Guys please help me to do these!!!! im tryin’ my best but it’s so hard

Answer:Step-by-step explanation:1.∠ACB=6 x°∠DCE=30°6x=30°-----------------divide both sides by 6x=30°/6= 5°∠BCE=180°-30°=150°----------------sum of angles on a straight line2.∠ABD=(4+5x)°∠CBD=(x+2)°(4+5x)°+(x+2)°=180°-----------sum of angles on a straight line4+5x+x+2=180°----------------collect like terms6x+6=180°6x=180°-6°6x=174°------------divide by 6 both sidesx=174°÷6= 29°∠ABD=4+5x=4+(5*29)= 4+145=149°∠CBD= x+2=29°+2°=31°3.BE bisects ∠ABE=(3x+1)°m∠DBA=(8x-14)°1/2 (m∠DBA)° = m∠ABE1/2(8x-14)°=(3x+1)°(4x-7)°=(3x+1)°4x-7=3x+14x-3x=7+1-------------collecting like termsx=8°m∠ABE=(3x+1)°=(3*8+1)=24+1=25°m∠DBA=(8x-14)°= (8*8-14)°=(64-14)=50°4.∠ADE=∠CDG50+3x-y=x+2x-16------------collect like terms50+16-y=x+2x-3x66-y=3x-3x66°=y∠ADC +∠ADB+∠BDE90°+50°+(3x-y)°=180°140°+(3x-y)°=180°3x-66°=180°-140°3x=40°+66°3x=106°-----------divide both sides by 3x=106°÷3= 35.33°∠FDG=(2x-16)°= (2×35.33° - 16° )= 54.67°∠BDE = (3x-y)°= (3×35.33°-66°)= 105.99-66=39.99°5.∠ABD+∠DBC=90°(6x+4)°+32°=90°(6x+4)°=90°-32°6x+4=58°6x=58°-4°6x=54°x=54°÷6=9°∠ABD= (6x+4)°+32°∠ABD= (6×9 +4)°+32°∠ABD= (54°+4°) + 32°∠ABD=58°+32°=90°6.∠AED=∠CEB(3x+5)°=(4y-15)°------------------form equation of equality3x+5=4y-155+15=4y-3x20=4y-3x------------------------------(1)∠AEC=∠DEB(y+20)°=(x+15)°y+20=x+1520-15=x-y5°=x-y5+y=x----------------------------(2)Use equation (2) in equation (1)20=4y-3x20=4y-3(5+y)20=4y-15-3y20+15=4y-3y35°=ySolve for xx=5+y=5°+35°=40°∠AED=(3x+5)°=(3×40 +5 )=120+5=125°∠AEC= (y+20)°= 35° + 20° =55°