- Think of absolute value as a distance on the number line. - How many points are a positive distance from 0, and on which sides of 0 can they lie?

The absolute value \(\lvert x\rvert\) is the distance from \(x\) to 0. A number that is 6 units from 0 can lie on either side of 0, so \(x=6\) or \(x=-6\).

\(x=-6\) or \(x=6\). The solutions are the two numbers 6 units from 0.

- Sketch the three numbers in order on a line, with equal spacing between whole numbers. - Moving left and moving right can cover the same distance.

1. \(|1-3|=|-2|=2\) and \(|5-3|=|2|=2\). 2. The numbers \(1\) and \(5\) are each \(2\) units from \(3\).

Both equal \(2\). Each number is \(2\) units from \(3\).

- Where is the center of the distance in the expression? - Check each candidate's distance from that center.

1. \(-1\) is \(3\) units left of \(2\), so \(|-1-2|=3\). 2. \(5\) is \(3\) units right of \(2\), so \(|5-2|=3\). 3. \(0\) is only \(2\) units from \(2\), so it does not work.

\(-1\) and \(5\): each is \(3\) units from \(2\). The number \(0\) is only \(2\) units away.

- Try \(x=4\) and then a number on each side of \(4\). What do the absolute values look like? - What is the smallest possible value of the expression on the left?

1. \(|x-4|\) is the distance between \(x\) and \(4\). 2. A distance cannot be negative, so no real number solves the equation.

No solution; a distance cannot be negative.

- Try putting a positive number and its opposite between absolute-value bars. - The bars apply to the whole \(x+1\), not just to \(x\).

1. The expression inside the absolute value may be \(4\) or \(-4\). 2. The two equations are \(x+1=4\) and \(x+1=-4\).

\(x+1=4\) and \(x+1=-4\)

- Identify the center of the distance described by the expression inside the absolute value. - A positive distance can be reached in two opposite directions from that center. - Check each candidate by measuring its distance from 4.

The equation means that \(x\) is 7 units from 4. Therefore, \(x-4=7\) or \(x-4=-7\). Solving gives \(x=11\) or \(x=-3\). These two values lie 7 units to the right and left of 4.

\(x=-3\) or \(x=11\). Both solutions are 7 units from 4.

- Recall the range of possible values of an absolute value. - Ask what must be true of the inside expression when an absolute value equals 0. - Decide whether a distance can ever be negative.

a) An absolute value equals 0 only when the expression inside it equals 0. Thus \(2x-6=0\), so \(x=3\). b) Absolute value is never negative. Therefore, \(\lvert 2x-6\rvert=-4\) has no solution. c) A target of 0 is possible because an expression can have distance 0 from 0. A negative target is impossible because distance cannot be negative.

a) \(x=3\) b) No solution c) Absolute value can equal 0 but cannot equal a negative number.

- Isolate the absolute-value expression before creating two cases. - Once the absolute value equals a positive number, what two equations can represent that distance? - Check each candidate in the original equation, not only in the isolated form.

1. Subtract \(1\): \(2|x-3|=8\). 2. Divide by \(2\): \(|x-3|=4\). 3. The two cases are \(x-3=4\) and \(x-3=-4\), giving \(x=7\) and \(x=-1\). 4. For each value, \(|x-3|=4\), so \(2|x-3|+1=9\). Both solutions check.

\(x=-1\) or \(x=7\)

- Try a positive and a negative value of \(2x\) and compare their distances from zero. - Keep the factor \(2\) attached to \(x\) until you have set up your cases. - A check uses the expression with absolute-value bars, not just one case equation.

1. The two cases are \(2x=10\) and \(2x=-10\). 2. Dividing by \(2\) gives \(x=5\) and \(x=-5\). 3. Both check: \(|2\cdot5|=10\) and \(|2(-5)|=10\).

Cases: \(2x=10\) gives \(x=5\), and \(2x=-10\) gives \(x=-5\). Checks: \(|2\cdot5|=10\) and \(|2(-5)|=10\).

- What number has the same distance from each marked position? - Once you find that number, how far is it from each marker? - How do those two quantities fit the given equation form?

1. The two markers are at \(-1\) and \(5\). Their midpoint is \(h=\frac{-1+5}{2}=2\). 2. Each marker is \(3\) units from the midpoint, so \(r=3\). 3. The equation is \(|x-2|=3\).

\(h=2\), \(r=3\); \(|x-2|=3\)

- Focus on what the principal square root returns when the quantity inside the square is negative before squaring. - After the required rewrite, interpret the absolute value as a distance from \(4\). - The resulting distance equation has two cases.

1. For every real \(u\), \(\sqrt{u^2}=|u|\) because the principal square root is nonnegative even when \(u\) is negative. 2. Therefore, \(\sqrt{(x-4)^2}=|x-4|\), so the equation becomes \(|x-4|=9\). 3. Thus, \(x-4=9\) or \(x-4=-9\). 4. Solving gives \(x=13\) or \(x=-5\). 5. Both values make the original square root equal to \(9\).

\(\sqrt{(x-4)^2}=|x-4|\), because a principal square root is nonnegative. Therefore, \(x=13\) or \(x=-5\).

- Test a positive and a negative input inside an absolute value. Can either produce \(11\)? - Keep track of every possibility for the expression inside the bars. - Your check should use the original equation, not only a simplified one.

1. The expression inside the absolute value can equal \(11\) or \(-11\). 2. \(3x+2=11\) gives \(3x=9\), so \(x=3\). 3. \(3x+2=-11\) gives \(3x=-13\), so \(x=-\frac{13}{3}\). 4. Check: \(\lvert3\cdot3+2\rvert=11\), and \(\left\lvert3\left(-\frac{13}{3}\right)+2\right\rvert=11\).

\(x=3\) or \(x=-\frac{13}{3}\). Checks: \(\lvert3\cdot3+2\rvert=11\) and \(\left\lvert3\left(-\frac{13}{3}\right)+2\right\rvert=11\).

- Absolute value can represent distance between two positions on a number line. - Identify the center location and the fixed distance from that center. - A positive fixed distance from one point usually gives two possible locations.

1. The distance from mile marker \(m\) to \(18.5\) is \(|m-18.5|\). 2. Set that distance equal to \(6.2\): \(|m-18.5|=6.2\). 3. Solve the two cases: \(m-18.5=6.2\) or \(m-18.5=-6.2\). 4. The mile markers are \(m=24.7\) and \(m=12.3\).

\(|m-18.5|=6.2\); the call boxes can be at mile markers \(12.3\) and \(24.7\).

- The two solutions of a positive-distance absolute-value equation are symmetric about its center. - Find the number halfway between the two given solutions. - After finding the center, compare its distance to either solution.

The center \(a\) must be halfway between the two solutions. Their midpoint is \(\frac{-5+7}{2}=1\), so \(a=1\). The distance from 1 to either solution is 6, so \(b=6\). Therefore, the equation is \(\lvert x-1\rvert=6\). Indeed, \(-5\) and 7 are both 6 units from 1.

\(a=1\), \(b=6\), so the equation is \(\lvert x-1\rvert=6\).