Question

1. $2 + \sqrt{3p + 7} = 6$
2. $5 = \sqrt{2g - 7}$

Explanation:

Problem 4: \(2 + \sqrt{3p + 7} = 6\)

Step 1: Isolate the square root term

Subtract 2 from both sides of the equation to get the square root by itself.
\(2 + \sqrt{3p + 7}-2 = 6 - 2\)
Simplifying both sides, we have \(\sqrt{3p + 7}=4\)

Step 2: Eliminate the square root

To eliminate the square root, we square both sides of the equation. Remember that \((\sqrt{x})^2=x\) for \(x\geq0\).
\((\sqrt{3p + 7})^2 = 4^2\)
This simplifies to \(3p+7 = 16\)

Step 3: Solve for \(p\)

Subtract 7 from both sides:
\(3p+7 - 7=16 - 7\)
\(3p=9\)
Then divide both sides by 3:
\(\frac{3p}{3}=\frac{9}{3}\)
\(p = 3\)

Step 4: Check the solution

Substitute \(p = 3\) back into the original equation to verify.
Left - hand side: \(2+\sqrt{3\times3 + 7}=2+\sqrt{9 + 7}=2+\sqrt{16}=2 + 4=6\)
Right - hand side: \(6\)
Since the left - hand side equals the right - hand side, \(p = 3\) is a valid solution.

Problem 6: \(5=\sqrt{2g - 7}\)

Step 1: Eliminate the square root

Square both sides of the equation. Using the property \((\sqrt{x})^2=x\) (\(x\geq0\)) and \(a^2=a\times a\)
\(5^2=(\sqrt{2g - 7})^2\)
Simplifying, we get \(25=2g - 7\)

Step 2: Solve for \(g\)

Add 7 to both sides of the equation:
\(25 + 7=2g-7 + 7\)
\(32 = 2g\)
Divide both sides by 2:
\(\frac{32}{2}=\frac{2g}{2}\)
\(g = 16\)

Step 3: Check the solution

Substitute \(g = 16\) back into the original equation.
Right - hand side: \(\sqrt{2\times16-7}=\sqrt{32 - 7}=\sqrt{25}=5\)
Left - hand side: \(5\)
Since the left - hand side equals the right - hand side, \(g = 16\) is a valid solution.

Answer:

s:
For problem 4: \(p=\boldsymbol{3}\)

For problem 6: \(g=\boldsymbol{16}\)