Pre-Algebra II: Systems of Linear Equations A system of equations is a set of two or more equations that are related. One thing we often want to do for a system of equations is to find the places where the graphs of the equations in the system intersect. This is equivalent to finding the values of x for which the equations generate the same y-value. A system of linear equations will often be combined with a brace. This will look like: ìï y = 3x + 2 ìï 2x + 5y = 10 or í í ïî y = 2x + 1 ïî 3x - 4y = 12 We have a couple of methods available to us for solving systems of equations. The first method is to solve by graphing. This is sometimes impractical to do by hand, but the first system above actually has a nice solution that can be found by graphing, so let’s go for it.
Solving by Graphing ìï y = 3x + 2 Problem: Graph each of the lines in the system í to find the solution. ïî y = 2x + 1
Click here for the solution to this system. Now, you have—in your possession at all times during class—a device that is practically designed to solve systems by graphing. It’s your graphing calculator! Make sure you know how to solve systems by graphing them on your calculator. That is almost certainly one of the most common ways you’ll be doing it.
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ìï 2x + 5y = 50 Example: Solve the system í by graphing on your calculator. ïî 3x - 4y = 96 Solution: Before you can graph the equations on your calculator, you have to first solve each of the equations for y. Let’s do that here. For the first equation in the system: Step 1: Start with the original equation.
2x + 5y = 50
Step 2: Subtract 2x from both sides.
5y = 50 - 2x
Step 3: Divide both sides by 5.
y=
50 - 2x 5
For the second equation in the system: Step 1: Start with the original equation.
3x - 4y = 96
Step 2: Subtract 3x from both sides.
-4y = 96 - 3x
Step 3: Divide both sides by -4 .
y=
96 - 3x -4
ì 50 - 2x ï y= ï 5 So the system we’re solving by graphing is really this: í . I’m not ï y = 96 - 3x ïî -4 particularly interested in even remotely trying to simplify the right sides of those equations because the calculator is going to do the work for us anyway. Now, let’s solve this system by graphing it on our TI-Nspires. Click the image below to watch a YouTube video of me solving this system on a TI-Nspire.
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ìï 19x + 2y = 100 Example: Solve the system í by graphing on your calculator. ïî 23x - 3y = 210 Solution: Before you can graph the equations on your calculator, you have to first solve each of the equations for y. Let’s do that here. For the first equation in the system: Step 1: Start with the original equation.
19x + 2y = 100
Step 2: Subtract 19x from both sides.
2y = 100 -19x
Step 3: Divide both sides by 2.
y=
100 -19x 2
For the second equation in the system: Step 1: Start with the original equation.
23x - 3y = 210
Step 2: Subtract 23x from both sides.
-3y = 210 - 23x
Step 3: Divide both sides by -3 .
y=
210 - 23x -3
ì 100 - 19x ï y= ï 2 So the system we’re solving by graphing is really this: í . Again, I ï y = 210 - 23x ïî -3 don’t particularly care to simplify those since I’m just going to graph to find the solution anyway. Now, let’s solve this system on our TI-Nspires. Click the image below to watch a YouTube video of me solving this system on a TI-Nspire.
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To successfully solve a system by graphing on your calculator you’ll need to know how to:
Solve each equation for y; Graph each equation; Change the window; and Find intersection points.
Make sure you know how to do those things! If you watch the videos for both of the systems above, and follow along with your calculator, you’ll know how to do each of those things. You need to practice it a few times to make sure you can go through each step without having to wonder what to do next.
Now Do This Solve each of the following systems by graphing on your TI-Nspire. Round your answers to three decimal places. ìï -3x + 5y = 40 ìï 4x + 5y = -140 1. í (Sol.) 2. í (Sol.) ïî 2x + 3y = -100 ïî 5x + 2y = -20
ìï x + 2y = 20 3. í (Sol.) 4x y = 380 îï
Pre-Algebra II: Solving Systems
ìï 4x - y = 700 4. í (Sol.) x 3y = -50 îï
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Solving Systems by Substitution Solving by graphing is nice, but it doesn’t always get us the exact solution and sometimes we want that. There are several other methods we can use. The first one I’ll highlight is called substitution. Substitution is a great method to use when one of the equations is already solved for a one of the variables—it’s even better when both of them are solved for the same variable. Reading not your thing? If you’d rather watch YouTube videos than read, click here.
