6) $$ m \angle R Q G=6+5 x, m \angle G Q P=121^{\circ} $$ and $$ m \angle R Q P=17 x+7 $$. Find $$ x $$

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To solve for $$ x $$ in the given angles, we can use the fact that the sum of the angles in triangle $$ RQP $$ is $$ 180^\circ $$. The angles given are:

- $$ m \angle R Q G = 6 + 5x $$
- $$ m \angle G Q P = 121^\circ $$
- $$ m \angle R Q P = 17x + 7 $$

We can set up the equation:

$$
m \angle R Q G + m \angle G Q P + m \angle R Q P = 180^\circ
$$

Substituting the expressions for the angles, we have:

$$
(6 + 5x) + 121 + (17x + 7) = 180
$$

Now, let's simplify this equation step by step:

1. Combine like terms:
   $$
   6 + 121 + 7 + 5x + 17x = 180
   $$
   $$
   134 + 22x = 180
   $$

2. Subtract 134 from both sides:
   $$
   22x = 180 - 134
   $$
   $$
   22x = 46
   $$

3. Divide by 22:
   $$
   x = \frac{46}{22}
   $$
   $$
   x = \frac{23}{11}
   $$

Thus, the value of $$ x $$ is $$ \frac{23}{11} $$.
$$ x = \frac{23}{11} $$

6) \( m \angle R Q G=6+5 x, m \angle G Q P=121^{\circ} \) and \( m \angle R Q P=17 x+7 \). Find \( x \)

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