# Solving a System of Equations using Multiplication

## Presentation on theme: "Solving a System of Equations using Multiplication"— Presentation transcript:

Solving a System of Equations using Multiplication
Chapter 6-4

Solving Systems using Multiplication
NCSCOS 4.03 Students will know how to solve a system of equations by using multiplication

Solving Systems using Multiplication
Key Concept: 3 = 3 Let’s look at a simple example: Can I multiply both sides? Yes! As long as I do the same thing on both sides, it will still be equal! Notice that the numbers are bigger, but it’s still equal 6 = 6 1.

Solving Systems using Multiplication
Key Concept: Are we allowed to multiply an equation like the one on the right? As long as we do the same thing on the right as we do on the left we can! If we multiply both sides by 2 it will still be the same equation but it changes the numbers 2 2x + 3y = 4 4x + 6y = 8 2. 3.

Solving Systems using Multiplication
Example 1: Find the solution to the following system of equations: x + y = -2 and 2x + y = 6. In order to solve a system of equations we have to eliminate one of the variables First look to see if you can add the equations to eliminate a variable If you can’t, you’re going to have to multiply one of the equations first 4. 5. 6.

Solving Systems using Multiplication
-1 x + y = -2 Let’s multiply the first equation by -1 Remember to multiply all the numbers by the -1 The new equation is still the same as the first, but the numbers are different We can now use addition to solve this system 2x + y = 6 7. -x – y = 2 8. 9.

Solving Systems using Multiplication
Add the two equations to eliminate the y’s Now that we know what x is, we can plug it into either equation to solve for y It doesn’t matter which equation we use 2x + y = 6 + -x – y = 2 ________ x = 8 10. 11.

Solving Systems using Multiplication
x + y = -2 Plug 8 in for x Solve for y Now that we know y and x we can put them together for our answer! That’s the point where the two lines intersect 8 + y = -2 y = -10 13. (8, -10) 12.

Solving Systems using Multiplication
Solve the following system of equations using multiplication. 2. 3x + y = 5 2x + y = 10 1. x + y = -3 x + 2y = 1 3. 3x – 5y = -35 2x – 5y = -30 4. 13x + 5y = -11 13x + 11y = 7 5. 4x + 5y = 7 8x + 5y = 9 Practice 1

Solving Systems using Multiplication
Solve the following system of equations using multiplication. 2. 3x + y = 5 2x + y = 10 1. x + y = -3 x + 2y = 1 (-7, 4) (-5, 20) 3. 3x – 5y = -35 2x – 5y = -30 4. 13x + 5y = -11 13x + 11y = 7 (-5, 4) (-2, 3) 5. 4x + 5y = 7 8x + 5y = 9 Practice 1 (1/2, 1)

Solving Systems using Multiplication
Example 2: Use elimination to solve the system of equations. -5x – 2y = 4 First we have to see if we can add them together to eliminate one of the variables NO! So we’ll have to find a different way to solve 3x + 4y = 6 14.

Solving Systems using Multiplication
-5x – 2y = 4 To eliminate a variable they must add up to zero One has to be negative and one has to be positive Since the y values are opposite signs and 4 is a multiple of 2 we will eliminate the y values first 15. 3x + 4y = 6 16. 17.

Solving Systems using Multiplication
-5x – 2y = 4 2 In order to get rid of the y values we have to multiply the first equation by 2 This doesn’t change the actual equation, just the numbers 3x + 4y = 6 -10x – 4y = 8

Solving Systems using Multiplication
3x + 4y = 6 Now we can add the equations together to eliminate the y’s Divide by -7 Now that we know x, plug it back into any equation to find y + -10x – 4y = 8 ___________ -7x = 14 ___ __ x = -2 19.