ìï y = 5x - 3 Example: Solve the system í using substitution. ïî y = 2x + 1 Solution: Notice how both 5x - 3 and 2x +1 are equal to y? That means, by the transitive property that they are equal to each other. (Humans innately seem to understand transitivity. It’s why we think that people should be treated the same— we’re applying transitivity of treatment to equal people. Anyway…), So we have… Subtract 2x from both sides: Add 3 to both sides: Divide both sides by 3:
5x - 3 = 2x +1
3x - 3 = 1 3x = 4 x=4 3
Now we need to find the y-coordinate. Since y = 2x +1 and x = 4 3 , we have: 8 8 3 11 æ 4ö y = 2 ç ÷ +1 Þ y = +1 Þ y = + Þ y = è 3ø 3 3 3 3
So the solution to the system is the ordered pair ( 4 3,11 3) . You should check that by graphing on your calculator. The figure to the left is the graph of both lines and their intersection point that I created using GeoGebra on my computer. You can see that it would be tough to really estimate that point. In this case you might get it right, but only because you’re familiar with estimating thirds of things. But you can see the intersection is right where we thought it would be. Make sure to check on your calculator!
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ìï 2x + 5y = 8 Example: Solve the system í using substitution. ïî y = 2x + 1 Solution: Notice how the second equation is already solve for y? It’s telling us that anywhere we see y in the first equation we can replace it with 2x +1. Let’s do that and solve! Replace y in the first equation: Distribute the 5: Collect like terms on the left: Subtract 5 from both sides:
2x + 5 ( 2x +1) = 8 2x +10x + 5 = 8 12x + 5 = 8
Divide both sides by 12: Simplify:
x = 3 12 x =1 4
12x = 3
So we know that x = 1 4 is the x-coordinate, but we need the y-coordinate. Clearly the second equation is easier to use to find y, so let’s do that:
1 1 2 3 æ 1ö y = 2 ç ÷ +1 Þ y = +1 Þ y = + Þ y = è 4ø 2 2 2 2 æ 1 3ö So the solution to our system is x = 1 4 and y = 3 2 , or the ordered pair ç , ÷ . è 4 2ø You should check that by graphing on your calculator. So, here’s the graph I got using GeoGebra on my computer. You can see that the point is right where we expected it to be. You should make sure that you can do this on your calculator! Don’t just read and accept what I’m doing here.
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ìï 3x + 5y = 6 Example: Solve the system í using substitution. [You can use your ïî 5x - 2y = 9 calculator for arithmetic on this problem.] Solution: Sometimes it’s easier to just solve both equations for y and go from there. I’m going to do that in this problem because, well, it’s not really obvious that solving either equation is better than the other. Doing this will turn this example into something really similar to the first example. Working on First Equation 3x + 5y = 6
Working on Second Equation 5x - 2y = 9
5y = 6 - 3x
-2y = 9 - 5x
y=
6 - 3x 5
y=
9 - 5x -2
So the original system now looks like this:
ì 6 - 3x ï y= ï 5 í ï y = 9 - 5x ïî -2 Since both equations are solved for y, I can set them equal to each other using transitivity. Let’s do that and then solve for x.
6 - 3x 9 - 5x = 5 -2 -2 ( 6 - 3x ) = 5 ( 9 - 5x )
Setting both equations equal: Cross multiply:
-12 + 6x = 45 - 25x
Distribute on both sides: Add 25x to both sides:
-12 + 31x = 45
Add 12 to both sides:
31x = 57
Divide both sides by 31:
x = 57 31
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Now that we know what x is, we have to solve for y. I’m going to use a calculator to find that, but I’ll use both equations to show that it doesn’t matter which you use. Using the first equation æ 57 ö 6 - 3ç ÷ è 31 ø y= 5 3 y= 31
Using the second equation æ 57 ö 9 - 5ç ÷ è 31 ø y= -2 3 y= 31
So the solution to our system is x = 57 31 and y = 3 31 , or the ordered pair ( 57 31, 3 31) . This problem was a lot of work, but there’s a tremendous feeling of satisfaction when you’re able to follow through so many steps and get the answer correct. If you’re not experiencing that success, keep practicing! Trust me. It’s worth it. Here’s the graph I made on GeoGebra. You should be checking on your calculator. You can see the intersection point is where we expected it to be (x is a little less than 2 and y is close to 0), but there’s no way we’d have solved this by graphing by hand.
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ìï 4x - 7y = 14 Example: Solve the system í using substitution. [You can use your ïî -3x + 4y = 8 calculator for arithmetic on this problem.] Solution: We look at this and there’s no obvious choice for what to solve for y, so we just go right ahead and solve both equations for y. Working on First Equation 4x - 7y = 14
Working on Second Equation -3x + 4y = 8
-7y = 14 - 4x
4y = 8 + 3x
y=
14 - 4x -7
y=
8 + 3x 4
Since both equations are now set equal to y, we can set them equal to each other, using the transitive property, and solve for x. Let’s do it!
