Solving systems of equations with 2 variables
This presentation is the property of its rightful owner.
Sponsored Links
1 / 10

Solving systems of equations with 2 variables PowerPoint PPT Presentation


  • 61 Views
  • Uploaded on
  • Presentation posted in: General

Solving systems of equations with 2 variables. Word problems (Number Problems). 1) The sum of two numbers is 72. Their difference is 40. Find the numbers. The sum of two numbers is 72. x + y = 72 Their difference is 40.

Download Presentation

Solving systems of equations with 2 variables

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Solving systems of equations with 2 variables

Solving systems of equations with 2 variables

Word problems

(Number Problems)


1 the sum of two numbers is 72 their difference is 40 find the numbers

1) The sum of two numbers is 72. Their difference is 40. Find the numbers.

The sum of two numbers is 72.

x + y = 72

Their difference is 40.

x – y = 40


1 the sum of two numbers is 72 their difference is 40 find the numbers1

1) The sum of two numbers is 72. Their difference is 40. Find the numbers.

x + y = 72

x – y = 40

Which method should be used to solve this system of equations?

a) Substitution Method

b) Elimination (Addition) Method


1 the sum of two numbers is 72 their difference is 40 find the numbers2

1) The sum of two numbers is 72. Their difference is 40. Find the numbers.

Solve using the Elimination (Addition) Method

x + y = 72

x – y = 40

2x = 112

x = 56

Back substitute

56 + y = 72

56 + y + (-56) = 72 + (-56)

y = 16

The numbers are 56 and 16.


2 the sum of two numbers is 21 their difference is 13 find the numbers

2) The sum of two numbers is 21. Their difference is 13. Find the numbers.

Solve using the Elimination (Addition) Method

x + y = 21

x – y = 13

2x = 34

x = 17

Back substitute

17 + y = 21

17 + y + (-17) = 21 + (-17)

y = 4

The numbers are 17 and 4.


3 the sum of two numbers is 27 one number is three more than the other find the numbers

3) The sum of two numbers is 27. One number is three more than the other. Find the numbers.

The sum of two numbers is 27.

x + y = 27

One number is 3 more than the other.

y = x + 3


3 the sum of two numbers is 27 one number is three more than the other find the numbers1

3) The sum of two numbers is 27. One number is three more than the other. Find the numbers.

x + y = 27

y = x + 3

Which method should be used to solve this system of equations?

a) Substitution Method

b) Elimination (Addition) Method


3 the sum of two numbers is 27 one number is three more than the other find the numbers2

3) The sum of two numbers is 27. One number is three more than the other. Find the numbers.

Solve using the Substitution Method

x + y = 27

y = x + 3

x + (x + 3) = 27

2x + 3 = 27

2x + 3 + (-3) = 27 + (-3)

2x = 24

x = 12

The numbers are 12 and 15.

Back substitution

y = x + 3

y = 12 + 3

y = 15


Solving systems of equations with 2 variables

4) A 36-ft rope is cut into two pieces. One piece is three times the other. Find the length of each piece.

Solve using the Substitution Method

x + y = 36

y = 3x

x + (3x) = 36

4x = 36

x = 9

The pieces are 9 ft and 27 ft.

Back substitution

y = 3x

y = 3(9)

y = 27


Solving systems of equations with 2 variables

5) The difference between two numbers is 3. The larger number is one more than twice the smaller number. Find the numbers.

Solve using the Substitution Method

L – S = 3

L = 2S + 1

(2S + 1) – S = 3

S + 1 = 3

S + 1 + (-1) = 3 + (-1)

S = 2

The numbers are 2 and 5.

Back substitution

L = 2S + 1

L = 2(2) + 1

L = 5


  • Login