Solving Systems Solving Systems by Graphing by Solving Systems by - - PDF document

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Algebra I

System of Linear Equations

2015-11-02 www.njctl.org

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Solving Systems by Graphing Solving Systems by Substitution Solving Systems by Elimination Choosing your Strategy Writing Systems to Model Situations

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Standards

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Solving Systems by Graphing

Return to Table

• f Contents

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A system of linear equations is comprised of 2 or more linear equations. The solution of the system will be the values of the variables which make all the equations true.

System of Equations Slide 6 / 176

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y y = 2x - 4 Example: The graph of the line that represents the solutions to the above equation is shown. It represents all the points whose x and y values make the above equation true. The line is easy to find from the equation since the equation is in slope- intercept form.

Solve by Graphing

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y Example: Similarly, this graph is of the line that represents the solutions to this equation. It represents all the points whose x and y values make the above equation true. y = –x + 5

Solve by Graphing Slide 8 / 176

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y y = 2x - 4 y = –x + 5 Example: Here are the lines that represent the solutions to both those equations. Each line shows the infinite set of solutions for each equation. What must be true about the point at which they cross? DISCUSS.

Solve by Graphing Slide 9 / 176

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y y = 2x - 4 y = –x + 5 Example: At the point they cross, both equations must be true, since that point is on both lines. They appear to cross at (3, 2). Let's check that in both equations.

Solve by Graphing Slide 10 / 176

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y Substitute x = 3 and y = 2 into both equations and see if both equations are true. y = 2x - 4 (2) = 2(3) - 4 2 = 2 correct y = –x + 5 (2) = -(3) + 5 2 = 2 correct Example:

Solve by Graphing Slide 11 / 176

Not all systems have solutions...and some have an infinite number of solutions. Let's see how to figure out whether there are solutions, how many, and what they are.

System of Equations

Click here to watch a music video that introduces what we will learn about systems.

Slide 12 / 176 The Number of Solutions

When graphing two lines there are three possibilities. · They meet in one point: the point of intersection. · They never meet: they are parallel. · They meet at all their points: they are the same line.

Slide 13 / 176 The Number of Solutions

So, systems of equations can have either: · 1 solution, if the lines meet at one point · 0 solutions, if they never meet · Infinite solutions, if they are the same line

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The two lines intersect in exactly ONE place. The solution is the point at which they intersect. The slopes of the lines must be different, or they would never cross.

Type 1: One Solution

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y = 2x - 4 y = –x + 5 This is the example we started with. As we confirmed there is one solution to this system of equations: (3, 2).

Type 1: One Solution

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y The lines never meet. There is no solution true for both lines. The lines are parallel. They must have the same slope, since they are parallel. But, they must have different intercepts, or they would be the same line.

Type 2: No Solution Slide 17 / 176

Both are written in slope intercept form y = mx + b to make it easy to compare slopes and y-intercepts. The slope for both lines is 2 (the coefficient of x). So, the lines are parallel. The y-intercepts are different, +6 and +2, so the lines never cross. y = 2x + 6 y = 2x + 2

Type 2: No Solution

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The lines overlap at all points. They are different equations for the same line. The lines are parallel. So, they must have the same slopes. The intercepts are the same, since all their points are the same.

Type 3: Infinite Solutions

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Both are written in slope intercept form y = mx + b to make it easy to compare slopes and y- intercepts. The slope for both lines is 2 (the coefficient of x). So, the lines are parallel. The y-intercepts for both lines are +2, so the lines

• verlap everywhere.

y = 2x + 2 y = 2x + 2

Type 3: Infinite Solutions

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In slope intercept form, the fact that these are the same line is obvious. But, if the equations were written as below, it would be less obvious: 2y - 4x = 4

• 6x = -3y + 6

That's why it's always a good idea to put equations into slope-intercept form...they're easier to read, graph and compare. y = 2x + 2 y = 2x + 2

Type 3: Infinite Solutions

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Slide 21 / 176 The Number of Solutions

First, put the equations into slope-intercept form by solving for y. Then, decide on the number of solutions. After that, solutions can be found in three different ways.

Slide 22 / 176 How can you quickly decide the number of solutions a system has?

1 Solution Different slopes Different lines No Solution Same slope Different y-intercept Parallel Lines Infinitely Many Same slope Same y-intercept Same Line

Math Practice

Slide 23 / 176 Solving both Equations for y

Let's solve this system of equations y = -5x + 4 10x + 2y = 6 The equation on the left is in slope-intercept form. Do you see that the slope is -5 and its y-intercept is +4? The equation on the right is not in slope-intercept form, so we can't see it's slope or y-intercept. So, we can't tell yet how many solutions will satisfy both equations. Let's solve the second equation for y.

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10x + 2y = 6 Subtract 10x from both sides

• 10x -10x

2y = -10x + 6 Divide both sides by 2 2y = -10x + 6 y = -5x + 3 This is now in slope-intercept form. We can see the slope and y-intercept m = -5 b = 3 2 2

Solving for y

Slide 25 / 176 Solving both Equations for y

y= -5x + 4 Original Equations 10x + 2y = 6 y= -5x + 4 Slope Intercept Form y = -5x + 3 m = -5 b = 4 Slopes and Intercepts m = -5 b = 3 The slopes are the same but the y-intercepts are different. How many solutions are there?

Slide 26 / 176 Solving both Equations for y

Let's solve this system of equations y = 2x + 5 6x + 2y = 4 The equation on the left is in slope-intercept form and we can see the slope is +2 and the y-intercept is +5. The equation on the right is not in slope-intercept form, let's solve that equation for y.

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6x + 2y = 4 Subtract 6x from both sides

• 6x -6x

2y = -6x + 4 Divide both sides by 2 2y = -6x + 4 y = -3x + 2 This is now in slope-intercept form. m = -3 b = +2 2 2

Solving for y Slide 28 / 176 Solving both Equations for y

y= -5x + 4 Original Equations 6x + 2y = 4 y= -5x + 4 Slope Intercept Form y = -3x + 2 m = -5 b = 4 Slopes and Intercepts m = -3 b = +2 The slopes are different. How many solutions are there?

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1 How many solutions does this system have: A 1 solution B no solution C infinitely many solutions y = 2x - 7 y = 3x + 8

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2 How many solutions does this system have: A 1 solution B no solution C infinitely many solutions 3x - y = -2 y = 3x + 2

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3 How many solutions does this system have: A 1 solution B no solution C infinitely many solutions 1 3 3x + 3y = 8 y = x

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4 How many solutions does this system have: A 1 solution B no solution C infinitely many solutions y = 4x 2x - 0.5y = 0

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5 How many solutions does this system have: A 1 solution B no solution C infinitely many solutions 3x + y = 5 6x + 2y = 1

Slide 34 / 176 Consider this...

Suppose you are walking to school. Your friend is 5 blocks ahead of you. You can walk two blocks per minute and your friend can walk one block per minute. How many minutes will it take for you to catch up with your friend?

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Time (min.) Friend's distance from your start (blocks) Your distance from your start (blocks) 5 1 6 2 2 7 4 3 8 6 4 9 8 5 10 10 First, make a table to represent the problem.

Solution Slide 36 / 176

Next, plot the points on a graph. 10 5 5 Blocks Time (min.) 10 Time (min.) Friend's distance from your start (blocks) Your distance from your start (blocks) 5 1 6 2 2 7 4 3 8 6 4 9 8 5 10 10

Solution Continued

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10 5 5 Blocks Time (min.) 10 The point where the lines intersect is the solution to the system. (5, 10) is the solution In the context of this problem this means after 5 minutes, you will meet your friend at block 10.

Solution Continued Slide 38 / 176

Solve this system of equations graphically:

Example

y = 2x - 3 y = x - 1 5 10 5 10

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Solve the system of equations graphically:

Example

y = -3x + 4 y = x - 4 5 10 5 10

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y y = -3x - 1 y = 4x + 6

Checking Your Work

Given the graph below, what is the point of intersection?

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y = 4x + 6 (2) = 4(-1) + 6 2 = -4 + 6 2 = 2 y = -3x - 1 (2) = -3(-1) - 1 2 = 3 - 1 2 = 2 (-1, 2) Now take the ordered pair we just found and substitute it into the equations to prove that it is a solution for BOTH lines.

Checking Your Work Slide 42 / 176

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6 Solve the following system by graphing: A (3, 1) B (1, 3) C (-1, 3) D (1, -3)

Click for answer choices AFTER students have graphed the system

y = -x + 4 y = 2x + 1

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7 Solve the following system by graphing: A (0,-1) B (0,0) C (-1, 0) D (0, 1) y = x – 1 1 2 y = – x – 1 1 2

Click for answer choices AFTER students have graphed the system

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8 Solve the following system by graphing: A (0, 4) B (-4, 2) C (5, 6) D (2, 5) y = x + 3 y = x + 4 1 2

Click for answer choices AFTER students have graphed the system

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Slide 45 / 176 Graphing Quickly

Transforming linear equations into slope-intercept form usually saves time in the end. It also makes it easy to check your work.

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• x + y = 2

+x +x y = x + 2 2x + y = 5

• 2x -2x

y = -2x + 5

Example

Step 1: Rewrite the linear equations in slope-intercept form Solve the following system of linear equations by graphing: 2x + y = 5

• x + y = 2

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y y = -2x + 5 y-intercept = (0, 5) slope = -2 slope= (down 2, right 1) y = x + 2 y-intercept = (0, 2) slope = 1 slope= (up 1, right 1) Step 2: Plot the y-intercept and use the slope to plot the second point

Solution Continued Slide 48 / 176

y = -2x + 5 (3) = -2(1) + 5 3 = -2 + 5 3 = 3 y = x + 2 (3) = (1) + 2 3 = 3 Step 3: Locate the point of intersection and check your work:

Solution Continued

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Solve the system of equations graphically: 2x + y = 3 x - 2y = 4 Step 1: Rewrite in slope-intercept form

Example

x - 2y = 4

• x -x
• 2y = -x + 4
• 2
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y = x - 2 2 1

• 2y = -x + 4

2x + y = 3

• 2x -2x

y = -2x + 3

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y-intercept = (0, 3) slope = -2 slope= (down 2, right 1) y-intercept = (0, -2) slope = slope= (up 1, right 2) 1 2

Solution Continued

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y = x – 2 (–1) = (2) – 2 –1 = 1 – 2 –1 = –1 1 2 1 2 y = -2x + 3 (-1) = -2(2) + 3

• 1 = -4 + 3
• 1 = -1

Step 3: Locate the Point of Intersection and check your work:

Solution Continued

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9 What is the solution of the system of linear equations provided on the graph? A (0, 1) B (1, 0) C (2, 3) D (3, 2)

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10 Which graph below represents the solution to the following system of linear equations:

• x + 2y = 2

3y = x + 6 A B C D

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11 Solve the following system by graphing: A (3, 4) B (9, 2) C infintely many D no solution x – 3y = 3 y = x – 7

Click for answer choices AFTER students have graphed the system

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Solve the system of equations graphically: Step 1: Rewrite in slope-intercept form

Example

9x - 3y = -18

• 9x -9x
• 3y = -9x -18
• 3
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• 3y = -9x -18

y = 3x + 6 y = 3x + 6 y = 3x + 6 9x - 3y = -18

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y-intercept = (0, 6) slope = 3 slope= (up 3, right 1) y-intercept = (0, 6) slope = 3 slope= (up 3, right 1) y = 3x + 6 y = 3x + 6

Solution Continued

Step 2: Plot y-intercept and use slope to plot second point

Slide 57 / 176 Solution Continued

Step 3: Locate the Point of Intersection and check your work: infinite amount of points: infinite solutions 9x - 3y = -18 y = 3x + 6

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Solve the system of equations graphically: 4x - 2y = 10 8x - 4y = 12 Step 1: Rewrite in slope-intercept form

Example

8x - 4y = 12

• 8x 8x
• 4y = -8x + 12
• 4y = -8x +12
• 4 -4

y = 2x - 3 4x - 2y = 10

• 4x -4x
• 2y = -4x + 10
• 2y = -4x + 10
• 2 -2

y = 2x - 5

Slide 59 / 176 Solution Continued

Step 2: Plot y-intercept and use slope to plot second point y-intercept = (0, -5) slope = 2 slope= (up 2, right 1) y-intercept = (0, -3) slope = 2 slope= (up 2, right 1) y = 2x - 5 y = 2x -3

Slide 60 / 176 Solution Continued

Step 3: Locate the Point of Intersection and check your work: no point of intersection: no solution y = 2x - 5 y = 2x - 3

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12 Solve the this system by graphing: A (2, 4) B (0.4, 2.2) C infinitely many solutions D no solution y = 3x + 4 4y = 12x + 12

Click for answer choices AFTER students have graphed the system

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13 Solve the this system by graphing: A (3,4) B (-3,-4) C infinitely many D no solution y = 3x + 4 4y = 12x + 16

Click for answer choices AFTER students have graphed the system

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Solving Systems by Substitution

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Solve the system of equations graphically.

Example

Why was it difficult to solve this system by graphing? y = x + 6.1 y = -2x - 1.4 5 10 5 10

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Graphing can be inefficient or approximate. Another way to solve a system of linear equations is to use substitution. Substitution allows you to create a one variable equation.

Substitution Explanation Slide 66 / 176 Solving by Substitution

Step 1: If you are not given a variable already alone, find the EASIEST variable to solve for (get it alone) Step 2: Substitute the expression into the other equation and solve for the variable Step 3: Substitute the numerical value you found into EITHER equation and solve for the other variable. Write the solution as (x, y)

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Solve the system using substitution: Step 1: Choose an equation from the system and substitute it into the other equation

Example

y = x + 6.1 First Equation y = -2x - 1.4 Second Equation y = x + 6.1 y = -2x - 1.4 x + 6.1 = -2x - 1.4 Substitute First Equation into Second Equation

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x + 6.1 = -2x - 1.4 +2x +2x Add 2x to both sides 3x + 6.1 = - 1.4

• 6.1
• 6.1

Subtract 6.1 from both sides 3x = - 7.5 3x = - 7.5 Divide both sides by 3 3 x = -2.5 This is the value of x for our solution...now we have to find y. Step 2: Solve the new equation

Solution Continued Slide 69 / 176

y = x + 6.1 y = (-2.5) + 6.1 y = 3.6 y = -2x - 1.4 y = -2(-2.5) - 1.4 y = +5 - 1.4 y = 3.6 The solution to the system of linear equations is (-2.5, 3.6). We only had to plug the x value into one of the equations to get this. The second one just provides a check. If it comes out the same, our solution must be correct. Step 3: Substitute the solution x = -2.5 into either equation and solve.

Solution Continued Slide 70 / 176 Good Practice

y = -2x - 1.4 (3.6) = -2(-2.5) - 1.4 3.6 = 5 - 1.4 3.6 = 3.6 y = x + 6.1 (3.6) = (-2.5) + 6.1 3.6 = 3.6 CHECK: See if (-2.5, 3.6) satisfies both equations If your checks end in true statements, the solution is correct. After you evaluate the solution, it is good practice is to double check your work by substituting the solution into both equations.

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2x - 3y = -1 y = x - 1 2x - 3(x - 1) = -1 Solve the system using substitution:

Example

2x - 3y = -1 y = x - 1 Step 1: Substitute one equation into the other equation. Since one equation is already solved for y, I'll substitute that into the other equation.

