How do I solve this algebra problem?

2n/5+7=3

I am kind of stuck on it.

Comments

  • Subtract 7: 2n/5 = –4

    Multiply by 5: 2n = –20

    Divide by 2: n = –10

  • the answer is -10. First subract 7 from both sides to get negative 4 on the right sides of the equation. Then multiply each sides by 5. The 5 will cancel out with the 5 in the denominator of 2n/5 and give you -20 on the other side of the equation. Then divide each side by 2 and then you get -10.

  • First, move the 7 to the other side of the equation by subtracting 7:

    2n/5 + 7 = 3

    2n/5 = -4

    Now set it up as a cross multiplication problem:

    2n/5 = -4/1

    Now cross multiply:

    2n = -20

    Now divide by 2 to solve for n:

    2n/2 = -20/2

    n = -10

  • 2n/5+7=3

    I assume ( I know that's scary) that it's

    (2n/5)+7=3

    so multitply both sides by 5 to get rid of the fraction (keeps it an equation)

    2n + 35 = 15

    subtract 35 from both sides

    2n = -20

    and I'm tired - you have it from here

    all the best

  • Do the same thing to both sides to solve it: 2n/5+7=3

    Subtract 7 from both sides: -7 -7

    You get: 2n/5 = -4

    Then, multiply both sides by 5: x5 x5

    You get: 2n = -20

    Then, divide both sides by 2: n = -10

  • Multiply both sides by 5 :~ 2n + 35 = 1

    Subtract 35 from each side :~ 2n = -20

    Divide both sides by 2 :~ n = -10

  • First get your term with n by itself. So you get:

    2n/5=3-7

    2n/5=-4

    Multiply both sides by 5 to get:

    2n=-20

    Divide both sides by 2 to get:

    n=-10

  • 2n/5+7=3

    2n/5 = 3-7

    2n/5 = -4

    2n = -20

    n = -20/2

    n = -10

  • n = -10

    2n/5 + 7 = 3

    2n/5 = 3-7

    2n = (3-7)*5

    n = -20/2

    n = -10

  • 2n/5 + 7 = 3

    2n/5 = -4

    2n = -20

    n = -10

    Best of luck!

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