14 - 4x 8 + 3x = -7 4 4 (14 - 4x ) = -7 ( 8 + 3x )
Setting both equations equal: Cross multiply:
56 -16x = -56 - 21x
Distribute on both sides:
56 + 5x = -56
Add 21x to both sides:
5x = -112 x = -112 5
Subtract 56 from both sides: Divide both sides by 5:
Now we know that x = -112 5 in our solution, but we need to find y. I’ll use both equations and my calculator to find y—remember: it doesn’t matter which equation I use. I should get the same y-value. Using the First Equation æ 112 ö 14 - 4 ç è 5 ÷ø y= -7 74 y=5
Using the Second Equation æ 112 ö 8 + 3ç è 5 ÷ø y= 4 74 y=5
The solution to our system is x = -112 5 and y = -74 5 , or the ordered pair ( -112 5,-74 5 ) .
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In this case, the graph is useful if I can use some built in function to find the intersection point (which obviously the TI-Nspire will let me do), but otherwise not so useful to me. It does confirm the solution is in the third quadrant, though, and that’s reassuring.
Steps to solving a system by substitution: 1. Solve one of the equations for whichever variable is easiest to solve for. 2. Substitute this into the other equation. 3. Solve for the only variable left in your equation. 4. Substitute back into one of the original equations to find the value of the other variable. (This is sometimes called back solving, which is kind of a cool name.) You hopefully just followed closely along with several examples of this method. That’s not nearly enough to become good at it. You should really try as many as you have time for to become a beast at these. Video Examples Click on each of the images to go to a YouTube video of me working through an example of solving a linear system by substitution.
Now go back and read through all of the written examples! The more you do the better. Click here to do that.
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Now Do This Solve each of the following systems by using substitution. You should check every system you solve on your calculator. ìï y = 3x + 6 ìï y = -5x + 4 1. í (Sol.) 2. í (Sol.) y = 14 x 4 y = 2x + 15 îï îï
ìï 4x + 3y = 16 3. í (Sol.) y = 5x 3 îï
ìï 8x - 3y = -36 4. í (Sol.) 6x + 5y = 44 îï
ìï 4x - 7y = 28 5. í (Sol.) ïî 5x - 12y = 30
ìï 9x - 3y = 20 6. í (Sol.) ïî 6x + 7y = -25
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Solving Systems by Linear Combinations Another method of solving systems of linear equations is called linear combinations. This method is kind of a big deal because it’s very easy to program a computer to do it. Also, your book calls it elimination. Linear combinations is a great method to use when both of the equations are in standard form, which you might recall looks like Ax + By = C . Reading not your thing? If you’d rather watch YouTube videos than read, click here.
ìï 3x + 5y = 12 Example: Solve the system í using linear combinations. 5x 5y = 28 îï Solution: If you look at the two equations in the system, you’ll notice that they have equal but opposite coefficients for y. That means that if we add the two equations together, the y terms will cancel out. Let’s do that!
3x + 5y = 12 + 5x - 5y = 28 8x + 0y = 40 From here we can solve the equation 8x = 40 by dividing both sides by 8 to get x = 5 . Once we know that x = 5 we can use either of the equations to solve for y since neither looks easier to use. Using the first equation 3( 5 ) + 5y = 12
Using the second equation 5 ( 5 ) - 5y = 28
15 + 5y = 12 5y = -3
25 - 5y = 28 -5y = 3
y = -3 5
y = -3 5
Notice that we found the same y-coordinate from both of the equations when we substituted x = 5 ? That better happen or we know we’re wrong. So the solution to the system is x = 5 and y = -3 5 , or the ordered pair ( 5,-3 5 ) . You should check that by graphing on your calculator.
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ìï 6x - 3y = 4 Example: Solve the system í using linear combinations. ïî 2x + y = 5 Solution: This problem is a little different from the previous problem because neither variable has equal and opposite coefficients. We’re going to force that to happen. Notice that if I multiply everything in the second equation by 3, the coefficient of y would be 3? Then we’d have equal but opposite coefficients for y. Let’s do that.
3( 2x + y ) = 3( 5 ) 6x + 3y = 15 So now our system looks like this:
ìï 6x - 3y = 4 í ïî 6x + 3y = 15 Now if we add the equations we’ll eliminate the y variable, so let’s do it.
6x + 6x
- 3y = 4 + 3y = 15
12x + 0y = 19 We can solve the equation that we’re left with on for x by dividing both sides by 12 to get x = 19 12 . Now we need to solve for y. If you look back at the original equations, the second one is definitely going to be easier to use.
