{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

K_Algebra_Review_Day_1 - 4 2 y y 5x 12 7x-1 = 3 2 5 3 = 7x...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
K Algebra Review Day 1 Name_________________________________ Period______ Date __8/22_____ Simplify: 1. 2. 3. 4. 2 5 7 6 2 3 125 5 2 3 4 2 10 4 6 2 2 3 3 2 7 20 5 Multiply: 5. 6. 7. 9x(2x + 3) (x - 2)(x + 6) (3x + 7)(5x - 2) Factor: 8. 9. 10. 11. y 2 – 24y + 144 y 2 + 13y +36 2x 2 – 15 – x 2x 2 + 5x – 42 12. 13. 14. 15. 5y 2 + 34y – 7 9x 2 + 4 – 12x 20y 2 + 15y 4y 2 - 9 Solve for the variable: ( Leave answers in exact form) 16. 17. 18. 3x - 10 = 5x + 2 5x + 3 = 12x – 8.2 3x + 7 = 8(x + 2) + 1 19. 20. 21. 2x + 5 = 4(x + 3) –5 + 3(x – 2) = 15 – (2x – 4) 10 3 x = 13 – x 22. 23. 24. 1 1 8 4 17 4 8 4 2 y y
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4 2 y y 5x+12 7x-1 = 3 2 5 3 = 7x 2x-1 25. 26. 27. 6x 2 + x - 12 = 0 12x 2 = 3x 2x 2 + 7x +3 = 0 28. 29. 30. 3x 2 + 8 = -10x 5x 2 + 4 = 21x 6x 2 + 7x = 5 31. 32. 33. 20x 2 + 15x = 0 x+12 22 = 2x 3x-1 4x+1 x+1 = 3 3x-5 Solve the systems: Use the substitution method to solve the system. 34. y x 6 2x 3y 22 Use the elimination method to solve the systems. 34 . 35. 36. 8 6 42 2 6 36 x y x y 2 21 9 x y x y 4 2 8 5 +10 5 x y x y 37. 38. 4 28 2 +3 24 x y x y 8 3 32 7 + 9 28 x y x y...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern