Free math worksheets

- 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.

- 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 \(x=7\)

- 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 \(x=-5\).

- A graph of \(|x-h|\) has its vertex at \((h,0)\). - A horizontal line has a constant y-value. - The solutions to the equation are the x-coordinates where the two graphs intersect.

1. The V-shaped graph has vertex \((2,0)\), so it is the graph of \(f(x)=|x-2|\). 2. The horizontal graph is at \(y=1\), so \(g(x)=1\). 3. The equation \(f(x)=g(x)\) is \(|x-2|=1\). 4. The graphs intersect at x-values \(1\) and \(3\), so the solutions are \(x=1\) and \(x=3\).

a) Vertex \((2,0)\); horizontal level \(y=1\) b) \(f(x)=|x-2|\) and \(g(x)=1\) c) \(|x-2|=1\), with \(x=1\) or \(x=3\)

- A positive absolute value can come from two opposite values of the inside expression. - Write one equation for each possible sign of the inside expression. - Substitute both candidates into the original absolute-value equation when checking.

The expression inside the absolute value can equal 11 or \(-11\). 1. \(3x+2=11\) gives \(3x=9\), so \(x=3\). 2. \(3x+2=-11\) gives \(3x=-13\), so \(x=-\frac{13}{3}\). Checking: \(\lvert 3\cdot3+2\rvert=\lvert11\rvert=11\), and \(\left\lvert3\cdot\left(-\frac{13}{3}\right)+2\right\rvert=\lvert-11\rvert=11\).

\(x=3\) or \(x=-\frac{13}{3}\)

- 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 \(\lvert x-1\rvert=6\).