Slide 72 / 176 Solution Continued

2x - 3y = -1 2(4) - 3y = -1 8 - 3y = -1

• 3y = -9

y = 3 (4, 3) (4, 3) y = x - 1 y = (4) - 1 y = 3 You end with the correct answer with either equation you use for this step. Step 2: Solve the new equation 2x - 3(x - 1) = -1 2x - 3x + 3 = -1 x = 4 Step 3: Substitute the solution into either equation and solve

Slide 73 / 176 Example Continued

Check: See if (4, 3) satisfies both equations The ordered pair satisfies both equations so the solution is (4, 3) 2x - 3y = -1 2(4) - 3(3) = -1 8 - 9 = -1

• 1 = -1

y = x - 1 (3) = (4) - 1 3 = 3

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14 Solve by substitution: A (4, 9) B (-4, -9) C (4, 1) D (1, 4) y = x - 3 y = -x + 5

Click for answer choices AFTER students have solved the system

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15 Solve by substitution: A (2, -8) B (-3, 2) C infinitely many solutions D no solutions y = - x y = -3x – 7 2 3

Click for answer choices AFTER students have solved the system

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16 Solve by substitution. A (4, 5) B (5, 4) C infintely many solutions D no solutions y = 4x - 11

• 4x + 3y = -1

Click for answer choices AFTER students have solved the system

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17 Solve by substitution. A (-2, -2) B (-2, 2) C (2, -2) D (2, 2) y = 8x + 18 3x + 3y = 0

Click for answer choices AFTER students have solved the system

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18 Solve by substitution. A (-8 , 5) B (7, 5) C (-3, 5) D (-7, 5) 8x + 3y = -9 y = 3x + 14

Click for answer choices AFTER students have solved the system

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Examine each system of equations. Which variable would you choose to substitute? Why?

Choosing a Variable

y = 4x - 9.6 y = -2x + 9

• y + 4x = -1

x - 4y = 1 2x + 4y = -10

• 8x - 3y = -12

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19 Examine this system of equations. Which variable could quickly be solved for and substituted into the other equation? A x B y y = -2x + 5 2y = 10 - 4x

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20 Examine this system of equations. A x B y Which variable could quickly be solved for and substituted into the other equation? 2y - 8 = x y + 2x = 4

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21 Examine this system of equations. A x B y Which variable could quickly be solved for and substituted into the other equation? x - y = 20 2x + 3y = 0

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y = 3x - 5 2x + 5y = -8

Rewriting

Which letter is the easiest to solve for? The "y" in the first equation because there is only a "-1" as the coefficient. Sometimes you need to rewrite one of the equations so that you can use the substitution method. For example: The system: 3x - y = 5 2x + 5y = -8 Solve for y: 3x - y = 5

• 3x 3x
• y = -3x + 5
• 1
• 1

y = 3x - 5

• 1

So, the original system is equivalent to: Click to discuss which letter Click to see

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y = 3x - 5 2x + 5 y = -8

Solution Continued

Now Substitute and Solve: 2x + 5(3x - 5) = -8 2x + 15x - 25 = -8 17x - 25 = -8 17x = 17 x = 1

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Substitute x = 1 into one of the equations. 3x - y = 5 3(1) - (-2) = 5 3 + 2 = 5 5 = 5 2x + 5y = -8 2(1) + 5(-2) = -8 2 - 10 = -8

• 8 = -8

Solution Continued

The ordered pair (1, -2) satisfies both equations in system. 2(1) + 5y = -8 2 + 5y = -8 5y = -10 y = -2

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22 Solve using substitution. A (-6 , 2) B (6 , -2) C (-6 , -2) D (2, -6) 6x + y = 6

• 3x + 2y = -18

Click for answer choices AFTER students have solved the system

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23 Solve using substitution. A (6, -1) B (-6, 5) C (5, 5) D (-6, -1) 2x - 8y = 20

• x + 6y = -12

Click for answer choices AFTER students have solved the system

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24 Solve using substitution. A (-3, -7) B (-7, 3) C (3, 7) D (7, 3)

• 3x - 3y = 12
• 4x - 7y = 7

Click for answer choices AFTER students have solved the system

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Your class of 22 is going on a trip. There are four drivers and two types of vehicles, vans and cars. The vans seat six people, and the cars seat four people, including drivers. How many vans and cars does the class need ?

Example

Let v = the number of vans and c = the number of cars

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Solve the system by substitution:

Set Up the System

Drivers: v + c = 4 People: 6v + 4c = 22

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v + c = 4 6v + 4c = 22 solve for v substitute v = -c + 4 6(-c + 4) + 4c = 22

• 6c + 24 + 4c = 22
• 2c + 24 = 22
• 2c = -2

substitute v = -(1) + 4 c = 1 v = 3 then check: c = 1; v = 3 (3) + (1) = 4 6(3) + 4(1) = 22 4 = 4 22 = 22

Substitute, Solve and Check Slide 92 / 176

Solve this system using substitution:

Example

x + y = 6 5x + 5y = 10 x + y = 6 x = 6 - y 5(6 - y) + 5y = 10 30 - 5y + 5y = 10 30 = 10

• solve the first equation for x
• substitute 6 - y for x in 2nd equation
• solve for y
• This is FALSE!

Since 30 = 10 is a false statement, the system has no solution. Answer: NO SOLUTION

Slide 93 / 176 Example

Since -6 = -6 is always a true statement, there are infinitely many solutions to the system. Answer: Infinite Solutions Solve the following system using substitution: x + 4y = -3 2x + 8y = -6 x + 4y = -3 - solve the first equation for x x = -3 - 4y 2(-3 - 4y) + 8y = -6 - sub. -3 - 4y for x in 2nd equation

• 6 - 8y + 8y = -6 - solve for y
• 6 = -6 - This is ALWAYS TRUE!

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25 Solve the system by substitution: A (-10, -4) B (-4, 2) C (2, -4) D (10, 4) y = x - 6 y = -4 Click for answer choices AFTER students have solved the system

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26 Solve the system by substitution: A (1, 20) B (1, 18) C (8, -2) D (-8, 2) y + 2x = -14 y = 2x + 18 Click for answer choices AFTER students have solved the system

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27 Solve the system by substitution: A (6, 6.5) B (5, 6) C (4, 5) D (6, 5) 4x = -5y + 50 x = 2y - 7 Click for answer choices AFTER students have solved the system

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FT - m1g = m1a

• FT + m2g = m2a

solve both for FT FT = m1g + m1a FT = m2g - m2a substitute m1g + m1a = m2g - m2a

• 2c + 24 = 22
• 2c = -2

substitute v = -(1) + 4 c = 1 v = 3 then check: c=1; v=3 (3) + (1) = 4 6(3) + 4(1) = 22 4 = 4 22 = 22

Solving for a Slide 98 / 176

c) Find the equations for the tension force FT

Problem 3 - Tension Force

We have two equations (one for each mass) and two unknowns (FT and a). This means we can combine the equations together to solve for each variable! FT - m1g = m1a FT - m1g = m1a FT = m1g + m1a

• FT + m2g = m2a
• FT + m2g = m2a

FT = m2g - m2a Solve each for FT: Now we can set them equal to one another: m1g + m1a = m2g - m2a

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Solve each for FT: FT - m1g = m1a FT - m1g = m1a FT = m1g + m1a

• FT + m2g = m2a

FT = m2g - m2a

• FT + m2g = m2a

Problem 3 - Tension Force

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c) Find the equations for the tension force FT We have two equations (one for each mass) and two unknowns (FT and a). This means we can combine the equations together to solve for each variable! Now we can set them equal to one another: m1g + m1a = m2g - m2a

Slide 100 / 176 Problem 3 - Tension Force

c) Find the equation for the acceleration Now we can combine the tension equations m1g + m1a = FT FT = m2g - m2a There is only one unknown (a) here. Solve for a: m1g + m1a = m2g - m2a m1a + m2a = m2g - m1g a(m1 + m2) = m2g - m1g a = m2g - m1g m1 + m2 Add m2a and subtract m1g from both sides: factor out 'a' : (remember factoring is just the opposite of distributing) divide by (m1 + m2):

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28 Solve the system by substitution: A (6, 5) B (-7, 5) C (42, -103) D (6, -5) y = -3x + 23

• y + 4x = 19

Click for answer choices AFTER students have solved the system

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29 Solve the system using substitution. A (-4, 5) B (4, -1) C infinitely many solutions D no solutions Click for answer choices AFTER students have solved the system 3 4 x + y = 2 6x + 8y = 16

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30 Solve using substitution. A (-3, -1) B No Solution C Infinite Solutions D (-1, -3) 16x + 2y = -5 y = -8x - 6 Click for answer choices AFTER students have solved the system

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Solving System by Elimination

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Slide 105 / 176

Recall that the Standard Form of a linear equation is: Ax + By = C When both linear equations of a system are in standard form the system can be solved by using elimination. The elimination strategy adds or subtracts the equations in the system to eliminate a variable.

Standard Form Slide 106 / 176 Additive Inverses

Let's talk about what's happening with these numbers

• 2 + 2 =

3 + (-3) =

• 5x + 5x =

9x + (-9x) =

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How do you decide which variable to eliminate? First: Look to see if one variable has the same or

• pposite coefficients. If so, eliminate that variable.

Choosing a Variable Slide 108 / 176

If the variables have the same coefficient, subtract the two equations to eliminate the variable. If the variables have opposite coefficients, add the two equations to eliminate the variable. 3x 3x 3x

• (3x)

0x 3x

• 3x

3x + -3x) 0x { Same Coefficients { Subtract { Opposite Coefficients { Add

Addition or Subtraction

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Solve the following system by elimination: 5x + y = 44

• 4x - y = -34

Step 1: Choose which variable to eliminate The y in both equations have opposite coefficients so they will be the easiest to eliminate Step 2: Add the two equations 5x + y = 44

• 4x - y = -34

x + 0y = 10 x = 10

Example Slide 110 / 176 Solution Continued

Step 3: Substitute the solution into either equation and solve 5x + y = 44 5(10) + (-6) = 44 50 - 6 = 44 44 = 44

• 4x - y = -34
• 4(10) - (-6) = -34
• 40 + 6 = -34
• 34 = -34

x = 10 5(10) + y = 44 50 + y = 44 y = -6 The solution to the system is (10, -6) Check:

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Solve the following system by elimination: 3x + y = 15

• 3x - 3y = -21

Step 1: Choose which variable to eliminate The x in both equations have opposite coefficients so they will be the easiest to eliminate Step 2: Add the two equations

Example

3x + y = 15

• 3x - 3y = -21
• 2y = -6

y = 3

Slide 112 / 176 Solution Continued

Step 3: Substitute the solution into either equation and solve 3x + y = 15 3(4) + 3 = 15 12 + 3 = 15 15 = 15

• 3x - 3y = -21
• 3(4) - 3(3) = -21
• 12 - 9 = -21
• 21 = -21

The solution to the system is (4, 3) Check: y = 3 3x + 3 = 15 3x = 12 x = 4

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31 Solve the system by elimination: A (5, 1) B (-5, -1) C (1, 5) D no solution x + y = 6 x - y = 4 Click for answer choices AFTER students have solved the system

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32 Solve the system by elimination: A (-2,1) B (-1,-2) C (-2,-1) D infinitely many 2x + y = -5 2x - y = -3 Click for answer choices AFTER students have solved the system

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33 Solve using elimination. A (-2, 3) B (4, -6) C (-6, 4) D (3, -2)

• 2x - 8y = 10

2x - 6y = 18 Click for answer choices AFTER students have solved the system

Slide 116 / 176 Multiple Methods

There are 2 ways to complete the problem below using elimination. 5x + y = 17

• 2x + y = -4

Step 1: Choose which variable to eliminate The y in both equations have the same coefficient so they will be the easiest to eliminate Step 2: Add or Subtract the two equations First Method: Multiply one equation by -1 then add equations Second Method: Subtract equations keeping in mind that all signs change

Slide 117 / 176 Solution Continued

Second Method 5x + y = 17

• (-2x + y = -4)

7x = 21 x = 3 First Method

• 1(-2x + y = -4) = 2x - y = 4

5x + y = 17 2x - y = 4 7x = 21 x = 3 Why do both methods produce the same solution?

Slide 118 / 176 Solution Continued

Step 3: Substitute the solution into either equation and solve x = 3

• 2(3) + y = -4
• 6 + y = -4

y = 2 The solution to the system is (3, 2) Check: 5x + y = 17 5(3) + 2 = 17 15 + 2 = 17 17 = 17

• 2x + y = -4
• 2(3) + 2 =
• 4
• 6 + 2 = -4
• 4 = -4

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34 Solve the system by elimination: A (-4, 2) B (3, 5) C (4, 2) D infinitely many 2x + y = -6 3x + y = -10 Click for answer choices AFTER students have solved the system

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35 Solve the system by elimination: 3x + 6y = 48

• 5x + 6y = 32

A (2, -7) B (2, 7) C (7, 2) D infinitely many Click for answer choices AFTER students have solved the system

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Sometimes, it is not possible to eliminate a variable by simply adding or subtracting the equations. When this is the case, you need to multiply one or both equations by a nonzero number in order to create a common coefficient before adding or subtracting the equations.

Common Coefficient Slide 122 / 176

Solve the following system using elimination: 3x + 4y = -10 5x - 2y = 18 The y would be the easiest variable to eliminate because 4 is a common coefficient. Multiply second equation by 2 so the coefficients are opposites. 2(5x - 2y = 18) The y coefficients are opposites, so solve by adding the equations 3x + 4y = -10 10x - 4y = 36 13x = 26 x = 2 +

Example Slide 123 / 176

Solve for y, by substituting x = 2 into one of the equations.

Example Continued

3x + 4y = -10 3(2) + 4(-4) = -10 6 + -16 = -10

• 10 = -10

5x - 2y = 18 5(2) - 2(-4) = 18 10 + 8 = 18 18 = 18 3x + 4y = -10 3(2) + 4y = -10 6 + 4y = -10 4y = -16 y = -4 (2, -4) is the solution Check:

Slide 124 / 176 Choosing Variable to Eliminate

In the previous example, the y was eliminated by finding a common coefficient of 4. Creating a common coefficient of 4 required one additional step: Multiplying the second equation by 2 3x + 4y = -10 5x - 2y = 18 Either variable can be eliminated when solving a system of equations as long as a common coefficient is utilized.

Slide 125 / 176

Solve the same system by eliminating x. 3x + 4y = -10 5x - 2y = 18 Multiply the first equation by 5 and the second equation by 3 so the coefficients will be the same Now solve by subtracting the equations.

• 15x + 20y = -50

(15x - 6y = 54) 26y = -104 y = -4 5(3x + 4y = -10) 15x + 20y = -50 3(5x - 2y = 18) 15x - 6y = 54

Example Slide 126 / 176

Solve for x, by substituting y = -4 into one of the equations.

Example Continued

3x + 4y = -10 3(2) + 4(-4) = -10 6 + -16 = -10

• 10 = -10

5x - 2y = 18 5(2) - 2(-4) = 18 10 + 8 = 18 18 = 18 3x + 4y = -10 3x + 4(-4) = -10 3x + -16 = -10 3x = 6 x = 2 (2, -4) is the solution. Check:

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Examine each system of equations. Which variable would you choose to eliminate? What do you need to multiply each equation by? 2x + 5y = -1 x + 2y = 0 3x + 8y = 81 5x - 6y = -39 3x + 6y = 6 2x - 3y = 4

System of Equations Slide 128 / 176

36 Which variable can you eliminate with the least amount of work in the system below? 2x + 5y = 20 3x - 10y = 37 A x B y

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37 Solve the following system of equations using elimination: 2x + 5y = 20 3x - 10y = 37 A (1, 57) B (1, 77) C D infinitely many solutions (11, - ) 2 5 Click for answer choices AFTER students have solved the system

