1 / 24

MA 1128: Lecture 03 – 1/25/2011

MA 1128: Lecture 03 – 1/25/2011. Solving Linear Equations. Expressions. We will use the equal sign in two slightly different ways. When we are simplifying expressions , we’ll start with an expression like x ( x + 2 2 ) And change the form of the expression somehow.

mrinal
Download Presentation

MA 1128: Lecture 03 – 1/25/2011

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MA 1128: Lecture 03 – 1/25/2011 Solving Linear Equations

  2. Expressions We will use the equal sign in two slightly different ways. When we are simplifying expressions, we’ll start with an expression like x(x + 22) And change the form of the expression somehow. For example, since 22 = 4, replacing 22 with 4 results in another expression that is equal x(x + 22) = x(x + 4) Going one step further, we can distribute the first x over the x + 4, x(x + 22) = x(x + 4) = x2 + 4x and we get a third expression that is equal to the first two. When changing expressions, the equal sign indicates that the expressions are equivalent. Next Slide

  3. Equivalent Expressions Specifically, we will say that two expressions are equivalent, if they are equal for all values of x (or whatever the variable letter is). For example, if we put 2 in for the x, the first expression x(x + 22) is equal to (2)((2) + 22) = 2(2 + 4) = 2(6) = 12 Putting 2 into the last expression x2 + 4x gives us the same result (2)2 + 4(2) = 4 + 8 = 12. If we were to put a 5 in for the x, we would get 45 from each expression. Next Slide

  4. Practice Problems for Equivalent Expressions • What do the first and last expressions equal if you put a 3 in for the x? • Are the numbers you get the same? • The distributive property applied to x(x + 1) results in what? Note that the distributive property will give you an equivalent expression. • Put x = 2 into the expressions x(x + 1) and x2 + x. What do you get? • Put x = 1 into both expressions from problem 4. What do you get? Click for answers: 1) 3 [[ (3)((3) + 22) = (3)(1) = 3 and (3)2 + 4(3) = 9  12 = 3]]; 2) Yes; 3) x2 + x; 4) 6; 5) 0. Next Slide

  5. Equations We will also use the equal sign in an equation. Here we will put an equal sign between two non-equivalent expressions, and we’ll be asking the question, for what values of x are the expressions equal. In solving an equation, we will start with an equation like x + 4 = 2x Here, the expressions x + 4 and 2x are not equivalent. For most of the values we might put in for the x, the results won’t be equal. Next Slide

  6. Practice Problems • Put a 3 in for the x in both expressions. What do you get? • Put a 4 in for the x in both expressions. What do you get? Click for answers 1) 7 for one and 6 for the other. 2) 8 for both of them. When solving this equation, x = 8 is what we’ll be looking for. Next Slide

  7. Solving Equations When we solve an equation, we are looking for those values for x that make the two expressions equal. These values are called solutions. The solutions are obvious in some equations. For example, in the equation x = 4, The only way x and 4 can be equal is if x is 4. In the equation, as another example, x – 1 = 10, We can see pretty easily that x must be 11. Next Slide

  8. Continued For the equation x2 = 4 We can see that x can be 2 or 2. In this case, we’ll usually write something like x = 2,2 to indicate that these are the solutions. Again, look at how the two expressions x2 and 4 are in no way equivalent. The expression 4 is always equal to 4, and x2 can be any non-negative number. Next Slide

  9. Practice Problems Find all of the solutions for the following equations. • x = 10. • x2 = 9. • x + 3 = 5. • x – 4 = 2. Click for answers. 1) x = 10; 2) x = 3,3; 3) x = 2; 4) x = 6. Next Slide

  10. Special cases. In this class, most of our equations will only have one or two solutions. BUT there are a few special cases. For example, in the equation 3x = x + 2x The two expressions are equal for any value of x (the two expressions are equivalent). For the solution, we’ll say that x can be any real number. We’ll also see equations like x = x + 2. The expression on the right is always 2 larger than the one on the left, So in this case, there are no solutions. Next Slide

  11. Basic Techniques Getting back to how we solve equations, we will follow the following Basic Principle Given an equation to solve, we will try to transform the equation into one where the solutions are obvious. Example. Consider the equation 3x + 7 = 1. You may be able to see the solution, but let us try to transform this into a simpler equation (where the solution is even more obvious). We can add 7 to both sides (or subtract 7 from both sides). 3x + 7 – 7 = 1 – 7 3x + 0 = 6 3x = 6 This last equation is simpler than the original, but we can get simpler. Next Slide

