2 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES 1a Solving Equations Solving Equations: One Step Let s start with the definition of an equation. Equation. An equation is a mathematical statement that equates two mathematical expressions. The key difference between a mathematical expression and an equation is the presence of an an equals sign. So, for example, 2x +3, x (3 2x), and 2(y +3) 3(1 y) are mathematical expressions, while 2x +3=0, x (3 2x) =, and 2(y +3) 3(1 y) = 11 are equations. Note that each of the equations contain an equals sign, but the expressions do not. Next we have the definition of a solution of an equation. What it Means to be a Solution. A solution of an equation is a numerical value that satisfies the equation. That is, when the variable in the equation is replaced by the solution, a true statement results. Show that 2 is a solution of the equation 2y + 3 = 7. EXAMPLE 1. Show that 8 is a solution of the equation x 12 =. Solution. Substitute 8 for x in the given equation and simplify. x 12 = The given equation. 8 12 = Substitute 8 for x. = Since the left- and right-hand sides of the last line are equal, this shows that when 8 is substituted for x in the equation a true statement results. Therefore, 8 is a solution of the equation. Equivalent Equations
1A. SOLVING EQUATIONS 3 Equivalent Equations. Two equations are equivalent if they have the same solution set. EXAMPLE 2. Are the equations x 3=6andx = 9 equivalent? Are the equations x =5and Solution. The number 9 is the only solution of the equation x 3 = 6. x 7 = 10 equivalent? Similarly, 9 is the only solution of the equation x =9. Thereforex 3=6and x = 9 have the same solution sets and are equivalent. Answer: No. Operations that Produce Equivalent Equations We now turn to operations that will produce equivalent equations. Adding the Same Quantity to Both Sides of an Equation. Adding the same quantity to both sides of an equation does not change the solution set. That is, if a = b, then adding c to both sides of the equation produces the equivalent equation a + c = b + c. Subtracting the Same Quantity from Both Sides of an Equation. Subtracting the same quantity to both sides of an equation does not change the solution set. That is, if a = b, then subtracting c from both sides of the equation produces the equivalent equation a c = b c. Wrap and Unwrap Suppose that you are wrapping a gift for your cousin. You perform the following steps in order. 1. Put the gift paper on. 2. Put the tape on.
MODULE 1. LINEAR EQUATIONS AND INEQUALITIES 3. Put the decorative bow on. When we give the wrapped gift to our cousin, he politely unwraps the present, undoing each of our three steps in inverse order. 1. Take off the decorative bow. 2. Take off the tape. 3. Take off the gift paper. This seemingly frivolous wrapping and unwrapping of a gift contains some deeply powerful mathematical ideas. Consider the mathematical expression x +. To evaluate this expression at a particular value of x, we would start with the given value of x, then 1. Add. Suppose we started with the number 7. If we add, we arrive at the following result: 11. Now, how would we unwrap this result to return to our original number? We would start with our result, then 1. Subtract. That is, we would take our result from above, 11, then subtract, which returns us to our original number, namely 7. Addition and Subtraction as Inverse Operations. Two extremely important observations: The inverse of addition is subtraction. If we start with a number x and add a number a, then subtracting a from the result will return us to the original number x. In symbols, x + a a = x. The inverse of subtraction is addition. If we start with a number x and subtract a number a, then adding a to the result will return us to the original number x. In symbols, x a + a = x. Solve for x: EXAMPLE 3. Solve x 7=12forx. x 6=
1A. SOLVING EQUATIONS 5 Solution: To undo the effect of subtracting 7, we add 7 to both sides of the equation. x 7 = 12 x 7+7=12+7 x =19 Adding 7 to both sides of the equation produces an equivalent equation. On the left, adding 7 undoes the effect of subtracting 7 and returns x. On the right, 12 + 7 = 19. Therefore, the solution of the equation is 19. Check: To check, substitute the solution 19 into the original equation. x 7 = 12 19 7=12 Substitute 19 for x. 12 = 12 The fact that the last line of the check is a true statement guarantees that 19 is a solution of x 7=12. Answer: 10 In the solution of Example 3, we use the concept of the inverse. If we start with x, subtract 7, then add 7, we are returned to the number x. In symbols, x 7+7=x. We are returned to x because subtracting 7 and adding 7 are inverse operations of one another. That is, whatever one does, the other undoes.
