Solving Algebraic Equations

Solving Algebraic Equations Following are the primary tools required for solving algebraic equations, as well as some common pitfalls to avoid. Tools 1. Distributive Property: The distributive property states that a(b + c) = ab + ac Thus, we can use the distributive property to either multiply terms through, or factor terms out. For instance, if we are confronted with 13x + 4x, we can use the distributive property to factor out the x, and find that 13x + 4x = (13 + 4)x = 17x This works exactly the same for a more complicated example, such as 13yx + 4x. We can apply the distributive property to find that 13yx + 4x = (13y + 4)x Similarly, if we have 12(x + 2), the distributive property tells us that we can multiply through by 12, and find that 12(x + 2) = 12x + 24 Although it is not usually required, we can apply the distributive property twice to find (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd 2. Arithmetic Operators: The arithmetic operators are +, −, ×, ÷ (addition, subtraction, mulitiplication, and division. If we have an equation, say x = y, then we can perform any arithmetic operation involving variables or constants (other than dividing or multiplying by 0), as long as we perform it on both sides of the equation. Thus, it follows that y+2−p x+2−p = 5 5 In general, however, we do not wish to use the arithmetic operations to complicate the equation; we want to use them in a way to isolate the variable we are solving for. For instance, if we have 2x + 1 = 2, we can first subtract 1 from both sides to find 2x = 1 Then we can divide both sides of the equation by 2 to find x= 1 2 3. Recognizing Constants: When solving an equation, it is of the utmost importance to be cognizant of what variable you are solving for. Suppose you have the equation x + 2y − z = 14. The solution to such an equation depends on the variable being solved for. If you are solving for x, then you can consider y and z to be constants (but be careful, because you can only divide by y or z if you know that they are constants that do not equal 0). Similarly, if you were solving for y, then x and z would be constants, with the same disclaimer as before. Now what about a situation with more complicated looking constants? Suppose 2x − 1 − eατ = 32 y+1 is the equation you need to solve, and you need to solve for x. Since you are solving for x, that means that α, τ, and y are all constants. As long as you are careful not to divide by 0, you can treat these constants the same as any number. Similarly, any expression that is ατ a combination of these constants, such as 1−e y+1 can simply be treated as a single constant. Upon recognizing that fact, the equation we are faced with is just 2x − (some constant) = 32 Now we can simply use our arithmetic operators, and add “some constant” to both sides of the equation, to find 1 − eατ 2x = 32 + y+1 Afterwards, all we need to do is divide both sides by 2, to find the result 1 1 − eατ x = (32 + ) 2 y+1 ατ As you can see, once you recognize that the very complicated looking term 1−e y+1 is just a constant that can be manipulated as a single entity, the problem becomes relatively simple. Just as we were able to manipulate the complicated constant term as a single term, we can manipulate complicated terms consisting of variables as a single entity as well. In the following example 1 − eατ 2x = 32 + x y+1 we can subtract by the term 1−eατ y+1 x to find 2x − 1 − eατ x = 32 y+1 Now we apply the distributive property to factor our the x, yielding (2 − 1 − eατ )x = 32 y+1 Finally we can divide by the even more complicated term, and find x= 32 2− 1−eατ y+1 Similarly, if we have a term consisting of multiple variables such as x = p(2x + 1)x we can manipulate it as a single entity yielding x − p(2x + 1)x = 0 Now applying the distributive property we find x(1 − p(2x + 1)) = 0 This is an equation with two solutions x = 0, and 1 − p(2x + 1) = 0, or x = 21 ( p1 − 1). It is left to the reader to verify these are indeed solutions. 4. Multiplying by 1: 1 is the multiplicative identity, which means that for any number a, we find a·1=a Thus, unlike the arithmetic operators, which we must apply to both sides of the equation, we can multiply a single entity by 1 without changing our equation at all. For instance, if we need to solve x x + =1 7 2 we can multiply both of the fraction terms by 1, in order to create a common denominator. Thus x x x 2 x 7 2x 7x 2x + 7x x x + = ·1+ ·1= · + · = + = 7 2 7 2 7 2 2 7 14 14 14 Up to this point, we’ve used the trick of multiplying by 1 in order to create a common denominator. Now we can apply the distributive property to factor out the x, and then perform arithmetic operations to find a solution. Thus 2x + 7x 14 (2 + 7)x 14 9x 14 9x = 1 = 1 = 1 = 14 14 x = 9 Pitfalls to Avoid 1. Dividing by the Variable: In order to solve the vast majority of algebraic equations, you do not need to divide by the variable you are solving for (if you think you’ve found a special case where you do, think very carefully before you procede in doing so). If you are faced with an equation that looks like x(2x + 3) = x it may seem very tempting to divide both sides of the equation by x. However, this is not a valid operation if x = 0. Thus, dividing both sides of the equation by x carries the assumption that x 6= 0. Dividing both sides of the equation by x we find 2x + 3 = 1, so our only solution is x = −1. This is untrue. While x = −1 is a solution, it is not the only solution. (x = 0 is also solution). Thus, when you divide both sides of the equation by x, you lose one of the solutions to the equation, so you find a single solution rather than 2, which is incorrect (because you are assuming x 6= 0, but x = 0 is actually a solution). In general, if a·b=0 then one of the following is true: a = 0, b = 0, or a = b = 0. Thus, when one is faced with an equation like x(x + 2) = 0 there are two solutions: x = 0, and x + 2 = 0 which implies x = −2. 