19 19 30 19 11 æ 19 ö 2ç ÷ + y = 5 Þ + y = 5 Þ y = 5 - Þ y = - Þy= è 12 ø 6 6 6 6 6 So the solution is x = 19 12 and y = 11 6 , or the ordered pair (19 12,11 6 ) . You should check that by graphing on your calculator.
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ìï 5x + 4y = 12 Example: Solve the system í by linear combinations. ïî 3x + 2y = 10 Solution: In this case if we multiply the second equation by a positive number and add the equations nothing will cancel since everything already has a positive coefficient. So instead, let’s multiply it by a negative number. I like killing off the yvariable when I do this, so let’s multiply the second equation by -2 .
-2 ( 3x + 2y ) = -2 (10 ) -6x - 4y = -20 So our system now looks like this:
ìï 5x + 4y = 12 í ïî -6x - 4y = -20 Now if we add down, we’ll be able to solve for x. Let’s do it.
5x
+ 4y = 12
+ -6x - 4y = -20 -x
+ 0y = -8
From the equation that we’re left with you can see that x = 8 . Looking at the two original equations, I’m not sure it matters which we go with to solve for y, so I’ll show the work for both. Using the first equation 5 ( 8 ) + 4y = 12
Using the second equation 3( 8 ) + 2y = 10
40 + 4y = 12
24 + 2y = 10
4y = -28 y = -7
2y = -14 y = -7
No matter which original equation you use, you end up with y = -7 . So the solution to the system is x = 8 and y = -7 , or the ordered pair ( 8,-7 ) .
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ìï 9x - 4y = -2 Example: Solve the system í by linear combinations. ïî 5x + 3y = 14 Solution: In this case there’s no easy fix to make the coefficients of one of the variables equal but opposite so we’ll have to modify both of the original equations. This is not a problem because there’s a sort of easy way to do this. Suppose we want to eliminate y. I will multiply the top equation by 3, the coefficient of y in the bottom equation and I will multiply the bottom equation by 4, the absolute value of the coefficient of y in the top equation. (I want the coefficients I end up with to have opposite signs, so I’ve planned accordingly.) Top Equation Work 3( 9x - 4y ) = 3( -2 )
Bottom Equation Work 4 ( 5x + 3y ) = 4 (14 )
27x -12y = -6
20x +12y = 56
Our revised system looks like this:
ìï 27x - 12y = -6 í ïî 20x + 12y = 56 Now if we add down, we’ll eliminate the y variable. Let’s do it.
27x - 12y = -6 + 20x + 12y = 56 47x + 0y
= 50
We divide both sides of the remaining equation to get x = 50 47 . I’m not excited about substituting that back into either of the original equations, but to prove it doesn’t matter, I’ll do it for both—with a calculator. Here goes… Using the first equation æ 50 ö -2 - 9 ç ÷ è 47 ø y= -4 136 y= 47
Using the second equation æ 50 ö 14 - 5 ç ÷ è 47 ø y= 3 136 y= 47
So we’ve found that the solution is x = 50 47 and y = 136 47 , or the ordered pair ( 50 47,136 47 ) .
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Video Examples: Click on each of the images to go to a YouTube video of me working through an example of solving a linear system by linear combinations (or elimination, if you prefer).
Now that you’ve finished watching the examples, go back and read the other examples! The more you do, the better off you’ll be! Click here to go back and read. Steps to solving a system of equations by linear combinations: 1. Decide which variable you’d like to eliminate and look at the coefficients. 2. Multiply one or both equations by a number that forces the coefficients of the variable you’re eliminating to be equal but opposite. 3. Add the two equations. This will “kill” the variable you’re eliminating. 4. Solve for the remaining variable. 5. Back solve to find the other variable’s value. I’ve done four examples of this…that’s not nearly enough to master it. Find more and do them! Do tons and tons of them…
Now Do This ìï x - y = 13 1. í (Sol.) 3x + 2y = 6 îï
ìï 7x + 3y = 12 2. í (Sol.) 4x + 9y = 8 îï
ìï 2x + 5y = 4 3. í (Sol.) 3x 2y = 6 îï
ìï 4x + 6y = 9 4. í (Sol.) 3x 5y = 4 îï
ìï 5x - 4y = -3 5. í (Sol.) 6x 5y = 12 îï
ìï -2x + 5y = 6 6. í (Sol.) 7x 3y = 14 îï
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Solving Systems by linSolve There is another way to solve a system of linear equations on your TI-Nspire. It has advantages and disadvantages though. This method involves using linSolve, the built-in linear equation solver.
ìï 3x + 25y = 8 Example: Solve the system í using linSolve. 12x -14y = 15 îï Solution: To watch a YouTube video of me solving this system using linSolve click the image below.