6 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Solve for x: x + 1 2 = 3 5 EXAMPLE. Solve x + 2 3 = 1 for x. 2 Solution: To undo the effect of adding 2/3, we subtract 2/3 from both sides of the equation. x + 2 3 = 1 2 x + 2 3 2 3 = 1 2 2 3 Subtracting 2/3 from both sides produces an equivalent equation. x = 3 6 6 x = 1 6 Therefore, the solution of the equation is 1/6. On the left, subtracting 2/3 undoes the effect of adding 2/3 and returns x. On the right, make equivalent fractions with a common denominator. 3 Subtract: 6 6 = 1 6 More Operations That Produce Equivalent Equations Here are two more operations that produce equivalent equations. Multiplying Both Sides of an Equation by a Nonzero Quantity. Multiplying both sides of an equation by a nonzero quantity does not change the solution set. That is, if a = b, and c 0, then multiplying both sides of the equation by c produces the equivalent equation ac = bc. Dividing Both Sides of an Equation by a Nonzero Quantity. Dividing both sides of an equation by a nonzero quantity does not change the solution set. That is, if a = b, and c 0, then dividing both sides of the equation by c produces the equivalent equation a c = b c.
1A. SOLVING EQUATIONS 7 Like addition and subtraction, multiplication and division are inverse operations. Multiplication and Division as Inverse Operations. Two extremely important observations: The inverse of multiplication is division. If we start with a number x and multiply by a number a, then dividing the result by the number a returns us to the original number x. In symbols, a x a = x. The inverse of division is multiplication. If we start with a number x and divide by a number a, then multiplying the result by the number a returns us to the original number x. In symbols, a x a = x. EXAMPLE 5. Solve 2.1x = 0.2 for x. Solve for x: Solution: To undo the effect of multiplying by 2.1, we divide both sides of the equation by 2.1. 3.6x =0.072 2.1x = 0.2 2.1x 2.1 = 0.2 2.1 x = 2 Dividing both sides by 2.1 produces an equivalent equation. On the left, dividing by 2.1 undoes the effect of multiplying by 2.1 and returns x. On the right, divide: 0.2/( 2.1) = 2. Therefore, the solution of the equation is 2. Check: To check, substitute the solution 2 into the original equation. 2.1x =0.2 2.1( 2) = 0.2 Substitute 2 for x. 0.2 = 0.2 On the left, multiply: 2.1( 2) = 0.2 The fact that the last line of the check is a true statement guarantees that 2 is a solution of 2.1x = 0.2. Answer: 0.02
8 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Solving Equations: Clearing Fractions and Decimals In this section we introduce techniques that clear fractions and decimals from equations. When clearing fractions from an equation, you will need to simplify products like the ones posed in the following examples. Example?? shows all of the steps involved in arriving at the answer. Again, the goal in this section is to perform this calculation mentally, so we just Divide 7 into 28 to get, then multiply by 6 to get 2. ( ) 6 28 7 x =2x Not only does this approach allow us to write down the answer without doing any work, the numerical calculations involve smaller numbers. You need to practice this mental calculation until you can write down the answer without writing down any steps. ( ) 5 EXAMPLE 6. Simplify: 27 9 x. Solution: Divide 9 into 27 to get 3, then multiply 3 by 5 to get 15. ( ) 5 27 9 x =15x Simplify: Answer: 27x ( ) 3 18 2 x Clearing Fractions Now that we ve done the required fraction work, we can now concentrate on clearing fractions from an equation. Clearing fractions from an equation. To clear fractions from an equation, multiply both sides of the equation by the least common denominator. EXAMPLE 7. Solve for x: x + 2 3 = 1 2. Solve for x: x 3 = 1 2
1A. SOLVING EQUATIONS 9 Solution: The common denominator for 2/3 and 1/2 is 6. We begin by multiplying both sides of the equation by 6. x + 2 3 = 1 ( 2 6 x + 2 ) ( ) 1 =6 3 2 ( ) ( ) 2 1 6x +6 =6 3 2 Multiply both sides by 6. On the left, distribute the 6. To simplify 6(2/3), you have two choices. You can multiply 6 and 2 to get 12, then divide 12 by 3 to get. Or you can divide 3 into 6 to get 2, then multiply 2 by 2 to get. Either way, 6(2/3) =. Similarly, 6(1/2) = 3. 6x +=3 Multiply: 6 ( ) ( ) 2 1 =,6 =3. 3 2 Note that the fractions are now cleared from the equation. To isolate terms containing x on one side of the equation, subtract from both sides of the equation. 6x + =3 6x = 1 Subtract from both sides. To undo multiplying by 6, divide both sides by 6. 6x 6 = 1 6 x = 1 6 Divide both sides by 6. r x: 3 7 x = 3 2 EXAMPLE 8. Solve for x: 5 x = 3. Solution: The common denominator for /5 and /3 is 15. We begin by multiplying both sides of the equation by 15. 15 ( 5 x 5 x = ) 3 =15 ( 3 ) Multiply both sides by 15.
