# 2.3 Solving Equations - Mathematical & Word

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1 2.3 Solving Equations - Mathematical & Word Problems 7.EE.4a Students will be able to use properties of numbers to correctly solve mathematical and word problems in one variable using fraction bars, word boxes, and geometry diagrams. Solving Two Step Equations (review) When working in the language of math, certain parts of the equation have specific names. Look at the equation below. Just like a sentence in English has nouns and verbs, a math equation has terms and coefficients. There are 2 terms in this equation. The first is 3x. The second is + 7. Always include the sign with the term. The coefficient of the x term is 3. When we know all the numbers in a math expression, we can solve by using order of operations. Solving for a variable in an equation means to work backward. We know the answer and work backward to find part of the math problem. Since we work backward, we use order of operations backward. Using order of operations backward to find part of the problem (that s the variable) is called the inverse operation. The chart shows how the operations are reversed when we solve for a variable. Here are three examples of two step equations. You can have any rational number in an equation. The steps for solving using inverse operations is the same. The first step in all three is adding or subtracting the inverse. The second step is multiplying or dividing by the coefficient of x. 54

2 Chapter 2. Unit 2: Expressions and Equations Example 1a, b, c is a video review of solving two step equations. It also has practice problems you can try with answers explained. It is a good idea to try this before going on to the next section. Solving Equations Using the Distributive Property If you were given the math expression -2(8-6), could you find the answer? Of course you could. 2(8 6) 2(2) 4 What if you were given the math expression -2(x - 2)? You could simplify it, but not come up with a number answer. Any time we have an expression with a variable, we can only simplify, not solve. However, once we set the expression equal to something, we will get one answer, no answer, or inifinte answers. One answer means that when we are done solving, the variable will equal one number. No answer means that when we are done solving, we will get an answer like 6 = 7. That doesn t work! So there is no answer. Infinite answers means that when we are done solving, we will get an answer like 5 = 5. This means it doesn t matter what the variable is. We will work with equations that have one answer. Here are the steps to solve using the distributive property. Solve for x: 5(x + 4)= 13 55

3 As we solve more complex equations, it is a good idea to simplify the expression(s) before using inverse operations. Here is a more complex model. We will solve for m. 1 2 (8m + 2) 2(m + 1)=17 It looks a lot harder, but it is only longer. To make algebra easier we break down any big problem into smaller parts. Just like running a mile - if you think about finishing one lap at a time, it makes running the whole mile seem easier to finish. Looking at the expression, do you notice that there are two parts that need distributing? We solve each piece separately and then put them together. (Hint: half of any number means to divide it by 2.) 1 (8m + 2) 2 2(m + 1) 1 2 (8m)+1 (2) 2 2(1m)+ 2(1) 4m + 1 2m + 2 Putting them together gives an equivalent equation. 4m + 1 2m + 2 = 17 Now combine like terms. Like terms are numbers and variables that are the same. All rational numbers can be combined together. All variables that are the same (including the exponent) can be combined. Combine means to add or subtract. 4m + 1 2m + 2 = 17 4m m + 2 = 17 4m + 2m = 17 2m + 1 = 17 additive inverse commutative property combine like terms At last we have a two step equation to solve. This is our goal with any complex equation. Simplify it down to a simple two step problem. In this model, we will show the two step equation solved across instead of vertical. Just like equivalent expressions, it looks different, but does the same thing. 56

4 Chapter 2. Unit 2: Expressions and Equations 2m + 17 = 17 2m = 17 1 inverse of adding -1 is subtracting -1 2m = 18 2m 2 = 18 2 inverse of multiplying by 2 is dividing by 2 m = 9 Example 2 2 Solve for x: 3 (x + 2)+3( 1 4 x 1 ) 6 = 2 3 Putting in fractions makes this problem look harder. It is still solved the exact same way. We use the same steps to solve with ALL rational numbers. Breaking the equation down into smaller parts makes solving easier. Our first goal is to simplify the equation down to a simple two step problem. Now that the equation has been simplified, follow the two steps for solving. Use inverse operations. Add or subtract first. Multiply or divide second. 57

5 Example 3 Solve for m: (m 4) (6m 5)= 34 Example 4 Solve for a: 3(a + 2)+4(a + 2)=63 58

6 Chapter 2. Unit 2: Expressions and Equations This is a unique problem. Look at the terms in the parentheses. They are the same! Because they are the same, we can make an equivalent equation. The steps below show you how. Now all that is left to do is distribute, and we have our two step equation. and http :// are video examples with step by step solutions. Word Problems and Algebraic Equations Word problems are like playing hide and seek with math problems. Hidden in the sentences are words that mean numbers, variables, and math operations. Our job is to find these and create a math equation. In any word problem, the very first step is to define the variable. Defining the variable means to decide what the problem wants us to find, and then make it into a variable. Some examples are: If you are finding a number, write: Let a number = n If you are finding an age, write: Let the age of the person = a If you are finding a width, write: Let the width = w It is helpful to use a variable that matches what you are looking for. It is also helpful to create a type of fraction bar to see how the problem fits together. Here is a model to help you. 59

