Question

Lines AB and CD at E. If m∠AEC=x^2+3x and m∠BED=6x+4 ,find m∠CEB.

Answer 1

$m\measuredangle CEB=152$

Explanation

Step 1

when two lines intersect, vertical angles that are equal are formed.Also two angles are Supplementary when they add up to 180 degrees

then

$\begin{gathered} m\measuredangle\text{AEC =}m\measuredangle BED \\ \text{replacing} \\ x^2+3x=6x+4 \end{gathered}$

and

$m\measuredangle\text{AEC}+\text{ m}\measuredangle CEB=180$

Step 2

solve for x,

$\begin{gathered} x^2+3x=6x+4 \\ x^2+3x-6x=+4 \\ x^2+3x-6x-4=0 \\ x^2-3x-4=0 \\ \text{factorize} \\ (x-4)(x+1)=0 \\ it\text{ means} \\ x-4=0 \\ x=4 \\ or \\ x+1=0 \\ x=-1 \end{gathered}$

we just take the positive number, because we are searching for an angle ( angles and distance are always positives)

then

$x=4$

Step 3

replace the value of x in the angle AEC

$\begin{gathered} m\measuredangle AEC=x^2+3x \\ m\measuredangle AEC=4^2+3\cdot4 \\ m\measuredangle AEC=16+12 \\ m\measuredangle AEC=28 \\ \end{gathered}$

replace the value of AEC in equation (2) to find CEB

$\begin{gathered} m\measuredangle\text{AEC}+\text{ m}\measuredangle CEB=180 \\ 28+m\measuredangle CEB=180 \\ \text{subtract 28 in both sides} \\ 28+m\measuredangle CEB-28=180-28 \\ m\measuredangle CEB=152 \end{gathered}$

I hope this helps you.