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How to Solve Algebraic Equations by Addition

Algebraic equations are mathematical statements in which the values on both sides of the equal sign are the same. In algebra, the terms or mathematical bit of information may contain a basic integer, or constant and a variable or letter which represents an unknown value. To solve algebraic equations, the terms move from one side of the equation to the other using inverse operations. Eventually, the variable is on one side of the equation with its value opposite the equal sign. The value is then substituted back into the equation to determine the validity of the statement.

Instructions

  1. Solving Equations with One Variable

    • 1

      Examine the addition equation, 5x + 15 = 30. The term 5x has a variable and a coefficient, meaning that the value of x is multiplied by five. 15 is positive and is added to the equation and together the product of 5 and x plus 15 should equal 30. Use the inverse operations to move terms.

    • 2

      Subtract 15 from both sides of the equation to keep the sides of the equation equal to each other and balanced. 5x + 15 -- 15 = 30 -- 15. Simplify the equation, 5x = 15.

    • 3

      Divide both sides of the equation by 5 to get x by itself and set equal to a value. 5x ÷ 5 = 15 ÷ 5. Simplify the equation x = 3.

    • 4

      Plug the value of 3 back in wherever there is an x, 5(3) + 15 = 30 and simplify, 15 + 15 = 30.

    • 5

      Examine the equation 2x + 3 = 7 + x. The product of two and x plus three should equal the sum of seven and x.

    • 6

      Move the lowest variable from one side of the equation to the other, 2x -- x + 3 = 7 + x -- x. Simplify, x + 3 = 7. Move 3 to the other side of the equation and simplify, x + 3 -- 3 = 7 -- 3 = x = 4.

    • 7

      Plug the value of four back into the equation, 2(4) + 3 = 7 + 4. Simplify, 8 + 3 = 7 + 4, to which both sums are 11.

    Solving Equation with Two Variables

    • 8

      Examine the equations 2x + y = 4 and 4x + 3y = 10. Now you must solve for both variables to check for accuracy.

    • 9

      Pick one variable to isolate. Since y is already by itself in the first equation, this will be the easier one. Subtract 2x to the other side of the equation, y = 4 -- 2x.

    • 10

      Substitute the value of y back into the second equation, 4x + 3(4 -- 2x) = 10. This is called the substitution method and is commonly used to find intercepts for graphing.

    • 11

      Solve for the parentheses by multiplying each inside term by the outside term, 3 x 4 = 12 and 3 x -2x = -6x. Simplify the equation, 4x + 12 -- 6x = 10.

    • 12

      Combine like terms and simplify, 4x -- 6x = -2x + 12 = 10. Move 12 to the other side and simplify, -2x = -2.

    • 13

      Divide both sides by -2 and simplify, x = 1.

    • 14

      Plug the value of x back into the simplest formula, which is the formula for y, y = 4 -- 2(1). Simplify for the value of y = 4 -- 2 = y = 2.

    • 15

      Plug the values of x and y back into any formula to check, 4(1) + 3(2) = 10. Simplify for 4 + 6 = 10, which is true.

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