7 Leonard wants to save \$100 in the next 2 months. He knows that in the second month he willbe able to save \$30 more than the first month. How much should he save each month? Since we are using money, we select m as our variable. So the first month Lenoard saved m. The second month he saved m and another \$20. Using the fraction bars, we can now make the math equation. m + m + 20 = 100 To solve, we combine like terms. Then we have a two step equation! Inverse operations - and we end up with the value of m. m + m + 20 = 100 2m + 20 = 100 2m = m = 80 2m 2 = 80 2 m = 40 Think you are done? Not yet! The tricky part of word problems is that the solution is not always the answer! What that really means is that once we get a value for the variable, we still have to answer the question. The question is; How much should he save each month? So to find this we look at the fraction bar. The first month is m. The second month is m The value of m is 40 or \$40. The first month is m, or \$40. The second month is m + 20, or \$40 + \$20. The second month is \$60. Now we are done. Always make sure you answer the question in any word problem. Your first answer may not really be the answer:). Watch the video at to really learn how this model can help make word problems easier. It even has guided practice for you tocheck your understanding. Another way to solve word problems is by using a word model and an algebra model. It is like the fraction bars. Here is a word problem to show you how. 60

8 Chapter 2. Unit 2: Expressions and Equations Randy spent \$89.98 on a new phone. After this, he had \$27.32 left. How much money did he have before he bought the phone? In any word problem, we can always use x. The word model shows what x represents - the original amount. Word models and algebra models help you see the words and their math numbers. Making the equation is easy using the algebra model. x = Using the inverse operation (addition), we find the value of x. x = x = In this word problem, the value of x is our final answer - as long as it has a label and is written as money. The label is money, so the final answer is \$

9 A specific type of algebra word problem is called a consecutive integer problem. Consecutive means one right after another. Examples of consecutive integers are -5, -4, -3 and 15, 16, 17. Consecutive integer word problems ask you to find a set of numbers that follow this pattern. If you were asked to give three consecutive integers beginning with 4, what would be your answer? That would be pretty easy; 4, 5, 6. How did you get from 4 to 5? By adding 1. How did you get from 4 to 6? By adding 2. How would you get from 4 to the next number in the series? By adding 3! Math is all about patterns. In algebra, the same pattern is used. If the first consecutive integer is x, how do you get to the next consecutive integer? By adding 1. How do you get to the second consecutive integer? By adding 2. Same pattern. Here is a model that shows how to find consecutive integers using algebra. Find three consecutive integers such that the sum of the second and third is 41. First, pick a variable to be the first integer. We will use n. Since the problem asks for three consecutive integers, we have to write all three in "algebra language". The first consecutive integer = n The second consecutive integer=n+1 The third consecutive integer=n+2 Now we can write the algebra equation. We will use a word model. Sum means to add. Now, solve for the variable. 62

10 Chapter 2. Unit 2: Expressions and Equations n n + 2 = 41 n + n = 41 2n + 3 = 41 2n = n = 38 2n 2 = 38 2 n = 19 commutative property combine like terms subtract three from both sides divide by 2 on both sides This is another word problem where the solution is not the final answer. We need three consecutive integers. Since 19 is the first consecutive integer, 20 must be the second, and 21 must be the third. is a video model of a consecutive integer problem. You can also browse the mathvids.com site for other examples on word problems. Here are some more examples for you to improve your word problem skills. Example 5 The length of a rectangle is 3 times its width. The perimeter of the rectangle is 72 meters. Find the dimensions of the rectangle. In any geometry problem, a diagram is helpful. A diagram also gives the algebra for the length and width. Perimeter is to add up all sides. The expression for the perimeter is 3w + w + 3w + w. The perimeter is 72 meters. The equation for the perimeter is 3w + w + 3w + w = 72. Now solve the equation for w. 3w + w + 3w + w = 72 8w = 72 8w 8 = 72 8 w = 9 We need to find the dimensions. That means the width AND the length. The length is 3 times the width. Multiply 9x3togetthelength. The length is

11 Finally we write the answer with labels. The width is 9 meters. The length is 27 meters. Example 6 On Wednesday, Harry sold 3 times as many bars of chocolate as on Tuesday. On Thursday, he sold 5 times as many bars as on Tuesday. In all he sold 90 bars of chocolate. Find the number of bars he sold each day. The number of chocloate bars Harry sold on Tuesday = c. (ALWAYS define the starting variable) Since we are adding up to make 90, we can use the modified fraction bars. Now make the equation and solve for c. c + 3c + 5c = 90 9c = 90 9c 9 = 90 9 c = 10 C is the number of candy bars sold on Tuesday. 3c is the number of candy bars sold on Wednesday. 5c is the number of candy bars sold on Thursday. Our algebra found that c = 10. Now we can find out how many candy bars were sold each day. 10 is the number of candy bars sold on Tuesday. 3(10) or 30 is the number of candy bars sold on Wednesday. 5(10) or 50 is the number of candy bars sold on Thursday. Need more help with word problems? Watch the video at: h-equation-in-one-variable/?id=941. commutative property, distributive property, coefficient, term, sum, equation, expression, variable, inverse operation, like terms, consecutive integer 64

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