x 1 y x 2 1 7 x y 2 7 x The system of equations - - PowerPoint PPT Presentation

DAY 78 – SOLVING SYSTEMS OF EQUATIONS USING SUBSTITUTION

WORK WITH A PARTNER TO EXPAND YOUR

UNDERSTANDING OF SOLVING SYSTEMS WITH SUBSTITUTION.

Explain: In the system of equations above, the first equation is “ ”. Explain what it means for two expressions to be equal.

1   x y

ANSWER KEY

Explain: In the system of equations above, the first equation is “ ”. Explain what it means for two expressions to be equal. They are the same even though they may look different

1   x y

OBSERVE AND DISCOVER

1   x y 7 2   y x

7 1 2    x x

The system of equations is written on the left. The first step of the substitution method is on the right. Explain WHAT was done and WHY it would be OK to do that.

ANSWER KEY

Explain WHAT was done and WHY it would be OK to do that. They y was replaced with x + 1. This is “OK” because y = x + 1. y and x + 1 are equivalent.

Using the equation from step 1 of the substitution method (also written below), solve for the variable. (Show your work!)

7 1 2    x x

ANSWER KEY

7 1 2    x x

7 1 3   x

1 1 

3 6 3 3  x

2  x

FINISH SOLVING

1   x y 7 2   y x

Select either equation from the system and write it here: ___________ Using your answer from the “Begin to Solve” step above, how could you find the value of y? Explain in words. Now actually do the work you just described and find the value of . (Remember to show your work.)

FINISH SOLVING

Select either equation from the system and write it here: ___________ Using your answer from the “Begin to Solve” step above, how could you find the value of y? Explain in words. I substituted the 2 in the place of the x and then calculated the result Now actually do the work you just described and find the value of . (Remember to show your work.)

1   x y 1 2  y 3  y

FINISH SOLVING

Let’s see if it would make a difference if we had chosen the

• ther equation from the system to find . Write down the

equation you didn’t pick … follow the same process … and find the value of again. Was your answer the same?

FINISH SOLVING

Let’s see if it would make a difference if we had chosen the

• ther equation from the system to find . Write down the

equation you didn’t pick … follow the same process … and find the value of again. Was your answer the same? YES

7 ) 2 ( 2   y 7 4   y 3  y

State the values you found for x and y as an ordered pair (x,y): _________________________ Substitute those values for x and y into both equations and simplify.

1   x y 7 2   y x

State the values you found for x and y as an ordered pair (x,y): _________________________ Substitute those values for x and y into both equations and simplify.

1   x y 7 2   y x

1 2 3   3 3 

7 7 7 3 ) 2 ( 2   

If the result for both equations is a true statement (examples: 5=5 … 24=24 … -2=-2 … etc.), that means our ordered pair IS a solution to the system because it satisfies (makes true) both equations.

LET’S GRAPH THE SYSTEM OF EQUATIONS TO

CONFIRM OUR SOLUTION FROM A GRAPHICAL PERSPECTIVE.

1   x y

Slope= ____ y-intercept= ____

7 2   y x

Solve for y to get in slope- intercept form Slope= ____ y-intercept= ____ Graph the 2 equations on the graph above. State the intersection point: _______ How does this compare to the solution you got using substitution?

LET’S GRAPH THE SYSTEM OF EQUATIONS TO

CONFIRM OUR SOLUTION FROM A GRAPHICAL PERSPECTIVE.

1   x y

Slope= 1 y-intercept= 1

7 2   y x

Solve for y to get in slope- intercept form Slope= -2 y-intercept= 7 Graph the 2 equations on the graph above. State the intersection point: (2,3) How does this compare to the solution you got using substitution?

SUBSTITUTION

Explain each step in solving this system of equation using substitution.

26 2 3 2 4     x y x y

ORIGINAL SYSTEM OF EQUATIONS

26 2 ) 2 4 ( 3    x x

Explain in words what’s been done in each step.

SUBSTITUTION

Explain each step in solving this system of equation using substitution.

26 2 3 2 4     x y x y

ORIGINAL SYSTEM OF EQUATIONS

26 2 ) 2 4 ( 3    x x

Substitute 4x + 2 in place of the y Explain in words what’s been done in each step.

SOLVE THE EQUATION FOR X

26 2 6 12    x x

26 2 6 12    x x 26 6 10   x

6  6 

10 20 10 10  x 2  x

Distribute Solve the equation for x

SUBSTITUTE THE X INTO

EITHER OF THE ORIGINAL EQUATIONS TO SOLVE FOR Y

2 4   x y 2 ) 2 ( 4   y 10  y 26 2 3   x y 26 ) 2 ( 2 3   y 26 4 3   y 4  4 

3 30 3 3  y

10  y

NOW YOUR TURN …

Solve these systems of equations using substitution

1)

x y x y 4 28 2 3   

28 2 ) 4 ( 3   x x

28 2 12   x x 14 28 14 14  x 2  x

) 2 ( 4  y 8  y

NOW YOUR TURN …

Solve these systems of equations using substitution

2)

y x y x     15 5 2 3

5 2 ) 15 ( 3    y y 5 2 3 45    y y 5 5 45   y

45  45 

5 40 5 5      y

8  y

8 15  x 7  x

WORD PROBLEM

Josh cannot find a publisher for a book he wrote, so he decides to publish the book

• himself. He calculates that his fixed costs

will be $2665 and that the unit cost of each book will be an additional $1.70. He intends to sell the book for $4.95. How many copies must he sell to break even? ("Break even" means that income equals expenses.)

ANSWER KEY

Let x represent the number of copies Josh must sell to break

• even. Let y represent the break-even amount. Write two

equations, one representing expenses and one representing income.

820 2665 25 . 3 2665 7 . 1 ) 95 . 4 ( 95 . 4 2665 7 . 1        x x x x x y x y

Expenses equation Income equation Substitution 4.95x for y in the expenses equation. Subtract 1.7x from both sides. Divide both sides by 3.25.

SOLVE BY THE SUBSTITUTION METHOD:

        1 3 5 4 2 y x y x

Since the x in equation 1 has a coefficient of -1, it will probably be simpler to solve this equations for x

Equation 1 Equation 2

4 2    y x y x 2 4   

Multiply each side by (-1)

4 2   y x

1 2

1 3 5   y x

Replace x with (2y-4) Solve for y

1 3 ) 4 2 ( 5    y y 1 3 20 10    y y 21 7  y 3  y

3

ANSWER KEY

3  y

Replace y with 3 in Equation 1

3

4 2    y x

1

4 ) 3 ( 2    x

2  x

4

)} 3 , 2 {(

Solution set:

SOLVE THE SYSTEM BY SUBSTITUTION

        2 3 4 3 4 y x y x

Equation 1 Equation 2

The variable easiest to isolate is y in Equation 2

2 10 5 6 4 6 6 5 4 6 5 4 6 9 4 4 ) 2 3 ( 3 4 2 3 2 3                         x x x x x x x x x y y x

Equation 2 Solve for y. Substitute (-3x-2) for y in Equation 1. Use the distributive Property. Combine like terms. Add 6 to both sides. Simplify. Divide both sides by -5

Having found x, you can now substitute in Equation 2.

4 2 6 2 ) 2 ( 3          y y y

Substitute -2 for x Simplify. Solve for y.