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## Find the Quadratic Function That Models 3 Ordered Pairs

In a previous post about quadratic functions and the graphs, I discussed the standard form of a quadratic equation, $f(x)=ax^2+bx+c$. Because a parabola is symmetric, its equation can be found by solving a system of equations with three variables. I have only discussed solving equations with 2 variables previously.  If you know how to solve a system with 2 variables, then a system with 3 variables should be no problem. The process is rather simple. I will explain how to solve the system of three equations as well as a system of two equations in the examples.

### Example 1 Find the Quadratic Function That Models 3 Ordered Pairs

Before discussing how to find the solution, there should be a discussion about what is to be found. Ultimately, the solution should be an equation in the standard form of a quadratic, $f(x)=ax^2+bx+c$. To find this equation or function involves multiple parts and each part consists of several steps. Many times we get so caught up in the details, we forget the overall picture.

The answer is: $f(x)=x^2-6x+7$. Yes, I just gave you the answer. Now, the question is:

Find the quadratic function that models the given ordered pairs: (2,-1), (3,-2), (1,2). Here is how you turn 3 ordered pairs into a quadratic function.

First you must create a system of three equations. You create this system of equations by substituting the ordered pairs into the standard form of a quadratic equation $f(x)=ax^2+bx+c$. Using the reflexive and substitution properties of equality, I use the equation $ax^2+bx+c=y$. This will make solve the system of equations easier.

Substituting the ordered pairs gives the following equations:

(2,-1) gives $a(2^2)+b(2)+c=-1$

(3,-2) gives $a(3^2)+b(3)+c=-2$

(1,2) gives $a(1^2)+b(1)+c=2$

Simplifying these equations, maintaining the order from above and labeling them as equations A, B and C results with:

$4a+2b+c=-1 (A)$

$9a+3b+c=-2 (B)$

$a+b+c=2 (C)$

Now that you have a system of three equations, it is good to review the overall goal, to find the coefficients of a quadratic function that models the ordered pairs given. Okay, on with solving the system of 3 equations. You need to use the 3 equations and the elimination method to create a new system of equations with only 2 variables. You have to choose two pairs of equations and eliminate a variable. For this example, I will subtract C from A to create a new equation D and I will subtract C from B to create equation E.

$(A) 4a+2b+c=-1$

$\underline{-(C) a+b+c=2}$

$(D) 3a+b=-3$

$(B) 9a+3b+c=-2$

$\underline{-(C) a+b+c=2}$

$(E) 8a+2b=-4$

Equations D and E form a new system of equations.

$3a+b=-3 (D)$

$8a+2b=-4 (E)$

To solve this system, it would be easiest to subtract with E from equation D multiplied by 2.

$2(3a+b=-3) (D)$

$\underline{-(8a+2b=-4)} (E)$

$6a+2b=-6 (D)$

$\underline{-(8a+2b=-4)} (E)$

$-2a=-2$  Divide both sides by -2 and

$a=1$

You now have the coefficient of the quadratic term. Using a = 1, you substitute into either equation D or E. For this problem, I used equation D.

$a=1$

$3a+b=-3$

$3(1)+b=-3$  Substitute 1 for a.

$3+b=-3$  Simplify.

$b=-6$  Subtract 3 from each side.

This is the coefficient of the linear term. Now use both a = 1 and b = -6 to find the value of the constant. You must use one of the original equations A, B or C. I used equation C because it is very simple.

$a=1 and b=-6$

$a+b+c=2 (C)$

$1+(-6)+c=2$  Substitute 1 for a and -6 for b.

$-5+c=2$  Simplify.

$c=7$  Add 2 to both sides.

Finally, we have all of the coefficients a = 1, b = -6 and c = 7 and can write the quadratic function that models the given points.

$a=1, b=-6, c=7$

$f(x)=ax^2+bx+c$

$f(x)=x^2-6x+7$ This is the solution.

But how do you know it is correct? You should check your work by substituting the values a = 1, b = -6 and c = 7 into one of the equations A, B or C. Since I already used C I will use equation A.

$a=1, b=-6, c=7$

$4a+2b+c=-1 (A)$

$4(1)+2(-6)+7=-1$  Substitute.

$4+(-12)+7=-1$  Simplify.

$-8+7=-1$  Simplify.

$-1=-1$  Simplify and it checks.

This is just one example of finding the equation of a quadratic function from three points. Also, because the three original equations started with all three variables, it was a more complicated type.

### Example Find the Quadratic Function That Models 3 Ordered Pairs For You To Do

Find the quadratic function that models the given ordered pairs:

(-2,-17), (-1,10), (5,-10).

## 3.2 Solving Systems of Equations with Equivalent Systems

I have not blogged about equivalent equations, but to solve a system of equations sometimes requires you to change one or both of the given equations.  Below is an example of two equivalent equations.