10 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES To simplify 15(/5), you have two choices. Multiply 15 and to get 60, then divide 60 by 5 to get 12. Or you can divide 5 into 15 to get 3, then multiply 3 by to get 12. Either way, 15(/5) = 12. Similarly, 15( /3) = 20 12x = 20 Multiply. To undo multiplying by 12, we divide both sides by 12. 12x 12 = 20 12 x = 5 3 Divide both sides by 12. Reduce to lowest terms. Check: To check, substitute 5/3 for x in the original equation. 5 x = ( 3 5 ) = 5 3 3 20 15 = 3 3 = 3 Substitute 5/3forx. Multiply numerators and denominators. Reduce. The fact that the last line is a true statement guarantees the 5/3 is a solution of the equation 5 x = 3. Answer: x = 7 2 EXAMPLE 9. Solve for x: 2x 3 3 = 1 2 3x. Solve for x: Solution: The common denominator for 2x/3, 3/, 1/2, and 3x/ is 12. We begin by multiplying both sides of the equation by 12. 12 ( 2x 3 2x 3 3 = 1 2 3x ( 2x 12 3 3 ) ( 1 =12 2 3x ) ) ( ) ( ) 3 1 12 =12 12 2 ( ) 3x Multiply both sides by 12. Distribute the 12 on each side. 5x 9 2 3 = 5 9 3 2 To simplify 12(2x/3), you have two choices. Multiply 12 by 2x to get 2x, then divide 2x by 3 to get 8x. Or you can divide 3 into 12 to get, then
1A. SOLVING EQUATIONS 11 multiply by 2x to get 8x. Either way, 12(2x/3) = 8x. Simlarly, 12(3/) = 9, 12(1/2) = 6, and 12(3x/) = 9x. 8x 9=6 9x Multiply. Note that the fractions are now cleared from the equation. We now need to isolate terms containing x on one side of the equation. To remove the term 9x from the right-hand side, add 9x to both sides of the equation. 8x 9+9x =6 9x +9x 17x 9=6 Add 9x to both sides. To remove the term 9 from the left-hand side, add 9 to both sides of the equation. 17x 9+9=6+9 17x = 15 Add 9 to both sides. Finally, to undo multiplying by 17, divide both sides of the equation by 17. 17x 17 = 15 7 x = 15 17 Divide both sides by 17. Clearing Decimals from an Equation Multiplying by the appropriate power of ten will clear the decimals from an equation. However, first note the following: 10(1.235) = 12.35. place to the right. Multiplying by 10 moves the decimal point one 100(1.235) = 123.5. Multiplying by 100 moves the decimal point two places to the right. 1000(1.235) = 123.5. Multiplying by 1000 moves the decimal point three places to the right. Note the pattern: The number of zeros in the power of ten determines the number of places to move the decimal point. So, for example, if we multiply by 1,000,000, which has six zeros, this will move the decimal point six places to the right.
12 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Solve for x: EXAMPLE 10. Solve for x: 2.3x 1.25 = 0.0x. 1.3.5x =2.2 Solution: The first term of 2.3x 1.25 = 0.0x has one decimal place, the second term has two decimal places, and the third and final term has two decimal places. At a minimum, we need to move each decimal point two places to the right in order to clear the decimals from the equation. Consequently, we multiply both sides of the equation by 100. 2.3x 1.25 = 0.0x 100(2.3x 1.25) = 100(0.0x) Multiply both sides by 100. 100(2.3x) 100(1.25) = 100(0.0x) Distribute the 100. 230x 125 = x Multiplying by 100 moves all decimal points two places to the right. Note that the decimals are now cleared from the equation. We must now isolate all terms containing x on one side of the equation. To remove the term x from the right-hand side, subtract x from both sides of the equation. 230x 125 x =x x 226x 125 = 0 Subtract x from both sides. Simplify. To remove 125 from the left-hand side, add 125 to both sides of the equation. 226x 125 + 125 = 0 + 125 226x = 125 Add 125 to both sides. Finally, to undo multiplying by 226, divide both sides by 226. 226x 226 = 125 226 x = 125 226 Divide both sides by 226. Simplify.