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**Solving Systems of Equations**

3 Approaches Click here to begin Ms. Nong Adapted from Mrs. N. Newman’s PPT

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Method #1 Graphically POSSIBLE ANSWER: Answer: (x, y) or (x, y, z) Method #2 Algebraically Using Addition and/or Subtraction Answer: No Solution Answer: Identity Method #3 Algebraically Using Substitution

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In order to solve a system of equations graphically you typically begin by making sure both equations are in Slope-Intercept form. Where m is the slope and b is the y-intercept. Examples: y = 3x- 4 y = -2x +6 Slope is 3 and y-intercept is - 4. Slope is -2 and y-intercept is 6.

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**How to Use Graphs to solve Linear Systems.**

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**Looking at the System Graphs:**

If the lines cross once, there will be one solution. If the lines are parallel, there will be no solutions. If the lines are the same, there will be an infinite number of solutions.

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**Check by substitute answers to equations:**

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In order to solve a system of equations algebraically using addition first you must be sure that both equation are in the same chronological order. Example: Could be

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**Now select which of the two variables you want to eliminate. **

For the example below I decided to remove x. The reason I chose to eliminate x is because they are the additive inverse of each other. That means they will cancel when added together.

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**Now add the two equations together.**

Your total is: therefore

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**I decided to substitute 3 in for y in the second equation.**

Now substitute the known value into either one of the original equations. I decided to substitute 3 in for y in the second equation. Now state your solution set always remembering to do so in alphabetical order. [-1,3]

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Lets suppose for a moment that the equations are in the same sequential order. However, you notice that neither coefficients are additive inverses of the other. Identify the least common multiple of the coefficient you chose to eliminate. So, the LCM of 2 and 3 in this example would be 6.

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**Multiply one or both equations by their respective multiples**

Multiply one or both equations by their respective multiples. Be sure to choose numbers that will result in additive inverses. becomes

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**Now add the two equations together.**

becomes Therefore

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**Now substitute the known value into either one of the original equations.**

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**Now state your solution set always remembering to do so in alphabetical order.**

[-3,3]

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**In this example it has been done for you in the first equation.**

In order to solve a system equations algebraically using substitution you must have one variable isolated in one of the equations. In other words you will need to solve for y in terms of x or solve for x in terms of y. In this example it has been done for you in the first equation.

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**Choosing to isolate y in the first equation the result is :**

Now lets suppose for a moment that you are given a set of equations like this.. Choosing to isolate y in the first equation the result is :

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**Now substitute what y equals into the second equation.**

becomes Better know as Therefore

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**Lets look at another Systems solve by Substitution**

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**Step 5: Check the solution in both equations.**

y = 4x 3x + y = -21 Step 5: Check the solution in both equations. 3x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21 -21= -21 y = 4x -12 = 4(-3) -12 = -12

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**This concludes my presentation on simultaneous equations. **

Please feel free to view it again at your leisure.

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7.4. 5x + 2y = 16 5x + 2y = 16 3x – 4y = 20 3x – 4y = 20 In this linear system neither variable can be eliminated by adding the equations. In this linear.

7.4. 5x + 2y = 16 5x + 2y = 16 3x – 4y = 20 3x – 4y = 20 In this linear system neither variable can be eliminated by adding the equations. In this linear.

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