0.2 + r = 0.2r − 0.06 + 0.3r?

Solve for r

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  • 1 decade ago
    Favorite Answer

    0.2 + r = 0.2r − 0.06 + 0.3r

    Fist, let's simplify the right side side:

    r + 0.2 = 0.5r − 0.06

    Next, isolate r and simplify.

    (r + 0.2) - 0.5r - 0.2 = (0.5r − 0.06) -0.5r - 0.2

    r - 0.5r = -0.06 - 0.2

    0.5r = -0.26

    (0.5r) / 0.5 = (-0.26) / 0.5

    r = -0.52

  • 1 decade ago

    0.2 + r = 0.2r − 0.06 + 0.3r

    put like terms together first. Remember to change the sign when taking a term to the other side.

    -0.2r-0.3r+r = -0.06-0.2 ------> now just do the math

    0.5r = -.26 -------> to isolate the r we need to divide by its coefficient, which in this case would be 0.5.

    0.5r/0.5 = -.26/0.5 -------> remeber that what you do to the left you must do to the right, so divide both sides by 0.5.

    r = -0.52

  • 1 decade ago

    0.2 + r = 0.2r − 0.06 + 0.3r

    r - 0.5r = -0.06 - 0.2

    0.5r = -0.26

    ... 0.5 is 1/2, so lets double everything

    r = -0.52

  • Xiomy
    Lv 6
    1 decade ago

    0.2 + r = 0.2r - 0.06 + 0.3r

    0.2 + r = 0.5r - 0.06

    r - 0.5r = -0.06 - 0.2

    0.5r = -0.26

    r = -0.52

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  • Como
    Lv 7
    1 decade ago

    20 + 100r = 20r - 6 + 30r

    50r = - 26

    r = - 26 / 50

    r = - 13 / 25

  • 1 decade ago

    0.2+r=0.2r-0.06+0.3r

    0.2+r=0.5r+0.06

    .26= -0.5r

    -0.52=r

    Some people can do simple math.

  • 1 decade ago

    0.2 + r = 0.2r - 0.06 + 0.3r

    1.5 r = - 0.26

    r = - 0.26 / 1.5 = - 0.17333 (to 5 dp)

  • 4 years ago

    Call some friends over and urinate over your clothes and make it some sort of event/accomplishment in your life.

  • Anonymous
    1 decade ago

    1.) combine like terms ------ .2+ r = 0.5r-.06

    2.) get variables to one side and constants to the other -----

    .26= -0.5r

    3.) solve for r .26/-0.5 = r

    r = -.52

  • 1 decade ago

    There should def be a limit on what can be asked...Why dont you do YOUR OWN HOMEWORK instead of having the "real smart people" answer it for you.

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