WEBVTT
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Find the solution set of the equation the modulus of π₯ plus three is equal to negative three π₯ plus seven.
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We could solve this equation graphically or algebraically.
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Weβre first gonna look at the graphical solution.
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The graph of the modulus of π₯ plus three intersects the π¦-axis at positive three and touches the π₯-axis at negative three as shown in the diagram.
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The line negative three π₯ plus seven intersects the π¦-axis at seven and the π₯-axis at positive seven-thirds.
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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.
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As this occurs in the first quadrant, where π₯ and π¦ are positive, we can solve the equation π₯ plus three equals negative three π₯ plus seven.
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Adding three π₯ to both sides of our equation gives us four π₯ plus three equals seven.
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We can then subtract three from both sides of the equation.
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This gives us four π₯ is equal to four.
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Finally, dividing by four gives us our answer or solution π₯ equal to one.
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Therefore, the solution set contains the single value π₯ equals one.
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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.
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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.
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Whilst this is not a solution in this question, it is worth showing how we could find it algebraically.
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The dotted line is an extension of the equation negative π₯ plus three.
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Therefore, weβre going to solve negative π₯ plus three is equal to negative three π₯ plus seven.
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Simplifying the left-hand side gives us negative π₯ minus three.
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Using our balancing method β to add three π₯ to both sides of the equation β gives us two π₯ minus three equals seven.
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Adding three to both sides of this equation gives us two π₯ equals 10.
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And dividing by two tells us that the two lines intersect at the point π₯ equals five.
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To prove that this is not a solution, we substitute π₯ equals five back into the initial equation.
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The modulus of five plus three is equal to negative three multiplied by five plus seven.
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Well, the modulus of eight is equal to eight.
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And negative three multiplied by five plus seven is equal to negative eight.
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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.