There are two main properties that we use when solving linear equations. Property #1: Additive Property of Equality

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Norman Mason
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1 Chapter 1.1: Solving Linear and Literal Equations Linear Equations Linear equations are equations of the form ax + b = c, where a, b and c are constants, and a zero. A hint that an equation is linear is that the highest exponent anywhere in the equation is one, and the equation consists only of a single variable There are two main properties that we use when solving linear equations. Property #1: Additive Property of Equality If A = B Then A + C = B + C Or A - C = B - C This property states that if two quantities are equivalent to each other, then if we add the same number to both sides of the equation, or subtract the same number from both sides of the equation, then the equality of the two equations is preserved. Similarly, we have property #2: Property #2: Multiplicative Property of Equality If A = B Then AC = BC Or (where C 0 ) = This property states that if two quantities are equivalent to each other, then if we multiply the same number to both sides of the equation, or divide the same number from both sides of the equation (except for zero), then the equality of the two equations remains preserved. To Translate these Properties: What ever you do to one side of the equation, in terms of operations, you must do to the other side of the equation. Otherwise, the equality of the equations is not preserved

2 There is a 5 step Process to Solving Linear Equations. We will list the steps here, then go through a few examples. To solve linear equations: 1. Use the distributive property where applicable 2. If the equation has a denominator, multiply both sides of the equation by the LCD of all of the denominators. 3. Combine like terms on either side of the equation 4. Bring all of your constant terms (algebraically) over to one side of the equation, and your variable terms over to the other side of the equation. 5. Divide or multiply both sides of the equation by the inverse operation of what ever number is attached to the coefficient. Consider the following examples: Solve the following: a. 2x - 4 = 8 1. No distributive property 2. No denominator 3. No like terms on either side 2x - 4 = 8 4. Bring the - 4 over (by adding it) +4 = +4 2x = 12 2x = Divide both sides of the 2 2 equation by 2 to isolate the x. (the inverse operation of multiplying by 2 is dividing by 2. x = 6 Therefore, x = 6. You can check your solution, by plugging it back into the equation, and showing that the equation is true, so that 2(6) 4 = 12 4 = 8

3 b. 5(y + 2) - 4(y + 6) = Distribute 5y y - 24 = No denominator y - 14 = Combine like terms (5y 4y = y) and (10 24 = - 14) y 14 = Bring the - 14 over, (by adding + 14 = ). y = You don t need to divide both sides of the equation by any number, because y is already isolated. y = 24 Therefore, y = 24 c. + = 4 1. No distributive property Note that this equation has denominators. We want to make our lives easier, and not have to deal with the denominators when solving the equation. One way to clear the equation of the denominators is to find the LCD of the whole equation, and multiply each term in the equation by the LCD. This guarantees that the numerator will contain a factor of the denominator, and will cancel out = 30 30(4) 2. Multiply each term in the equation by is the Least Common Denominator of 5, 2, and 3. Recall that when you multiply fractions, you multiply numerators and denominators together, so that 30 = " = " " + " = " 18a + 15 = - 20a Simplifying this, we get

4 18a + 15 = - 20a No like terms Bring the constant terms 18a = - 20a over to one side of the equation, +20a +20a and all of the variable terms over 38a = to the other side of the equation. 38a = Divide both sides of the equation by 38 to isolate a. a = "# " Linear Equations typically have one solution. Note that we use the word typically. It is possible for Linear Equations to have either no solutions, or an infinite amount of solutions. Lets look at the next two examples of each one of these cases: Examples: Solve the following d. 3(m 2) + 4m = 7m Distribute 3m m = 7m No Denominator 7m - 6 = 7m Combine like terms Bring the constant terms 7m = 7m + 14 over to one side of the equation, - 7m - 7m and all of the variable terms over 0 = 14 to the other side of the equation. Wow.What happened here? Our variable terms cancelled out. And, we are left with the equation 0 = 14. If it happens that we are solving a linear equation, and all of the variable terms cancel out, then we have 2 choices: Either our equation has no solutions, or our equation has all real numbers as solutions. So the question is, is the solution in this case no solution, or all real solutions? To determine this, we look at the equation that we are left with. Ask yourself, is this equation true? So, is it true that 0 = 14? NO This means that our equation has no solutions. There is no value for x that will make the left hand side and right hand side of the equation true. We call these types of equations CONTRADICTIONS.

