Video: Solving Linear Equations Involving Absolute Values

Find the solution set of the equation |π‘₯ + 3| = βˆ’3π‘₯ + 7.

03:17

Video Transcript

Find the solution set of the equation the modulus of π‘₯ plus three is equal to negative three π‘₯ plus seven.

We could solve this equation graphically or algebraically. We’re first gonna look at the graphical solution. The graph of the modulus of π‘₯ plus three intersects the 𝑦-axis at positive three and touches the π‘₯-axis at negative three as shown in the diagram. The line negative three π‘₯ plus seven intersects the 𝑦-axis at seven and the π‘₯-axis at positive seven-thirds.

As the two equations intersect once, we know that there will be one solution to the equation modulus π‘₯ plus three equals negative three π‘₯ plus seven. As this occurs in the first quadrant, where π‘₯ and 𝑦 are positive, we can solve the equation π‘₯ plus three equals negative three π‘₯ plus seven. Adding three π‘₯ to both sides of our equation gives us four π‘₯ plus three equals seven. We can then subtract three from both sides of the equation. This gives us four π‘₯ is equal to four.

Finally, dividing by four gives us our answer or solution π‘₯ equal to one. Therefore, the solution set contains the single value π‘₯ equals one. Whilst the only solution to this question is π‘₯ equal to one, it is also worth noting that there is a spurious solution that we might have otherwise come across. Extending the modulus of π‘₯ plus three below the π‘₯-axis, it is clear that it would intersect the line negative three π‘₯ plus seven again.

Whilst this is not a solution in this question, it is worth showing how we could find it algebraically. The dotted line is an extension of the equation negative π‘₯ plus three. Therefore, we’re going to solve negative π‘₯ plus three is equal to negative three π‘₯ plus seven. Simplifying the left-hand side gives us negative π‘₯ minus three. Using our balancing method β€” to add three π‘₯ to both sides of the equation β€” gives us two π‘₯ minus three equals seven. Adding three to both sides of this equation gives us two π‘₯ equals 10. And dividing by two tells us that the two lines intersect at the point π‘₯ equals five.

To prove that this is not a solution, we substitute π‘₯ equals five back into the initial equation. The modulus of five plus three is equal to negative three multiplied by five plus seven. Well, the modulus of eight is equal to eight. And negative three multiplied by five plus seven is equal to negative eight. As these two numbers are not equal, we can say that π‘₯ equals five is not a solution of the equation modulus π‘₯ plus three equals negative three π‘₯ plus seven.

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