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38 Which variable can you eliminate with the least amount of work in the system below? x + 3y = 4 3x + 4y = 2 A x B y

Slide 131 / 176

39 What will you multiply the first equation by in

• rder to solve this system using elimination?

x + 3y = 4 3x + 4y = 2

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40 Solve the following system of equations: x + 3y = 4 3x + 4y = 2 A B (-2, 1) C (-2, 2) D infinitely many solutions (-2, ) 2 3 Click for answer choices AFTER students have solved the system

Slide 133 / 176

Solve the following system using elimination: 9x - 5y = 4

• 18x + 10y = 10

The y would be the easiest variable to eliminate because 10 is a common coefficient. Multiply first equation by 2 so the coefficients are opposites. 2(9x - 5y = 4) The y coefficients are opposites, so solve by adding the equations 18x - 10y = 8

• 18x + 10y = 10

0 = 18 is this true? False, NO SOLUTION +

Example

Move for solution

Slide 134 / 176

Solve the following system using elimination:

• 4x - 10y = -22

2x + 5y = 11 The x would be the easiest variable to eliminate because 4 is a common coefficient. Multiply second equation by 2 so the coefficients are opposites. 2(2x + 5y = 11) The y coefficients are opposites, so solve by adding the equations

• 4x - 10y = -22

4x +10y = 22 0 = 0 is this true? True, INFINITE SOLUTIONS +

Example

Move for solution

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41 Solve the system by elimination: x - y = 5 x - y = -7 A (11, -4) B (4, 11) C (-4, -11) D no solution Click for answer choices AFTER students have solved the system

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42 Solve using elimination.

• 20x - 18y = -28

10x + 9y = 14 A (-8, -1) B infinite solutions C no solution D (-1, 8) Click for answer choices AFTER students have solved the system

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43 Solve using elimination. 9x + 3y = 27 18x + 6y = 30 A infinite solutions B (4, 7) C (-7, 4) D no solution Click for answer choices AFTER students have solved the system

Slide 138 / 176

Choose Your Strategy

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Slide 139 / 176

Systems of linear equations can be solved using any of the three methods we previously discussed. Before solving a system, an analysis of the equations should be done to determine the "best" strategy to utilize. Graphing Substitution Elimination

Choosing Strategy Slide 140 / 176

Altogether 292 tickets were sold for a basketball game. An adult ticket cost $3 and a student ticket cost $1. Ticket sales for the event were $470. How many adult tickets were sold? How many student tickets were sold?

Example Slide 141 / 176

Step 1: Define your variables Let a = number of adult tickets Let s = number of student tickets Step 2: Set up the system number of tickets sold: a + s = 292 money collected: 3a + s = 470

Example Continued Slide 142 / 176

Step 3: Solve the system a + s = 292 3a + s = 470

• 2a+ 0 = -178

a = 89

• (

)

Example Continued

Elimination was utilized for this example because the x had a common coefficient.

Note

Slide 143 / 176

a = 89 a + s = 292 89 + s = 292 s = 203 There were 89 adult tickets and 203 student tickets sold Check: a + s = 292 89 + 203 = 292 292 = 292 3a + s = 470 3(89) + 203 = 470 267 + 203 = 470 470 = 470

Example Continued Slide 144 / 176

44 What method would require the least amount of work to solve the following system: y = 3x - 1 y = 4x A graphing B substitution C elimination

Slide 145 / 176

45 Solve the following system of linear equations using the method of your choice: y = 3x - 1 y = 4x A (-4, -1) B (-1, -4) C (-1, 4) D (1, 4) Click for answer choices AFTER students have solved the system

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46 What method would require the least amount of work to solve the following system: 4s - 3t = 8 t = -2s -1 A graphing B substitution C elimination

Slide 147 / 176

D 47 Solve the following system of linear equations using the method of your choice: 4s - 3t = 8 t = -2s -1 A B C (-2, ) 1 2 ( , -2) 1 2 ( , 2) 1 2 (2 , -2) Click for answer choices AFTER students have solved system

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48 What method would require the least amount of work to solve the following system: 3m - 4n = 1 3m - 2n = -1 A graphing B substitution C elimination

Slide 149 / 176

49 Solve the following system of linear equations using

the method of your choice:

3m - 4n = 1 3m - 2n = -1 A (-2, -1) B (-1, -1) C (-1, 1) D (1, 1)

Click for answer choices AFTER students have solved the system

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50 What method would require the least amount of work to solve the following system: A graphing B substitution C elimination y = -2x y = x + 3 1 2

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51 Solve the following system of linear equations using the method of your choice: A (-6, 12) B (2, -4) C (-2, 2) D (1, -2) y = -x y = x + 3 1 2 Click for answer choices AFTER students have solved the system

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52 What method would require the least amount

• f work to solve the following system:

u = 4v 3u - 3v = 7 A graphing B substitution C elimination

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53 Solve the following system of linear equations using the method of your choice: (28, 7) (7, ) 7 4 ( , ) 7 9 28 9 ( , ) 7 9 28 9 A B C D u = 4v 3u - 3v = 7 Click for answer choices AFTER students have solved system

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54 A piece of glass with an initial temperature of 99° F is cooled at a rate of 3.5° F/min. At the same time, a piece of copper with an initial temperature of 0° F is heated at a rate of 2.5° F/min. Let m = the number of minutes and t = the temperature in °F. Which system models the given scenario? A B C t = 99 - 3.5m t = 0 + 2.5m t = 99 + 3.5m t = 0 - 2.5m t = 99 + 3.5m t = 0 + 2.5m