  12. More on techniques We can restate our Basic Principle like this. A mystery value for x will make both expressions equal. Therefore, if we do the same thing to both sides, then the same x will still make both sides equal. In particular, by adding or multiplying both sides of an equation by the same number, we can produce simpler equations with the same solutions. In this example, we can simplify our equation further by multiplying both sides by 1/3. x = 2 Next Slide

  13. Continued The last equation is so simple, we’ll use it to say what the solutions are, and we’ll actually call an equation like x = 2, a solution. Here’s another example. 7x – 3 = 5 7x – 3 + 3 = 5 + 3 7x = 8 Next Slide

  14. Practice Problems • Solve 5x + 3 = 23 • Solve 3x + 5 = 14 Click for answers. 1) x = 4; 2) x = 3. Next Slide

  15. Terms Consider the equation 2x – 3 + x = 9 In the order of operations, we go in the order exponents  multiplication  addition Exponents form the strongest bonds, and addition the weakest. In this equation, the weakest bonds, the additions, separate the expressions into 2x, 3, and x on the left side And just the 9 on the right. These four things are called terms. The 2x and x are called x-terms, And the 3 and the 9 are called constant terms. Next Slide

  16. Continued One Basic Principle we’ll use to determine what’s simpler is the following. Simpler equations have fewer terms. On the left of the equation we’re considering, we have 2x + x. That is, two x’s plus one x. Altogether, there are three x’s. This is the distributive property in action. 2x + x = (2 + 1)x = 3x. In the equation we have 2x – 3 + x = 9 3x – 3 = 9. We can always combine like terms to reduce the number of terms. And this makes the equation simpler. Next Slide

  17. Continuing Example We’re trying to reduce the number of terms, and we also know that we want to end up with an equation of the form x equals something. We can get rid of the constant term 3 on the left by adding 3 to both sides. 3x – 3 + 3 = 9 + 3 3x = 12 From here we can divide both sides by 3, and end up with x = 4. An equation with only x-terms and constant terms is called a linear equation. Given a linear equation, we can always find its solutions (except for the special cases mentioned earlier, there will be just one solution). Let’s finish off by looking at a few examples. Next Slide

  18. Example Consider the equation 4x– 2(3x – 7) = 2x – 6. We first want to make this look like a linear equation. The 2(3x – 7) is not an x-term or a constant term. The distributive property says multiply the 2 times the 3xand the 7. That is, 2(3x) and 2(7). If we do this, we get a linear equation. 4x– 6x + 14 = 2x – 6 Now combine like terms. 2x + 14 = 2x – 6 (continued) Next Slide

  19. Continued To finish this problem, we would like only an x-term on one side and only a constant term on the other. We can get this by adding terms to both sides. 2x + 14 – 2x = 2x – 6 – 2x 4x + 14 = 6 4x + 14 – 14 = 6 – 14 4x = 20 Now divide both sides by 4 (or multiply by 1/4). Next Slide

  20. Example Consider the equation 3(x – 4) = 6x– (4 – 5x) 3x – 12 = 6x– 4 + 5x 3x – 12 = 11x – 4 3x – 12 – 11x = 11x – 4 – 11x 8x – 12 = 4 8x – 12 + 12 = 4 + 12 8x = 8 Next Slide

  21. Practice Problems • Solve 8 – 3(2x – 4) = 5 + 3x – 4x • Solve 2 + 5(x + 1) = 3x + 3 Click for answers. 1) x = 3; 2) x = 2. Next Slide

  22. Example Given an equation with fractions, most people like to use the basic trick of multiplying both sides of the equation by a common denominator. For this equation, 12 works, since 4, 6, and 3 all divide 12. 24 works also, but you will get bigger numbers. Remember: By the distributive property, you must multiply every term. This gets you back to stuff we’ve already seen. Next Slide

  23. Example Consider the equation 3.9x – 5.7(2x – 3) = 9.3 Here we have decimal numbers. Do everything the same way as you would with whole numbers. You may want to use your calculator, and you may need to round. When you round, the more places you keep the better. Keep at least a couple of places past the decimal point. 3.9x – 11.4x + 17.1 = 9.3 7.5x + 17.1 = 9.3 7.5x = 7.8 Next Slide

  24. Practice Problems 1. Solve the equation 2. Solve the equation Click for answers: 1) x = 1; 2) x = 0. Note: An answer of x = 0 is fine. Zero is a perfectly good number, and can be a solution. This is very different from not having a solution, which means that no number for x will work. End

More Related