Solving Systems using Multiplication
It doesn’t matter which equation we use, so I’ll plug x into the second equation because the number are positive Now solve for y Multiply on the left Add 6 to both sides Divide by 4 3x + 4y = 6 3(-2) + 4y = 6 20. -6 + 4y = 6 4y = 12 __ __

Solving Systems using Multiplication
Now that we have y, plug the values of x and y into a point for your answer! 22. (-2, 3)

Solving Systems using Multiplication
Use elimination to solve each system of equations 1. -5x + 3y = 6 x – y = 4 2. x + y = 3 2x – 3y = 16 3. 2x + y = 5 3x – 2y = 4 4. 4x – 3y = 12 x + 2y = 14 5. 5x – 2y = -15 3x + 8y = 37 Practice 2

Solving Systems using Multiplication
Use elimination to solve each system of equations 1. -5x + 3y = 6 x – y = 4 2. x + y = 3 2x – 3y = 16 (5, -2) (-9, -13) 3. 2x + y = 5 3x – 2y = 4 4. 4x – 3y = 12 x + 2y = 14 (2, 1) (6, 4) 5. 5x – 2y = -15 3x + 8y = 37 Practice 2 (-1, 5)

Solving Systems using Multiplication
Example 3: Solve the following system of equations: 3x + 4y = -25 First check to see if you can just add these equations together to eliminate a variable Since we can’t do that, can we multiply one equation to eliminate a variable? No! 2x – 3y = 6 23. 24.

Solving Systems using Multiplication
Example 3: Solve the following system of equations: 3x + 4y = -25 What we’ll need to do is multiply both equations by a different number so that we can cancel out a variable 2x – 3y = 6 25. I would look at eliminating the y values since they have opposite signs already 26.

Solving Systems using Multiplication
3 3x + 4y = -25 Multiply the first equation by 3 Multiply the second equation by 4 Now we can add these equations together to eliminate the y’s! 4 2x – 3y = 6 9x + 12y = -75 28. 8x – 12y = 24

Solving Systems using Multiplication
Add the equations Divide by 17 Plug x into either of the original equations 9x + 12y = -75 + 8x – 12y = 24 ___________ 17x = -51 ___ __ x = -3

Solving Systems using Multiplication
It doesn’t matter if you use either of the original equations or the ones that you multiplied, but I will use the second equation since the numbers are smaller Solve for y Add 6 Divide by -3 2x – 3y = 6 2(-3) – 3y = 6 -6 – 3y = 6 – 3y = 12 ___ __

Solving Systems using Multiplication
Plug your x and y values into a point for your answer y = -4 (-3, -4) 31.

Solving Systems using Multiplication
Solve the following system of equations 1. 5x – 2y = -15 3x + 8y = 37 2. 4x – 7y = 10 3x + 2y = -7 3. 2x – 3y = 2 5x + 4y = 28 4. 4x + 3y = 19 3x – 4y = 8 5. 4x + 7y = 6 6x + 5y = 20 Practice 3

Solving Systems using Multiplication
Solve the following system of equations 1. 5x – 2y = -15 3x + 8y = 37 2. 4x – 7y = 10 3x + 2y = -7 (-1, -2) (-1, 5) 3. 2x – 3y = 2 5x + 4y = 28 4. 4x + 3y = 19 3x – 4y = 8 (4, 2) (4, 1) 5. 4x + 7y = 6 6x + 5y = 20 Practice 3 (5, -2)

Solving Systems using Multiplication
Quiz! Solve the following system of equations 1. x – y = -5 3x – y = 7 2. 3x – 3y = 6 3x + y = -10 3. 5x + 2y = 7 3x – y = 13 4. -5x + 2y = -18 3x + 4y = -10 5. 4x + 3y = -11 3x + 5y = 0

Solving Systems using Multiplication
Quiz! Solve the following system of equations 1. x – y = -5 3x – y = 7 2. 3x – 3y = 6 3x + y = -10 (-2, -4) (6, 11) 3. 5x + 2y = 7 3x – y = 13 4. -5x + 2y = -18 3x + 4y = -10 (3, -4) (2, -4) 5. 4x + 3y = -11 3x + 5y = 0 (-5, 3)