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55 Which method would you use to solve the system from the previous question? t = 99 - 3.5m t = 0 + 2.5m A graphing B substitution C elimination Click to Reveal System

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56 Solve the following system of linear equations: t = 99 - 3.5m t = 0 + 2.5m A m = 1 t = 2.5 B m = 1 t = 95.5 C m = 16.5 t = 6.6 D m = 16.5 t = 41.25 Click to Reveal System

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57 Choose a strategy and then answer the question. What is the value of the y-coordinate of the solution to the system of equations x − 2y = 1 and x + 4y = 7? A 1 B

• 1

C 3 D 4

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 158 / 176

Writing Systems to Model Situations

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Slide 159 / 176 Creating and Solving Systems

Step 1: Define the variables Step 2: Analyze components and create equations Step 3: Solve the system utilizing the best strategy

Slide 160 / 176 Example

A group of 148 peole is spending five days at a summer camp. The cook ordered 12 pounds of food for each adult and 9 pounds of food for each child. A total of 1,410 pounds of food was ordered. Part A: Write an equation or a system of equations that describe the above situation and define your variables. a = number of adults c = number of children a + c = 148 12a + 9c = 1,410

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 161 / 176 Example Continued

Part B: Using your work from part A, find (1) the total number of adults in the group (2) the total number of children in the group a + c = 148 12a + 9c = 1,410 (1) (2) c = -a + 148 12a + 9(-a + 148) = 1410 12a - 9a + 1332 = 1410 3a = 78 a = 26 a + c = 148 26 + c = 148 c = 122

Slide 162 / 176

Tanisha and Rachel had lunch at the mall. Tanisha

• rdered three slices of pizza and two colas. Rachel
• rdered two slices of pizza and three colas. Tanisha’s

bill was $6.00, and Rachel’s bill was $5.25. What was the price of one slice of pizza? What was the price of

• ne cola?

p = cost of pizza slice c = cost of cola 3p + 2c = 6.00 2p + 3c = 5.25

Example

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 163 / 176

3p + 2c = 6.00 2p + 3c = 5.25 Elimination: Multiply first equation by 2 Multiply second equation by -3 Cola: $0.75 Pizza: $1.50 6p + 4c = 12

• 6p - 9c = -15.75
• 5c = -3.75

c = 0.75 3p + 2c = 6.00 3p + 2(0.75) = 6 3p + 1.5 = 6 3p = 4.5 p = 1.5

Example Continued Slide 164 / 176

58 Your class receives $1,105 for selling 205 packages

• f greeting cards and gift wrap. A pack of cards

costs $4 and a pack of gift wrap costs $9. Set up a system and solve. How many packages of cards were sold?

You will answer how many packages of gift wrap in the next question.

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59 Your class receives $1105 for selling 205 packages

• f

greeting cards and gift wrap. A pack of cards costs $4 and a pack of gift wrap costs $9. Set up a system and solve. How many packages of gift wrap were sold?

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60 The sum of two numbers is 47, and their difference is 15. What is the larger number? A 16 B 31 C 32 D 36

Slide 167 / 176

61 Ramon rented a sprayer and a generator. On his first job, he used each piece of equipment for 6 hours at a total cost of $90. On his second job, he used the sprayer for 4 hours and the generator for 8 hours at a total cost of $100 What was the hourly cost for the sprayer?

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 168 / 176

62 You have 15 coins in your pocket that are either quarters or nickels. They total $2.75. How many quarters do you have?

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63 You have 15 coins in your pocket that are either quarters or nickels. They total $2.75. How many nickels do you have?

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64 Julia went to the movies and bought one jumbo popcorn and two chocolate chip cookies for $5.00. Marvin went to the same movie and bought one jumbo popcorn and four chocolate chip cookies for $6.00. How much does one chocolate chip cookie cost? A $0.50 B $0.75 C $1.00 D $2.00

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 171 / 176

65 Mary and Amy had a total of 20 yards of material from which to make costumes. Mary used three times more material to make her costume than Amy used, and 2 yards of material was not used. How many yards of material did Amy use for her costume?

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

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66 The tickets for a dance recital cost $5.00 for adults and $2.00 for children. If the total number of tickets sold was 295 and the total amount collected was $1220, how many adult tickets were sold?

From the New York State Education Department. Office of Assessment Policy, Development and

• Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 173 / 176

67 In a basketball game, Marlene made 16 field goals. Each

• f the field goals were worth either 2 points or 3 points,

and Marlene scored a total of 39 points from field goals. Part A Let x represent the number of two-point field goals and y represent the number of three-point field goals. Write a system of equations in terms of x and y to model the

• situation. When you finish, writing your answer, type the

number "1" into your Responder.

PARCC - EOY - Question #16 Calculator Section - SMART Response Format

Slide 174 / 176

68 In a basketball game, Marlene made 16 field goals. Each

• f the field goals were worth either 2 points or 3 points,

and Marlene scored a total of 39 points from field goals. Part B How many three-point field goals did Marlene make in the game?

PARCC - EOY - Question #16 Calculator Section

Slide 175 / 176

Standards

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Slide 176 / 176

Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of

• thers.

MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.