How to Solve Linear Equations of College Algebra

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When faced with linear equations for the first time, many people feel overwhelmed and confused by the complexity of mixing numbers and letters to solve the equations. With a few simple guidelines, however, you can learn these fundamental skills used in college algebra and higher mathematics. The methods used to solve one-variable and two-variable linear equations in college algebra are fairly simple.

Things You'll Need

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One-Variable Linear Equations

  • Recall inverse relationships, such as 1 and -1, and 1/3 and 3, to solve one-variable linear equations. The solutions require using the inverse relationships of addition and subtraction, and multiplication and division.

  • Isolate the variable "x" on one side of the equation. If x = y, then x + a = y + a. Based on this logic, use inverses to move values from one side of an equation to the other side of the equation.

    To isolate x by using the inverse of subtraction in the equation x - 5 = 8, add the inverse of -5, which is +5, to both sides of the equation. The result is: x - 5 + 5 = 8 + 5. The solution is: x = 13.

    To use the inverse of addition in the equation x + 9 = 12 to isolate x, subtract the inverse of +9 from both sides of the equation. The resulting equation is: x + 9 - 9 = 12 - 9. After subtracting 9 from both sides of the equation, you will find that x = 3.

    Using the inverse of division in the equation (1/2)x = 10 to isolate x requires multiplying the inverse of 1/2 by both sides of the equation. The resulting equation is: (1/2)(2) = 10(2). Multiplying both sides of the equation by 2 reveals that x = 20.

    To isolate x by using the inverse of multiplication in the equation 4x = 8, divide both sides of the equation by 4. The resulting equation is: 4x/4 = 8/4. The solution is: x = 2.

  • Check the solution. Plug the solution into the original equation to verify that its value is correct. If the original equation is x - 5 = 8 and you found that the value of x is 13, for instance, then check the solution by simply using the value 13 instead of x in the original equation. The equation then becomes 13 -- 5 = 8 or 8 = 8, which is the correct answer.

Two-Variable Linear Equations -- Addition/Elimination Method

  • Choose a variable to eliminate in a two-variable linear equation such as 4x -- 10y = 32 and 6x + 4y = 10. To eliminate "x," multiply the equations by common multiples to obtain equal but opposite values of x: 3(4x -- 10y = 32) and -2(6x + 4y = 10). The example will then look like this: 12x -- 30y = 96 and -12x -- 8y = -20.

  • Add the equations together to eliminate x. An example is:

    12x -- 30y = 96

    -12x -- 8y = -20


         -38y = 76
  • Solve for y in the equation -38y = 76. The process is:

    -38y/38 = 76/38

    -y = 2

    -y/-1 = 2/-1

    y = -2

  • Plug the value of y into the original equations, and find the value for x. The first original equation is 4x -- 10y = 32, and the solution process is:

    4x -- 10(-2) = 32

    4x + 20 = 32

    4x + 20 -- 20 = 32 -- 20

    4x = 12

    4x/4 = 12/4

    x = 3

    The second original equation is 6x + 4y = 10. Its solution process is:

    6x + 4(-2) = 10

    6x -- 8 = 10

    6x -- 8 + 8 = 10 + 8

    6x = 18

    6x/6 = 18/6

    x = 3

  • Check the solutions y = -2 and x = 3 for the original equations, 4x -- 10y = 32 and 6x + 4y = 10. The process for the first equation is:

    4(3) -- 10(-2) = 32

    12 +20 = 32

    32 = 32

    The process for the second equation is:

    6(3) + 4(-2) = 10

    18 -- 8 = 10

    10 = 10

    Two-variable linear equations can have one solution, no solution or many solutions. That is why it is very important to check solutions in the original equations.

References

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