there’s two options. Our values of π‘₯ can be greater than or equal to negative one or less than negative one. The equation is going to change depending on which of these inequalities our values of π‘₯ satisfy. So we’ll do these separately.

Let’s start with π‘₯ being less than three and π‘₯ is greater than or equal to negative one. We can now use these inequalities to rewrite our equation. First, since π‘₯ is less than three, π‘₯ minus three is negative. So the absolute value of π‘₯ minus three will be negative one times π‘₯ minus three. However, since our values of π‘₯ are greater than or equal to negative one, π‘₯ plus one is nonnegative. So the absolute value of π‘₯ plus one is equal to π‘₯ plus one. Therefore, we can rewrite our equation as negative one times π‘₯ minus three minus π‘₯ plus one is equal to four.

We can distribute the negatives over the left-hand side of our equation and simplify. We get negative two π‘₯ plus two is equal to four. And we can then solve this equation for π‘₯. We subtract two from both sides and divide through by negative two. We get π‘₯ is equal to negative one. And we do need to check that this value of π‘₯ satisfies both of our original inequalities. We know that negative one is greater than or equal to negative one, and negative one is less than three. Therefore, π‘₯ is equal to negative one is a solution to this equation. And we could justify this by substituting it back into our original equation.

But we’re not done yet. We still need to check the values of π‘₯ less than three and less than negative one. And, of course, if π‘₯ is less than negative one, then it’s automatically less than three. So we only need the inequality π‘₯ is less than negative one. We’re going to rewrite our equation in the same way. This time, since our value of π‘₯ is less than one, π‘₯ minus three is negative and π‘₯ plus one is also negative. So when we take the absolute value of these expressions, we’re going to need to multiply them by negative one. Therefore, the left-hand side of this equation simplifies to give us negative one times π‘₯ minus three minus negative one times π‘₯ plus one.

We can continue to simplify the left-hand side of this equation. We can distribute the negative over the parentheses. We get negative π‘₯ plus three plus π‘₯ plus one, which we can then simplify further. Negative π‘₯ plus π‘₯ is equal to zero and three plus one is four. So the left-hand side of this equation is the constant value of four. And we can see this is exactly equal to the right-hand side of this equation. In other words, if π‘₯ is less than negative one, the equation is always satisfied. Therefore, all values of π‘₯ less than negative one are solutions to the equation. Therefore, we found all the solutions to the equation, π‘₯ is negative one or π‘₯ is less than negative one.

We can combine these into a single inequality as π‘₯ is less than or equal to negative one. However, the question wants us to write this as a solution set. And the set of all values less than or equal to negative one can be written as the left-open, right-closed interval from negative ∞ to negative one. Therefore, we were able to show the solution set of the equation the absolute value of π‘₯ minus three minus the absolute value of π‘₯ plus one is equal to four is given by the left-open, right-closed interval from negative ∞ to negative one.

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