5 So, the solution set for d. is NO SOLUTION Let's look at a case where the solution set is all real numbers. e. 8 (p 1) = - p Distribute 8 p + 1 = - p No Denominator 9 - p = - p Combine like terms Bring constant terms over to - p = - p to one side of the equation, and all +p = +p the variable terms over to the 0 = 0 other side of the equation. So, note here, that once again the variable terms cancelled out. We are left with the equation, 0 = 0. Again, ask yourself is this true? YES It is true that 0 = 0. This means that our equation has solution set all real numbers. We call these types of equations IDENTITIES. So, the solution set for e. is ALL REAL NUMBERS If all variables cancel and we end up with a false statement, like 0 = 2, we have a contradiction. It is false that there is a solution. Thus, there is no solution. If all variables cancel and we end up with a true statement, like 2 = 2, we have an identity. It is always true that there is a solution. Thus, all real numbers are the solution.

Does the following equation have one solution, no solution, or infinite solutions? 5xx 8xx + 20 = 2xx + 20 3xx + 20 = 2xx + 20 +2xx +2xx 1xx + 20 = 20 20 20 1xx = 0 / 1 / 1 xx = 0 1.

6 Does the following equation have one solution, no solution, or infinite solutions? 5xx 8xx + 20 = 2xx xx + 20 = 2xx xx +2xx 1xx + 20 = xx = 0 / 1 / 1 xx = 0 1. No distributive property 2. No denominator 3. Combine like terms on the left side: 5xx 8xx = 3xx 4. Bring the -2x over by adding 2x 5. Since 20 is being added to x, do the inverse (opposite) operation: subtract 20 Since -1 is multiplying x, divide by -1. Note: 0 divided by -1 is 0. Even though the 20 and the -20 cancel out, the variables did not cancel out. We have one solution, x = 0. When we have 5xx 8xx on the left side, we combine like terms to get 5xx 8xx = 3xx, we do not add In the next step, we add 2xx to both sides because the 2xx and 3xx are on opposite sides. If two terms are on the same side, combine like terms. If two terms are on the opposite side, do the opposite operation to bring one term over to the other side.

7 Solving Literal Equations: Lets first define what we mean by a literal equation. A Literal Equation is an equation with more than one variable. We see literal equations in the real world a lot, for example, A = h (the area of a triangle is ½ times the base and height) P = 2L + 2W (the perimeter of a rectangle is 2 times the length plus two times the width) and many others. Oftentimes, when we are solving equations, we are interested in the characteristics of one variable, instead of the other variable. We may be asked to solve for a certain variable other than the one that is already solved for. When solving literal equations, undo whatever is being done to the variable you are solving for. For example, if you are solving for x, undo what is being done to x. You always want to hold onto the bit of the expression with the indicated variable last. Solve for the indicated variable a. y = mx + b, solve for x - b - b y - b = mx m m = STEP 1: Since b is being added to x, subtract b from both sides of the equation. STEP 2: Since m is multiplying x, divide both sides of the equation by m. b. 2x - 5y = 25, solve for x +5y +5y STEP 1: Since 5y is being subtracted from x, add 5y to both sides of the equation. Note: you can write y or 5y STEP 2: Since 2 is multiplying the x, Divide both sides of the equation by 2. STEP 3: Simplify if possible

8 c. 2x - 5y = 25, solve for y - 2x - 2x - 5y = - 2x + 25 /- 5 /-5 y = + " This time, since we are solving for y, we are undoing what is being done to y, instead of x. STEP 1: since 2x is positive, we can think of it as being added to the -5y, so we subtract 2x from both sides of the equation. STEP 2: Divide both sides of the equation by -5 STEP 3: Simplify if possible. y = 5 d. =, solve for c STEP 1: multiply both sides of the equation by 3 to clear out the denominator. 3 ( ) = 3 = 3T a + b + c = 3T - a - b - a - b c = 3T - a b STEP 2: subtract a and b from both sides of the equation. c = 3T - a - b

Chapter 2 Linear Equations and Inequalities in One Variable

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