Project 6: Solving Equations, Part I We can use the following properties to rewrite equations (to put them in a particular form like them, or to combine multiple equations into a single equation). Properties that apply to equations specifically: ↔ ↔ ↔ ⋅
↔
⋅ ↔ 0 ↔ 0
⋅
⋅
⋅
⋅
0
0
↔ ↔
↔ 0
0
0 ↔
0 0 ↔ ∗
∗
∗
0
,
⋅
Properties for expressions, that can also be used on expressions inside equations: ⋯
⋯
⋅ …⋅
⋯
⋯
0 1
0 ⋅ ,
√
√
0 ⋅…⋅
0 ⋯
0 √ √ √
√
0 ⋅…⋅
⋅ …⋅
⋯
,…,
0 0 , or to solve What does it mean to solve an equation for a specific variable? Often we think of solving an equation as a process that leads to a specific answer, and sometimes that happens when we solve an equation. But that is not really what it means. Solving an equation just means that we want to isolate the given variable until it is by itself on only one side of the equation. So, for example, this equation is NOT solved for : 2
6
because we have ’s on both sides of the equation, and also because the on the left side of the equation is not by itself. because the is by itself on one side of the equation. The On the other hand, this equation IS solved for : fact that the other side of the equation still contains a bunch of other variables instead of a number is irrelevant. Which of the following equations have been solved for , and which have not? Circle each one that has been solved for , and cross out any that have not yet been solved for . 1.
2.
3.
2 3
2
7 4.
5.
3
6.
9 4 7. 8
8.
9.
√5 How do we solve an equation for a specific variable? Just like with simplifying or rewriting expression, the process of solving an equation for a specific variable is really just a process of replacing an equation with a different equation that is equivalent to it. How can we tell if two equations are equivalent? Two equations (or inequalities, or systems of equations) are equivalent if they have the same solution set. The solution set is the set of all numbers that could be put in for the variables that would make the equation true. So, for example: 

49 14 are equivalent because they both have the same solution set The two equations 5 2 and 7. As it turns out, 7 is the only value that we can substitute in for that will make the first or the second equation true. Even though these two equations don’t look at all similar, it turns out they are actually equivalent because they have the same solution set. The two equations and 5
5 are equivalent because they both have the same solution set: any combination of values that we substitute in for , , in the first equation will also work in the second equation, since the second equation is the same as the first with only a 5 added to each side of the equation. But if is equal to for some set of values of , , , then 5 will also be equal to 5 for that same set of values of , , . To show that two equations are equivalent, we can NOT use the equals sign. Since equations already contain equals sign, adding a third equals sign between two equations would be confusing. Instead, we use the double‐arrow, like this: 

5
2 ↔ ↔ 49
14 5
5 So, to rewrite an equation, we simply need to replace it with an equivalent equation. The most systematic way to do this is to use properties (or identities) to substitute some or all of the equation for an equivalent equation. This is just like what we did with expressions, except with equations we do have to think carefully about where we put the equals sign. Example: Put the equation 3
values of and . 2
1 into the form (where , are any numbers), and identify the First, just like with simplifying expressions, I will rewrite all subtraction as adding a negative: 3
2
1 ↔ 3
2
1 Then, just like with simplifying expressions, I will try to rewrite the equation without parentheses: Using 3,
,
2 for yields: 3
2
1 ↔ 3
3
2
1 Multiplying 3 ⋅ 2 gives us: 3
3
2
1 ↔ 3
6
1 Now, we need this to be in the form (where , are any numbers). So we start by getting the side with the in it on the left side and the side with the in it on the right side by using the property ↔ : Using 3
6,
1 for ↔ yields: 3
6
1 ↔ 1
3
6 Now, I almost have what I need, but I need for the 1 that is on the left side to go elsewhere (because in the form , the is by itself on the left side). I look at the possibly properties that I could use, and I notice that if I were to add 1 to the 1 these would leave me with a zero on the left side (which when added to the would just give me on that side). I notice that I have the following property which tells me that I can add anything I want to one side of an equation as long as I also add it to the other side: ↔ So, using 1,
3
6 and choosing 1 for ↔ yields: 1
3
6 ↔ 1
1
3
6
1 If I think of each term as a single unit, I can see that all terms on the left are being added, and all terms on the right are being added, so I can rewrite the expressions on both the left and the right side of the equation without the parentheses: 1
1
3
6
1 ↔ 1
1
3
6
1 Now once again since all the terms on the left and all the terms on the right are being added, I can combine them in any 1 first on the left and 6
1 first on the right: order. So I can add 1
1
1
3
6
1 ↔ 0
3
5 I almost have what I need. I just need to rewrite the equation without the 0 on the left side. But I have an identity that says that 0
. So if I let , I can rewrite this equation one last time by replacing the expression 0 on the left side of the equation with the expression instead: 0
3
5 ↔ 3
5 Now the equation has the form , and we have 3,
5. Now you try! For each of the following equations, put them into the requested form: 10. 2
3
11.
