Question

solve for x.

1. triangle with vertices e, d, c; angle at e is 70°, segment ec is 2x + 12, segment cq is 12x + 12
2. triangle with vertices w, v, u; angle at u is 106°, side wv is 15x - 2, side vu is 11x + 4

Explanation:

Step1: Use the property of vertical angles and isosceles triangles

Since \( \angle ECD\) and \( \angle QCE\) are supplementary (\(\angle ECD + \angle QCE=180^{\circ}\)), and in \(\triangle DEC\), if \(DE = DC\) (assuming the triangle is isosceles based on the problem - like side - length expressions), we first find \(\angle ECD = 180^{\circ}- 70^{\circ}=110^{\circ}\). But wait, actually, if we assume \(DE = DC\) (from the side - length expressions \(2x + 12\) and \(12x+12\)), we can set up the equation \(2x + 12=12x + 12\).
Subtract \(2x\) from both sides:
\(2x+12-2x=12x + 12-2x\)
\(12 = 10x+12\)
Subtract \(12\) from both sides:
\(12-12=10x+12 - 12\)
\(0 = 10x\)
\(x = 0\) (This is wrong. Let's start over.

Wait, actually, if we consider the exterior - angle property. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. But if we assume \(DE = DC\) (because of the problem's structure with side - length expressions).
Set \(2x + 12=12x+12\)
\(12x-2x=12 - 12\)
\(10x=0\) (Wrong approach.

Wait, correct approach: Since \(\angle ECD = 180-(12x + 12)\) (linear pair) and in \(\triangle DEC\), if \(DE = DC\) (isosceles triangle), \(\angle D=\angle E\). But using the exterior - angle property of a straight line (\(\angle ECD = 180^{\circ}-(12x + 12)\) and \(\angle ECD\) is the exterior angle of \(\triangle DEC\) with \(\angle E = 70^{\circ}\) and \(\angle D\) (if \(DE = DC\), \(\angle D=\angle E = 70^{\circ}\)). Then \(\angle ECD=\angle D+\angle E=140^{\circ}\)
So \(180-(12x + 12)=140\)

Step2: Solve the equation

\(180-(12x + 12)=140\)
\(180-12x-12 = 140\)
\(168-12x=140\)
Subtract \(168\) from both sides:
\(-12x=140 - 168=-28\)
Divide both sides by \(- 12\): \(x=\frac{-28}{-12}=\frac{7}{3}\)

Answer:

\(x = \frac{7}{3}\)