Find the midpoint of the segment having the given endpoints.1-6, 52 and 1-6, 82

Lecture 3- Johnson Solve for x. 1. 2x +5 =7 2x = 12 x =6 2. 0 = 2/3x – 2 2 = 3/2x x =3 3. 6x – (3x – 5) = 17 6x – 3x – 5 = 17 3x = 12 x = 4 LEAST COMMON DENOMINATOR GETS RID OF THE FRACTION x−3 = 5 − x+5 4. 4 14 7 LCD = 2 · 2 · 7 = 28 28(x−3) 5∙28 28(x+5) = − →7 (−3 =)0−4 x+( →1) x=11→x=1 4 14 7 x=2+ x−25 →LCD=10→5x=20+2 x−50→3x=−30→ x=−10 5. 2 5 ZERO PRODUCT PROPERTY (x+5 )x−2 =0→ x=−5,2 6. 2 1 7. 2 x +x−1=0→ 2 (−1 x)( =0)x= ,−1 2 2 −b± √ −4ac 2 QuadraticFormula:x= 2a , where ax + bx +c = 0