Post a New Question

algebra

posted by .

pleae help solve: 3x-5y=7 and 5x-2y=-1

  • algebra -

    3x-5y=7 times 5 15x-25y=35
    5x-2y=-1times 3 15x -6y=-3
    subtract second from first
    0x -19y = 38
    y = -2
    3x-5(-2) = 7
    3x +10 =7
    3x = -3
    x = -1

    check
    3(-1) -5(-2) = 7 yes
    5(-1)-2(-2)=-1 yes

  • algebra 2 -

    25y to the second power times x to the second power plus 20xy to the second power plus 4x to the second power

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra 2

    Posted by Christian on Monday, May 7, 2007 at 11:34am. The area of a rectangular blanket in square centimeters is 40x^2+2x-65. The width is 4x-5 cm^2. Find the dimensions of the blanket in terms of x. For Further Reading Algebra 2 …
  2. Math, pleae help... just want to see if i got it

    -2(-6 + 4 x 3-3) would that be negative -30?
  3. english

    Pleae unscramble vrafos
  4. physics

    two point charges are 10.0 cm apart and have chargeso of 2.0 uC (micro)and -2 uC, respectively. What is the magnitude of the electric field at the midpoint between the two charges?
  5. math

    write an equation in slope-intercept formfor the line passing through each pair of points. (12,5) and (-4,1) pleae help me!!!! i don't understanf how to solve it
  6. Exceptional children

    Providing assistance to a child in self-care tasks a. always encourage helplessness b. is necessary for very young children c. should be left to the family d. should be as minimally intrusive as possible. I chose "d" for my answer. …
  7. math

    How can i subtract 3/4 by 1/8? Pleae explain answer.
  8. Math

    We have an important chemsitry test due and I want to make sure that I do not mess up on my algebra. I am dealing with the integrated rate law, but my question is purely algebraically in nature. I hope, somebody can help me. 1) ln(A) …
  9. chemistry Pleae

    C7H14O2 + NaOH What will this combination make?
  10. Math

    y-6¡Ý12 Solve this pleae.

More Similar Questions

Post a New Question