5x + 25y = -55

You can multiply the equation above by any number you wish. Keep in mind you can multiply by fractions to create an equivalent equation. Multiplying 5x + 25y = -55 by 1/5 gives

x + 5y = -11 .

When you encounter a system of equation that does not have the same or opposite coefficients, you will need to create one or two equivalent equations, thus the name of the lesson is Solving Systems of Equations with Equivalent Systems. Once you create your equivalent system, you proceed with the same steps outlined in my post on Solving Systems of Equations Using the Elimination Method.

3x + 7y = 15

## Example 1 Solving a System of Equations with an Equivalent System

4x + 6y = 36 (1)

-3x + 5y = 49 (2)

You should notice that neither of the variable terms have the same coefficient with the same or opposite signs. You should create 2 equivalent equations. There are several choices for this problem, but for this example the x-terms will be eliminated. To eliminate the x-terms, equation 1 should be multiplied by 3 and equation 2 should be multiplied by 4. The will give the equivalent system

12x + 18y = 108 (4),

-12x + 20y = 196 (5).

Now that the x-terms have the same coefficient with different signs, the equations will be added term by term to give:

38y = 304 (6).

The y-coordinate of this solution is found by dividing both sides of equation 6 by 38 to show

y = 8 (7).

At this point, any equation 1 through 4 could be used to find the value of the x-coordinate. Personally, I will choose one of the original equations. Substituting 8 in for the y in equation 1,

4x + 6(8) = 36 (8)   Substitution Property (=)

4x + 48 = 36 (9)    Simplify

4x = -12 (10)   Subtraction Property (=)

x = -3 (11)   Division Property (=)

So, this system has one solution of (-3,8). Even if you are confident in your math, you should check your solution by substituting the ordered pair and simplifying. Using equation 2,

-3(-3) + 5(8) = 49 (12)   Substitution Property (=)

9 + 40 = 49 (13)   Simplify

49 = 49 (14)   Simplify.

Since 49 indeed equals 49, this system is consistent and independent, thus it has one solution (-3,8).

## Example 2 Solving a System of Equations with an Equivalent System

Solving a System of Equations with an Equivalent System

In the image to the right the original system is on lines 1 and 2. The original system is change into an equivalent system by multiplying equation 1 by 5 and equation 2 by 3. The result is equations 3 and 4 on lines 3 and 4 respectively.

Because the y-terms have the same coefficient with opposite signs the equations will be add term by term to eliminate them. This give line 5 with only the x-term remaining. Dividing both sides of line 5 by 56, gives the x-coordinate as -1.

On line 8 and using equation 1, substitute the -1 in for the variable x and solve the equation by adding 6 to both sides and dividing both sides by 3 to give y = -3 (lines 8 – 13). The end result is the ordered pair of (-1,-3).

As always, checking the solution is an essential ingredient to solving any algebra problem. As the image show, the ordered pair does satisfy equation 2. Therefore, this system is consistent and independent.

## Solving Systems of Equations Using the Elimination Method

There are two ways to solve a system of linear equations. This post focuses on the elimination method. There are two examples of using the elimination method to solve a system of equations in this post.

When solving a system of two equations with elimination, you want look at the x and y terms. Be on the lookout for variables that have the same coefficient with the same or opposite signs. This method’s name describes it accurately. When you use elimination, you “eliminate” one of the variables by adding or subtracting the given equations together. The variable terms with the same coefficient is eliminated. The signs of the variable terms to be eliminated determine whether the equations are added or subtracted. If the signs are the same, you subtract the equations. You add the equations when the signs of the variable terms are different.

After eliminating one of the variables, you have a simple one step equation to solve. The resulting equation should involve multiplication of a real number. The method to solve the equation will depend on the type of coefficient. Once the value of the first variable is determined, you will need to substitute the value you just found into one of the original equations and solve for the other variable.

Now you have the values of the ordered pair and should be a solution to both of the original equations. You really should test the values to ensure an accurate answer.

### Example 1 Solve the System of Equations with the Elimination Method

x + 2y = 3  (1)

4x – 2y = 7  (2)

Looking at the equations, it is clear that the y-terms have the same coefficients with opposite signs. Thus, the equations 1 and 2 should be added term by term to get:

5x = 10  (3).

This simple equation only takes one step to solve. To isolate x, use the division property of the equality to give

x = 2  (4).

Use the result in equation 4 and substitute it into either equation 1 or 2. Using equation 1,

2 + 2y = 3  (5).

Subtracting 2 from both sides of equation 5 gives,

2y = 3  (6).
To finish up take equation 6 and divide it by 2 and

y = ½ or 0.5 (7).

The ordered pair that will satisfy both of the original equations is (2, 0.5).

Remember how this related to the graph of a system of equations. These equations have one unique solution, which means (2,0.5) is the point where the graphs will cross. To check the solution, choose one of the original equations and do the math. Using equation 1 and (2,0.5).