2
12.
5 6 1
13. 2
3
14.
5
15.
5 3 2
Put into Put into Put into Put into , , : Put into , , : Put into , , : form (where , are any numbers) and give the values of , : form (where , are any numbers) and give the values of , : form (where , are any numbers) and give the values of , : 0 form (where , , are any numbers) and give the values of 0 form (where , , are any numbers) and give the values of 0 form (where , , are any numbers) and give the values of Use the properties on the first page of the project to solve each of the following equations, by first simplifying each side of the equation or inequality, and then rewriting as needed to get the indicated variable by itself on one side. Then check your work by substituting the solution into the original equation. Original equation Example: Solve for the variable indicated Solve for : The fractions are making this equation more complicated than we would like, so my first goal is to rewrite the equation without them. Fractions are just division, so one way to do this is to multiply by the least common multiple of all the denominators—6 is a multiple of 2, and 3, and 6. So if I multiply both sides of the equation by 6, this should allow me to rewrite the equation without the fractions: ↔ 6 ⋅
↔6⋅
6⋅
6⋅
↔ ⋅
⋅
⋅
↔
⋅
⋅
⋅
⋅
⋅
⋅
↔
6⋅
↔
↔6
8 5 Now that we no longer have fractions to deal with, we proceed to try to get by itself on one side by thinking about what parts are in the way. The 8 needs to be rewritten: it is currently being added in the equation, so we add a negative 8 to both sides of the equation so that we get 0 on the left instead of 8: ↔ 6
8
8
5
8 ↔6
8
8
3 ↔6
0
3 Now we need to rewrite the 6 somehow—it is currently being multiplied in the equation, so we need to divide both sides of the equation by 6 so that we get 1 on the left instead of 6: ↔6
3 ↔
↔
↔1
↔
16. 2 7
6
2 Check your work and ↔ ↔ ↔ ↔ ⋅
⋅
↔ ↔ ↔ ⋅ ⋅
⋅
⋅
↔ ↔ Solve for : ⋅
⋅
TRUE! 17.
18. 3
6
4
19.
20. 6
2 8 1
5
2 3
3 Solve for Solve for Solve for Solve for : : : : Just like with the equations above, use the properties on the first page of this project to solve each of the following inequalities for the given variable, and check your answer by testing two sample values. Then graph the solution set. Solve for : Example: 10
10
10
10
10
10
10
10
10
10
10
10
2
3
3
3
1
21. 3 2
9 5 22. 5
3 3 23. 2
4 3 4
2
5 2
5
10 2
5
10 2⋅
2⋅5
10 2
10
10 2
10
10 10
2
0 0 2 2 2
2 0 0
3
0 0 Check answer by testing two values: Testing 1, which should be false, because 1 0: 10
1
2
1
5 10 1 2 4 11 8  FALSE! Testing 1, which should be true, because 1 0: 10
1
2 1
5 9 2 6 9 12  TRUE! 2 Graph the solution set 24. 12
2 3
1
5 2
For each of the equations, first put the equation into slope‐intercept form (
the intercept, and finally graph the line: Equation of the line: 25.
2
27. 2
28.
29.
4
1
2
2 4 2 2
y‐int: Graph 0, Is the slope positive, negative, zero, or undefined? 1 26.
Slope: ? Slope has magnitude less than one or greater than one? Slope intercept form: 2
1 , then identify the slope and For each of the pairs of equations, first put each equation into point‐slope form (
, then identify the slope and the intercept of each equation in the pair, and then graph the pair of lines. Finally, identify the point where the pair of lines intersect, or state that they are parallel or the same line: Equations of the lines: 1
30.
5
3
Slope intercept forms: Slope positive, negative, zero, or undefined?
Slope has Slope: y‐int: Graph magnitude less than 1 or greater than 1? Point of intersection:
Or if no intersection, parallel or same line? 2
3
3
31.
32.
33.
1
3
2
6
4
4
3
4
2
2
3
9
34.
35.
2
3
2
3
3
4
6
8
Solving Systems of Linear Equations: For each of the pairs of equations, pick a property that will allow you to combine the two equations into a single equation with only one variable. Then solve for that variable. After solving for the first variable, substitute that value into either original equation to find the value of the second value. Then check your answer. Properties: 1) ↔ 2) ↔ Equations of the lines: 1
36.
5
3
Which Use that property Solve to get a property is here to combine the number value for likely best to two equations into a the first variable: merge these single equation with two equations only one variable: into a single equation with For substitution, only one which equation variable? should be solved for which variable? Substitution or one of the If using the property properties #2 above, what above? should and be? Solve to get a number value for the second variable: Rewrite final solution here: Check that solution by substituting both the x and y values into both equations: 3 1
2
3
37.
38.
39.
3
2
6
4
3
4
2
4 2
2
3
9 3
2
3
3
4
6 8
40.
41.