This problem asks for an absolute value equation representing the distance from point $P$ at coordinate $x$ to the point 4. The distance formula on a number line is $|x-4|$, and if this distance equals 7, we have $|x-4|=7$. Solving this gives two cases: $x-4=7$ (so $x=11$) or $x-4=-7$ (so $x=-3$). The absolute value $|x-4|$ represents how far $x$ is from 4, regardless of direction. When setting up distance problems, remember that $|a-b|$ gives the distance between $a$ and $b$ on the number line.

To simplify $\sqrt{12}-\sqrt{27}$, we first simplify each radical separately. We have $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$ and $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$. Now we can subtract: $2\sqrt{3} - 3\sqrt{3} = -\sqrt{3}$. A common mistake is trying to combine the radicals before simplifying, such as writing $\sqrt{12-27} = \sqrt{-15}$, which would involve imaginary numbers. Always simplify radicals to their simplest form before combining like terms.

To solve $\sqrt{x+9}-\sqrt{x}=3$, we isolate one radical: $\sqrt{x+9} = 3 + \sqrt{x}$. Squaring both sides gives $x+9 = (3+\sqrt{x})^2 = 9 + 6\sqrt{x} + x$. This simplifies to $x+9 = 9 + 6\sqrt{x} + x$, or $0 = 6\sqrt{x}$, which means $\sqrt{x} = 0$, so $x = 0$. Let's verify: $\sqrt{0+9}-\sqrt{0} = 3-0 = 3$ ✓. The key is recognizing that when squaring $(a+b)^2$, we get $a^2+2ab+b^2$, not just $a^2+b^2$. Always check solutions in radical equations as squaring can introduce extraneous solutions.

To simplify $\frac{\sqrt{48}}{\sqrt{3}}$, we can use the property that $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for positive values. This gives us $\sqrt{\frac{48}{3}} = \sqrt{16} = 4$. Alternatively, we could simplify $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$, then $\frac{4\sqrt{3}}{\sqrt{3}} = 4$. A common error is trying to subtract or manipulate the numbers under the radicals incorrectly. When dividing radicals, you can combine them under one radical sign before simplifying.

To solve $|x+2|=|3x-4|$, we consider when the expressions inside the absolute values are equal or opposite. Case 1: $x+2 = 3x-4$ gives $6 = 2x$, so $x=3$. Case 2: $x+2 = -(3x-4) = -3x+4$ gives $4x = 2$, so $x = \frac{1}{2}$. Let's verify: for $x=3$, $|3+2|=|5|=5$ and $|9-4|=|5|=5$ ✓; for $x=\frac{1}{2}$, $|\frac{1}{2}+2|=|\frac{5}{2}|=\frac{5}{2}$ and $|\frac{3}{2}-4|=|-\frac{5}{2}|=\frac{5}{2}$ ✓. When absolute values are equal, the expressions inside can be equal or opposite in sign. Always check both possibilities and verify your solutions.

To simplify $\frac{\sqrt{6}}{\sqrt{2}}$, we use the quotient property of radicals: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. This gives us $\sqrt{\frac{6}{2}} = \sqrt{3}$. We can verify this is correct by checking that $\sqrt{3} \cdot \sqrt{2} = \sqrt{6}$. A common error is trying to subtract the numbers under the radicals (getting $\sqrt{4}$) or adding them, but division of radicals follows the quotient rule. When working with radical quotients, combine under one radical sign first, then simplify.

To rationalize $\frac{5}{2+\sqrt{3}}$, we multiply by the conjugate of the denominator. The conjugate of $2+\sqrt{3}$ is $2-\sqrt{3}$, so we multiply: $\frac{5}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{5(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$. The denominator becomes $(2)^2 - (\sqrt{3})^2 = 4-3 = 1$. Therefore, the answer is $\frac{5(2-\sqrt{3})}{1} = 5(2-\sqrt{3}) = 10-5\sqrt{3}$. A common mistake is multiplying only the denominator by the conjugate, but we must multiply both numerator and denominator. When the denominator simplifies to 1, don't forget to write the final answer without the fraction bar.

To simplify $\sqrt{18x^2}$, we need to factor out perfect squares and handle the variable carefully. We have $\sqrt{18x^2} = \sqrt{9 \cdot 2 \cdot x^2} = \sqrt{9} \cdot \sqrt{2} \cdot \sqrt{x^2} = 3\sqrt{2} \cdot |x|$. The crucial point is that $\sqrt{x^2} = |x|$, not just $x$, because the square root function always returns a non-negative value. A common error is writing $\sqrt{x^2} = x$, which is incorrect when $x$ is negative. Remember that for any real number $x$, we have $\sqrt{x^2} = |x|$.

To solve $|2x-7|=5$, we consider two cases based on the definition of absolute value. Case 1: $2x-7=5$ gives $2x=12$, so $x=6$. Case 2: $2x-7=-5$ gives $2x=2$, so $x=1$. Both solutions are valid since substituting back gives $|2(6)-7|=|5|=5$ ✓ and $|2(1)-7|=|-5|=5$ ✓. The key insight is that $|A|=k$ means either $A=k$ or $A=-k$ when $k>0$. Always solve both cases when dealing with absolute value equations.

To rationalize $\frac{3}{\sqrt{5}}$, we multiply both numerator and denominator by $\sqrt{5}$. This gives us $\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{3\sqrt{5}}{5}$. The key is that $\sqrt{5} \cdot \sqrt{5} = 5$, which eliminates the radical from the denominator. A common mistake is to multiply only the denominator by $\sqrt{5}$, forgetting that we must multiply both parts of the fraction by the same value. When rationalizing, always multiply by a form of 1 to maintain equality.