2. Not Performing an Operation to All Terms of an Equation: This pitfall occurs when one misapplies arithmetic operations to an equation. Suppose that η(x + 4) − x = γ In this case, one might try to divide both sides of the equation by η to find (x + 4) − x = γ η This is incorrect, because dividing both sides of the equation by η does not simply “move” η to the right side of the equation. Performing such an operation would actually yield x+4− x γ = η η which is not extremely useful in solving the equation. A better strategy would be to multiply through by η to find ηx + 4η − x = γ From there the x terms could be combined, yielding x(η − 1) + 4η = γ Next we would subtract 4η from both sides and find x(η − 1) = γ − 4η Finally, we would divide by (η − 1) completing the solution for x x= γ − 4η η−1 3. Not Carrying Through a Negative Sign: This pitfall occurs when one improperly applies the distributive property. Suppose we would like to expand −h(x + 1). It is important to carry the negative sign to both terms to find that −h(x + 1) = −hx − h Similarly, we find that h(x − 1) = hx − h where the second term is negative, because we have a negative 1. In order to avoid this pitfall, it is important to be conscious of the sign of the terms involved in the expansion using the distributive property. 4. Improperly Canceling Terms of a Fraction: The following pitfall occurs when dealing with fractions that have polynomial terms, such as x x+1 In this situation, one might see an x in the numerator and denominator of the function, and 0 . Both of these are incorrect. In order to feel tempted to cancel them, yielding either 12 , or 0+1 “cancel” an x from both the numerator and the denominator, it would have to be multiplying all of the terms in both the numerator and denominator, as follows 2x +x x2 Above we can factor x from the terms in the denominator, and then “cancel” a factor of x from the numerator and denominator as follows 2x x 2 2 2x = = · = 2 x +x x(x + 1) x x+1 x+1 Conversely, if we have a fraction multiplied by a constant, we need to distribute the constant to all terms within the numerator of the fraction 2(x + 1) 2x + 2 x+1 = = 2· x x x 5. Improperly Separating Fractions: Suppose we have the fraction x+1 x−3 One might be tempted to say that x+1 x+1 x+1 = − x−3 x 3 but this would not be correct, because we cannot separate the terms in the denominator. However, we can separate the terms in the numerator, so x 1 x+1 = + x−3 x−3 x−3 is a valid statement. General Strategy In order to solve algebraic equations, one must apply the given tools in the proper order, and avoid the common pitfalls along the way. The general strategy to solving an algebraic equation is as follows 1. Separate the variable and constant terms. The first step is to apply arithmetic operators in order to move all terms containing the variable being solved for to one side of the equation, and the remaining constant terms to the other. After this process is completed, if possible, simplify the expressions (such as canceling a 1 and −1 on the same side of the equation). 2. Factor the variable from constant terms, if necessary. If there are multiple terms containing the variable, use the distributive property to factor out the variable from the constants. Depending on the form of the equation, more complicated factoring schemes may need to be used (such as (a + b)(c + d) = ac + ad + bc + bd). 3. Divide by any constant terms that resulted from factoring, to isolate the variable. If you were able to factor out all instances of the variable from the constant terms in the previous step, then divide by the constant in order to isolate the variable. 4. If the equation has multiple terms multiplied together which equal 0, check for solutions to each of the terms equalling 0. It may turn out that your equation looks something like x(x − 2) = 0 after it is simplified. Remember that setting either of the terms on the left equal to zero provides a solution, so there are 2 solutions, not just one (x = 0 and x = 2). It should be emphasized that this is a general strategy, but it is not an algorithm that may be followed blindly in order to solve an equation. Thus, it is only with practice, and an understanding of how to utilize the tools presented above that one can master solving algebraic equations. In closing, if you are ever asked to verify a solution to an equation, you do not need to solve the equation to find the solution. Instead, you can substitute the value of the solution for the variable, and verify that the equation is still a true statement. You may be asked to verify a solution to a transcendental equation, which is a type of equation that cannot be solved explicitly for the variable! Having read the above examples, try to solve the following equations for x. p2 (2x + 4) + eατ = 1 p(2x − 17) = 4q + x x= x x + p(1 − x)