Solving for a Variable

Solving for a Variable

Solving for a Variable 150 150 Deborah

Overview:

In order to solve some algebraic equations, it is important to choose a variable within the equation and then solve the equation in terms of that variable. One strategy is to get the variable on one side of the equation and the known values on the other side of the equation, then solving the problem. Generally, equations can be solved in different ways depending on which variable is isolated, then solved.

Isolating a Variable

As a general rule, isolating a variable on one side of the equation is a matter of doing the same operation to both sides of the equation. The operation will cancel everything out on one side of the equation except for the variable that is chosen. The equation that is left is similar, but it is often easier to solve.

Using Addition or Subtraction

Suppose the equation is a general equation like x + y = z. If solving for x, isolate the x on one side of the equation by cancelling out the y, as in x + y -y =z -y, or x = z-y. Solving for y is similar, as x -x +y = z – x, so that the x is cancelled out and the y is isolated. It can be shown in a less general equation. For example, if x +6 = 9, then x = 9-6, or x =3. The equation can also be solved in terms of the 6, as 6 = 9-x.

Using Multiplication or Division

Suppose the equation is the equation of d = rt, where d is distance, r is rate, and t is time. Then, if solving for r, isolate the r on one side of the equation so that rt/t = d/t. If solving for t, isolate the t, so that rt/r = d/r. It can also be shown in a less general equation. Suppose 5y = 20. Then solving for y will isolate the y so that y = 20/5 (or 4). Similarly, 5 = 20/y. (Of course, y is not equal to zero, or else the problem is undefined.)

Using a Combination of Operations

Suppose a problem asks for a combination of operations. It can still be solved by isolating variables, but it will take more than one step. For example, the formula for the perimeter of a rectangle is P = 2l + 2w. Isolating l will take two steps. If 2l = P-2w, then l will equal (P-2w)/2.  Similarly, if 2w = P – 2l, then w will  equal (P-2l)/w.

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