The advantages of linSolve include the following: The system can be entered as it’s shown. There’s no need to solve for y. If the system has no decimals in it, linSolve will return an exact answer (meaning fractions). You don’t need to change the graphing window around searching for the solution. The disadvantages of linSolve include the following: It can only solve linear systems. Many systems you’ll end up solving are not linear. Graphing will always let you find approximate solutions to a system. Since you don’t get to see a picture of the system, you lose any intuitive sense of if you’ve entered the system correctly.
Now Do This: Solve each of the following systems using linSolve on your TI-Nspire. ìï 3x + 5y = 20 ìï 12x - 5y = 14 1. í (Sol.) 2. í (Sol.) 5x 8y = 46 15x + 21y = 6 îï îï
ìï 4x + 14y = 25 3. í (Sol.) ïî y = 90 - 34 x
Pre-Algebra II: Solving Systems
ìï y = 51- 32x 4. í (Sol.) ïî y = 44 + 18x
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Here’s a flowchart of how I tend to think about solving systems of linear equations. This isn’t the way to go about it, but it’s a way that’s been very effective for me over the years.
Feel free to implement this plan.
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Classifying Systems You know what we really need? More words. That was sarcasm… Sorry about this, but there’s more vocabulary that goes along with linear systems. On the other hand, the vocabulary sounds like it comes from an interrogation in a murder-mystery, so it’s not terribly difficult to remember what everything means. An inconsistent system has no solutions. In a plane, the only way for a system to be inconsistent is if the equations represent parallel lines. In a murder-mystery this would be like having two “witnesses” who didn’t see anything or are lying; there’s no way to solve the mystery by talking to them. A consistent system has at least one solution. Consistent systems can be further divided into two types: Independent—a system having only one solution. The linear equations in an independent system will have different slopes. In the murder-mystery this is like having two witnesses who can each share some of the story and you put that together to solve the crime. Dependent—a system having infinitely many solutions. I personally hate even calling this a system of equations…it’s just one equation written two different ways. Anyway, in our murder-mystery, this would be if one person told another person what to say: you don’t get any useful info out of the second person— they’re just repeating the same story.
ìï 5x + 4y = 12 Example: Determine if the system í is inconsistent or consistent. If 20x + 16y = 40 îï consistent determine if it is independent or dependent. Explain. Solution: I went ahead and used linSolve on this system to see what it would say. Here’s the screen shot.
Well, okay…the calculator says that the system has no solution. If a system has no solution, we say that the system is inconsistent. Why would that be true? Take a look at the left sides of each equation—they’re multiples of each other! If we multiply the top equation by 4, the resulting system will look like this: Pre-Algebra II: Solving Systems
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ìï 20x + 16y = 48 í 20x + 16y = 40 îï There’s no way those two lines are going to intersect! They’re parallel but have different y-intercepts. We could have figured this out without the calculator, but I wanted you to practice and to see what it will look like when you enter a system to which there is no solution.
ìï 5x + 4y = 12 Example: Determine if the system í is inconsistent or consistent. If ïî 20x + 16y = 48 consistent determine if it is independent or dependent. Explain. Solution: Here’s the screen shot again.
What is that mess? Well, that’s the result of the calculator trying to solve a dependent system! There are an infinite number of solutions to this system. In fact, both equations in the system are the exact same equation. If you multiply the first equation by 4 you’ll get the second equation. Every point on the first line is also a point on the second line—since they’re the same line. It’s sort of confusing the first time you see that output on the calculator, but what the calculator is telling you is this: if you pick a value for c1 you can determine the value of x and y—and the resulting ordered pair is always on both lines.
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By copying and pasting c1, I had the TI-Nspire evaluate the expressions for two values. You can confirm that both of those ordered pairs are, in fact, on both lines.
Now Do This Use linSolve on your TI-Nspire to classify the system as consistent (specify dependent or independent) or inconsistent. ìï 5x + 4y = 3 ìï 5x - 3y = 4 1. í (Sol.) 2. í (Sol.) 7x 5y = 8 10x = 6y + 8 îï îï
ìï y = 3x - 5 3. í (Sol.) y = 6x 10 îï
Pre-Algebra II: Solving Systems
ìï y = 4x - 3 4. í (Sol.) 2y = 8x 6 îï
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Solutions Solving by Graphing Solution: Here’s the correct graph of the two lines in the system. You can see that the two lines intersect at the point ( -1,-1). That’s the solution to the system. Another way of writing the solution is to say x = -1 and y = -1. Return
1. After solving each of the systems for y, I graphed them and found the intersection point. You can see the minimum and maximum x and y that I used for my window.