4(2) – 2(0.5) = 7  (8) –> substitute the x- and y-values into equation 1

8 – 1 = 7  (9) –>  multiply

7 = 7 (10) –>  subtract

In equation 10, 7 is equal to 7 and thus the ordered pair is the solution to the system of equations:

x + 2y = 3  (1)

4x – 2y = 7  (2)

### Example 2 Solve the System of Equations with the Elimination Method

5x + 3y = -19 (1)

8x + 3y = 25 (2)

## Systems of Equations and Problem Solving – How to Set Up a System of Equations

There are two ways to solve a system of equations algebraically: the elimination method and the substitution method. Though the elimination method can be used at anytime, there are certain problems that lend them self to the use of the substitution method. Now that you have seen some examples on solving systems of equations with substitution, I can move on to some problem solving. The topic of this post is setting up and solving a system of two equations with the substitution method. The problem that follows is the classic part + part = whole relationship. The two parts are the cost of a slice of pizza and the cost of a soda. The whole is the total. Since there are two variables in the problem, there must be two equations. Read more to find out how to set up and solve a system of linear equations.

Example 1

At Renaldi’s Pizza, a soda and two slices of the pizza of the day costs \$10.25. A soda and four slices of the pizza of the day costs \$18.75. Find the cost of each item.

As you can see, the solution begins with the defining of variables: s for the price of one soda and p for the price of 1 slice of pizza. This is a fairly easy system of linear equations to set up because of the simple part + part = whole relationship. The two parts are the cost of the soda and the cost of the pizza. The whole is represented by the total cost. Thus, equations A and B are written based on the first two sentences of the problem.

Problem Solving - System of Equations - Pilarski

Lines 3 and 4 are still equations A and B, but it is an equivalent system. The variable term in each equation was subtracted from each side to arrive at the new equations A and B. The equation in line 5 is from applying the substitution property to replace the variable s with the 10.25 – 2p from equation A in line 3. Line 6 is from by subtracting 4p and 10.25 from both sides of line 5. Finally, line 7 is the result of applying the division property of equality to line 6. The result: 1 slice of pizza (p) costs \$4.25.

This problem is not completely solved. The cost of one soda is still unknown. To find the cost of the soda, the cost of a slice of pizza must be substituted into any of the equations A or B. It appears the original equation A is used in the diagram on line 8. Line 9 shows the substitution of 4.25 into an original equation. The equation in line 10 is possible by simplifying 2 times 4.25 and the result in line 11 is from subtracting 8.50 from both sides of 10.

The problem finishes up with a brief sentence explaining the results. I try to embed a video on all of my posts. The video is the same problem, but if you are an auditory learner, it might help you more to hear me saying the things I write about in my posts. Until next time.

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## Solving Systems of Equations with Substitution

There are two ways to solve a system of equations algebraically: the elimination method and the substitution method. Though the elimination method can be used at anytime, there are certain problems that lend them self to the use of the substitution method. The topic of this post is solving a system of two equations with the substitution method. The problems covered in this post are basic, that is if you think algebra 2 is a basic course! By basic, I mean there is no problem solving. We are given the equations and told to solve them. Once solving the system is learned, then we can look at problem solving related to solving a system of equations.

Example 1 – Solve the system of equations using the substitution method.

As you can see in the figure below, I have the problem worked out. I have the given equations label A and B and I will refer to them as such as I solve the problem in paragraph format. This is one way to write mathematics. If your teachers are like me, then I am sure many of them spend time getting you to explain your answers. Here is what I think would be a viable way for you to write out how to solve a problem. You may notice the green numbers to the right of the problem. This is so you can follow along with needing to count the lines I am referring to in the problem. Let’s get started solving a system of two equations.

The Figure Above - Solving a System of Equations

In the problem to the right, lines 1 and 2 of the represent the given system of equations. I labeled the first equation A and the second equation B.

I could have solve equation A for x or y because each term had a coefficient of 1 and I could have solved equation B for y because it had a coefficient of 1. From my diagram, line 3 contains equation A, which I decided to solve for y and the equivalent equation A is on line 4.

Using the substitution property of equality I replace the y in equation B with the 10 – x (line 5) and the new equation B is contained in line 6. Simplifying the equation by combining the 2x and –x  gives line 7. The subtraction property of equality allows us to isolate the x with a value of 5 (line 8).

Since I was able to find an x value, I will be able to find a y value. In line 9, I used equation A to solve for y. I substituted 5 in for y , line 10. Subtracting 5 from both sides of the equation in line 10 will give the result for y. It happens to be 5.

Thus, this system is independent and consistent and has one solution of (5,5).

The video below covers the problem above and additional problem. Coming soon, parts 2 and 3 of this lesson on solving systems of equations, I will tackle to problem solving situations.

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