You can see that the solution to three decimal points is ( -32.631,-11.579 ). (Return)
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2. After solving each of the systems for y, I graphed them and found the intersection point. You can see the minimum and maximum x and y that I used for my window.
You can see that the solution to three decimal points is (10.588,-36.471) . (Return) 3. After solving each of the systems for y, I graphed them and found the intersection point. You can see the minimum and maximum x and y that I used for my window.
You can see that the solution to three decimal points is ( 86.667,-33.333) . (Return)
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4. After solving each of the systems for y, I graphed them and found the intersection point. You can see the minimum and maximum x and y that I used for my window.
You can see that the solution to three decimal points is (195.455,81.818 ) . (Return)
Solving by Substitution 1. This system is set up so that we can immediately set both equations equal to each other and solve. Let’s do that. Set both equations equal: 3x + 6 = 14x - 4 Subtract 3x from both sides: 6 = 11x - 4 Add 4 to both sides: 10 = 11x Divide both sides by 11: x = 10 11 Now substitute the x = 10 11 into either of the original equations:
30 66 96 æ 10 ö y = 3ç ÷ + 6 Þ y = + = . è 11 ø 11 11 11 The solution is x = 10 11and y = 96 11 , or the ordered pair (10 11,96 11) . (Return)
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2. This system is set up so that we can immediately set both equations equal to each other and solve. Let’s do that. Set both equations equal: Add 5x to both sides: Subtract 15 from both sides: Divide both sides by 7:
-5x + 4 = 2x +15 4 = 7x +15
-11 = 7x
x = -11 7
Now substitute the x = -11 7 into either of the original equations:
22 22 105 83 æ 11ö y = 2 ç - ÷ +15 Þ y = - +15 Þ y = - + Þy= . è 7ø 7 7 7 7 The solution is x = -11 7 and y = 83 7 , or the ordered pair ( -11 7,83 7 ). (Return) 3. This system is set up so that we can immediately substitute one equation into the other. Let’s do that.
4x + 3( 5x - 3) = 16
Substitute: Distribute: Collect like terms: Add 9 to both sides:
4x +15x - 9 = 16 19x - 9 = 16 19x = 25 x = 25 19
Divide both sides by 19:
Now substitute x = 25 19 into either of the original equations:
125 125 57 68 æ 25 ö y = 5ç ÷ - 3 Þ y = - 3Þ y = Þy= . è 19 ø 19 19 19 19 The solution is x = 25 19 and y = 68 19 , or the ordered pair ( 25 19,68 19 ) . (Return)
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4. This system isn’t set up for substitution, really. Let’s solve each equation for y, and then set them equal to each other. Working on the first equation 8x - 3y = -36
Work on the second equation 6x + 5y = 44
-3y = -36 - 8x
5y = 44 - 6x
y=
-36 - 8x -3
y=
44 - 6x 5
Now let’s solve:
-36 - 8x 44 - 6x = -3 5 5 ( -36 - 8x ) = -3( 44 - 6x )
Set both equations equal: Cross multiply:
-180 - 40x = -132 +18x -180 = -132 + 58x -48 = 58x
Distribute: Add 40x to both sides: Add 132 to both sides:
x = -48 58 x = -24 29
Divide both sides by 58: Simplify the fraction:
Now substitute x = -24 29 into either of the original equations. I’ll do both and use a calculator. Using the first equation æ 24 ö -36 - 8 ç - ÷ è 29 ø y= -3 284 y= 29
Using the second equation æ 24 ö 44 - 6 ç - ÷ è 29 ø y= 5 284 y= 29
So the solution is x = -24 29 and y = 284 29 , or the ordered pair ( -24 29,284 29 ) . (Return)
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5. This system isn’t set up for substitution, really. Let’s solve each equation for y, and then set them equal to each other. Working on the first equation 4x - 7y = 28
Work on the second equation 5x - 12y = 30
-7y = 28 - 4x
-12y = 30 - 5x
y=
28 - 4 x -7
y=
30 - 5x -12
Now let’s solve:
28 - 4x 30 - 5x = -7 -12 -12 ( 28 - 4x ) = -7 ( 30 - 5x )
Set both equations equal: Cross multiply:
-336 + 48x = -210 + 35x
Distribute:
-336 +13x = -210 13x = 126
Subtract 35x from both sides: Add 336 to both sides:
x = 126 13
Divide both sides by 13:
Now substitute x = 126 13 into either of the original equations. I’ll do both and use a calculator. Using the first equation æ 126 ö 28 - 4 ç è 13 ÷ø y= -7 20 y= 13
Using the second equation æ 126 ö 30 - 5 ç è 13 ÷ø y= -12 20 y= 13
So the solution is x = 126 13 and y = 20 13 , or the ordered pair (126 13,20 13) . (Return)
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6. This system isn’t set up for substitution, really. Let’s solve each equation for y, and then set them equal to each other. Working on the first equation 9x - 3y = 20
Work on the second equation 6x + 7y = -25
-3y = 20 - 9x
7y = -25 - 6x
y=
20 - 9x -3
y=
-25 - 6x 7
Now let’s solve:
20 - 9x -25 - 6x = -3 7 7 ( 20 - 9x ) = -3( -25 - 6x )
Set both equations equal: Cross multiply:
140 - 63x = 75 +18x 140 = 75 + 81x 65 = 81x x = 65 81
Distribute: Add 63x to both sides: Subtract 75 from both sides: Divide both sides by 81:
Now substitute x = 65 81 into either of the original equations. I’ll do both and use a calculator. Using the first equation æ 65 ö 20 - 9 ç ÷ è 81 ø y= -3 115 y=27
Using the second equation æ 65 ö -25 - 6 ç ÷ è 81 ø y= 7 115 y=27
So the solution is x = 65 81 and y = -115 27 , or the ordered pair ( 65 81,-115 27 ) . (Return)
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Solving Systems by Linear Combinations 1. The first thing I’ll do here is multiply the first equation by 2 so that the coefficients of y are equal but opposite.
2 ( x - y ) = 2 (13) 2x - 2y = 26 So now our system looks like this:
ìï 2x - 2y = 26 í 3x + 2y = 6 îï Now if we add the equations we’ll eliminate the y variable, so let’s do it.
2x - 2y = 26 + 3x + 2y = 6 5x + 0y = 32 We can solve the equation that we’re left with on for x by dividing both sides by 5 to get x = 32 5 . Now we need to solve for y. If you look back at the original equations, the second one is definitely going to be easier to use.
32 32 65 32 33 33 - y = 13 Þ -y = 13 Þ -y = Þ -y = Þy=5 5 5 6 5 5 So the solution is x = 32 5 and y = -33 5 , or the ordered pair ( 32 5,-33 5 ) . (Return)
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2. The first thing I’ll do here is multiply the first equation by -3 so that the coefficients of y are equal but opposite.
-3( 7x + 3y ) = -3(12 ) -21x - 9y = -36 So now our system looks like this:
ìï -21x - 9y = -36 í 4x + 9y = 8 îï Now if we add the equations we’ll eliminate the y variable, so let’s do it.
-21x + 4x
- 9y = -36 + 9y = 8
-17x + 0y = -28 We can solve the equation that we’re left with on for x by dividing both sides by -17 to get x = 28 17 . Now we need to solve for y. I’m going to use both equations and a calculator. Using the First Equation æ 28 ö 12 - 7 ç ÷ è 17 ø y= 3 8 y= 51
Using the Second Equation æ 28 ö 8 - 4ç ÷ è 17 ø y= 9 8 y= 51
So the solution is x = 28 17 and y = 8 51 , or the ordered pair ( 28 17,8 51) . (Return)
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3. I can’t easily multiply one equation by a number and kill a variable so I’m going to multiply both equations and work from there. I’ll multiply the top equation by 2 and the bottom equation by 5. This will lead to a system in which the coefficients of y are equal but opposite. Top Equation Work 2 ( 2x + 5y ) = 2 ( 4 )
Bottom Equation Work 5 ( 3x - 2y ) = 5 ( 6 )
4x +10y = 8
15x -10y = 30
Our revised system looks like this:
ìï 4x + 10y = 8 í 15x - 10y = 30 îï Now if we add down, we’ll eliminate the y variable. Let’s do it.
4x
+ 10y = 8
+ 15x - 10y = 30 19x + 0y
= 38
We divide both sides of the remaining equation by 19 to get x = 2 . I’ll substitute this back into both of the original equations to find the value of y, which should be the same regardless of which equation I use. Using the first equation 4 - 2 ( 2) y= 5 y=0
Using the second equation 6 - 3( 2 ) y= -2 y=0
So we’ve found that the solution is x = 2 and y = 0 , or the ordered pair ( 2,0 ) . (Return)
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4. I can’t easily multiply one equation by a number and kill a variable so I’m going to multiply both equations by something and work from there. I’ll multiply the top equation by 5 and the bottom equation by 6. This will lead to a system in which the coefficients of y are equal but opposite. Top Equation Work 5 ( 4x + 6y ) = 5 ( 9 )
Bottom Equation Work 6 ( 3x - 5y ) = 6 ( 4 )
20x + 30y = 45
18x - 30y = 24
Our revised system looks like this:
ìï 20x + 30y = 45 í 18x - 30y = 24 îï Now if we add down, we’ll eliminate the y variable. Let’s do it.
20x + 30y = 45 + 18x
- 30y = 24
38x + 0y
= 69
We divide both sides of the remaining equation by 38 to get x = 69 38 . I’ll substitute this back into both of the original equations and use a calculator to find the value of y, which should be the same regardless of which equation I use. Using the first equation æ 69 ö 9 - 4ç ÷ è 38 ø y= 6 11 y= 38
Using the second equation æ 69 ö 4 - 3ç ÷ è 38 ø y= -5 11 y= 38
So we’ve found that the solution is x = 69 38 and y = 11 38 , or the ordered pair ( 69 38,11 38 ) . (Return)
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5. I can’t easily multiply one equation by a number and kill a variable so I’m going to multiply both equations by something and work from there. I’ll multiply the top equation by 5 and the bottom equation by -4 . This will lead to a system in which the coefficients of y are equal but opposite. Top Equation Work 5 ( 5x - 4y ) = 5 ( -3)
Bottom Equation Work -4 ( 6x - 5y ) = -4 (12 )
25x - 20y = -15
-24x + 20y = -48
Our revised system looks like this:
ìï 25x - 20y = -15 í -24x + 20y = -48 îï Now if we add down, we’ll eliminate the y variable. Let’s do it.
25x
- 20y = -15
+ -24x + 20y = -48 x
+ 0y
= -63
This gives us that x = -63. I’ll substitute this back into both of the original equations and use a calculator to find the value of y, which should be the same regardless of which equation I use. Using the first equation -3 - 5 ( -63) y= -4 y = -78
Using the second equation 12 - 6 ( -63) y= -5 y = -78
So we’ve found that the solution is x = -63 and y = -78 , or the ordered pair ( -63,-78 ) . (Return)
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6. I can’t easily multiply one equation by a number and kill a variable so I’m going to multiply both equations by something and work from there. I’ll multiply the top equation by 3 and the bottom equation by5. This will lead to a system in which the coefficients of y are equal but opposite. Top Equation Work 3( -2x + 5y ) = 3( 6 )
Bottom Equation Work 5 ( 7x - 3y ) = 5 (14 )
-6x +15y = 18
35x -15y = 70
Our revised system looks like this:
ìï -6x + 15y = 18 í 35x - 15y = 70 îï Now if we add down, we’ll eliminate the y variable. Let’s do it.
-6x + 15y = 18 + 35x - 15y = 70 29x + 0y
= 88
Dividing both sides of the resulting equation by 29 gives us x = 88 29 . I’ll substitute this back into both of the original equations and use a calculator to find the value of y, which should be the same regardless of which equation I use. Using the first equation æ 88 ö 6 + 2ç ÷ è 29 ø y= 5 70 y= 29
Using the second equation æ 88 ö 14 - 7 ç ÷ è 29 ø y= -3 70 y= 29
So we’ve found that the solution is x = 88 29 and y = 70 29 , or the ordered pair ( 88 29, 70 29 ). (Return)
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Solving Systems with linSolve 1. This is mostly just pressing buttons so here’s a screen shot of the result.
You can see that the solution to the system is x = 390 49 and y = -38 49 , or the ordered pair ( 380 49,-38 49 ) . (Return) 2. This is mostly just pressing buttons so here’s a screen shot of the result.
You can see that the solution to the system is x = 108 109 and y = -46 109 , or the ordered pair (108 109,-46 109 ) . (Return)
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3. This is mostly just pressing buttons so here’s a screen shot of the result.
You can see that the solution to the system is x = 1235 472 and y = 245 236 , or the ordered pair (1235 472,245 236 ). (Return) 4. This is mostly just pressing buttons so here’s a screen shot of the result.
You can see that the solution to the system is x = 7 50 and y = 1163 25 , or the ordered pair ( 7 50,1163 25 ) . (Return)
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Classifying Systems 1. Here’s the screen shot of using linSolve on the system:
Since there is a solution the system is consistent, because a solution exists, and independent, because there’s only one solution. (Return) 2. Here’s the screen shot of using linSolve on the system:
This result is the TI-Nspire’s way of saying that there are an infinite number of solutions to this system. We classify this system as consistent, because a solution exists, and dependent, because an infinite number of solutions exist. (Return) 3. Here’s the screen shot of using linSolve on the system:
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We classify this system as consistent, because a solution exists, and independent, because only one solution exists. (Return) 4. Here’s the screen shot of using linSolve on the system:
This result is the TI-Nspire’s way of saying that there are an infinite number of solutions to this system. We classify this system as consistent, because a solution exists, and dependent, because only one solution exists. (Return)
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