Solving The System of Equations by Elimination Method

 Key Concepts

• Use elimination to solve a system of equations by adding
• Use elimination to solve a system by subtracting
• Use elimination to solve a system by multiplying

Solving System of Equations by Elimination 

1. Additive inverse of 5 is ………………. 

2. (-10) + 10 is equal to ………….. 

3. Identify and write the like terms in each of the following group………..
   • 5x, 9xy, –8xy, –6x, 3x 

4. Simplify the following   …………………………
   • 10 m²-9m +7m -3 m²-5m-8 

5. What is the additive inverse of ‘3x’……… why? 

6. Simplify and find the value of 4x+x-2x2x2 + x-1 when x = –1. 

7. Coefficient of – x is ……………. 

8. Use substitution to solve the system. y=x+1 and y = 5x – 3 is ………… 

9. Graph of the system and determine its solution. ………….. 

y=x+4 and y = -2x + 1 

10. Is there another way to solve system ………… 

Answers: 

1. –5  

2. 0 

3. 5x, –6x, 3x  and 9xy, –8xy 

4. 3m² – 7m – 8  

5. –3x 

For ‘3x’ there also exists ‘–3x’ such that 3x + (–3x ) = 0 

6. –9 

7. –1 

8. X=1 and y=2 

9. (–1, 3) 

10. I don’t know 

Ok let’s know about the new method. 

Solve and Discuss It!  

A list of expressions is written on the board. How can you make a list of fewer expressions that has the same combined value of those shown on the board? Write the expressions and explain your reasoning. 

Solution: 

The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms in addition.  

The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms. 

Use distributive property: 

1. 2y – 3 y  = (2-3)y = -1y 

2. 3x – 2x = (3-2)x=1x 

3. 5 – 3 = 2 

4. –5x + 5x = 0  

Thus, for every algebraic expression there exists another algebraic expression such that their sum is zero. These two expressions are called the additive inverse of the each other. 

Addition and subtraction of unlike terms:  

3x and 4y are unlike terms.  

Their sum can be written as 3x + 4y. However, ‘x’ and ‘y’ are different variables so we can not apply distributive law and thus cannot add them. 

Focus on math practices Reasoning Two expressions have a sum of 0. What must be true of the expressions? 

A set of equations is called a system of equations. The solutions must satisfy each equation in the system. 

Systems of linear equations:  

A solution to a system of equations is an ordered pair that satisfy all the equations in the system.  

A system of linear equations can have:  

1. Exactly one solution  

2. No solutions  

3. Infinitely many solutions 

There are four ways to solve systems of linear equations: 

1. By slope and intercept form  

2. By graphing  

3. By substitution  

4. By elimination  

Solve Systems by Elimination 

Elimination method for solving linear equations in two variables.  

The steps to solve linear equations in two variables by elimination method are given below:  

Step 1:  

Write both equations in standard form. If any coefficients are fractions, clear them. 

Step 2: 

Make the coefficients of one variable opposites. 

*Decide which variable you will eliminate. 

*Multiply one or both equations so that the coefficients of that variable are opposites. 

Step 3:  

Add the equations resulting from Step 2 to eliminate one variable. 

Step 4: 

Solve for the remaining variable. 

Step 5: 

Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. 

Step 6:  

Write the solution as an ordered pair. 

Step 7: 

Check that the ordered pair is a solution to both original equations.  

Solve a System of Equations by Adding 

Example 1:  

How can Remi use this system of equations to solve the riddle? 

2x + y = 8    and     4x – y = 4 

Solution: 

Step 1: Eliminate one variable. 

Step 2: Solve for the other Variable. 

∴The Solution is (2,4), so x = 2 , and y = 4. 

Solve a System of Equations by Subtracting: 

Example 2: 

Amelia is helping her younger brother learn to count money. She gave her brother a total of 20 coins, nickels, and pennis. Amelia’s brother said he counted $0.68. How can Amelia solve a system of Equations to tell whether he counted the money correctly? 

Solution: 

Step 1: 

Write a system of equations to relate the number of nickels and pennies. 

Let x = the number of pennies. 

Let y = the number of nickels. 

Step 2: 

Eliminate one variable. The difference of the coefficients of x is 0. Apply the Subtraction Property of equality to subtract the equations to eliminate x. 

Step 3: 

Solve for the other variable. 

x + y    = 20 

x + 12 = 20     (∵y=12) 

∴     x   = 8 

Amelia’s brother counted correctly if Amelia gave him 8 pennies and 12 nickels. 

Solve a System of Equations by Multiplying : 

Example 3:  

The difference of the lenght and width of the rectangle is 3 centimeters. What are the length and width of the rectangle? 

Solution: 

Step 1: 

Write a system of equations to relate the length and width. 

2l + 2w = 26  

   l  –  w  = 3 

Step 2: 

Eliminate one variable. The coefficients of l and w are not the same or opposites. Multiply one or both of the equations so that the variables are the same or opposites. 

Step 3 : 

Solve for the other variable. 

l – w = 3  

8 – w = 3      (∵l=8) 

      w = 5 

The Length of the rectangle is 8 centimeters and the width is 5 centimeters. 

Problems and solving. 

1. Solve each system by Elimination method. Describe the solutions.  

y – x = 28   and    y + x = 156 

Solution: 

Step 1: 

Write the given system of equations  

y + x  = 156 

y –  x  = 28  

Step 2: 

Eliminate one variable. The difference of the coefficients of x is 0. Apply the Subtraction Property of equality to subtract the equations to eliminate x. 

Step 3 : 

Solve for the other variable. 

y – x = 28 

y – 64 = 28   ( ∵ x= 64) 

y    = 92 

∴The x and y values are 64 and 92 

Let’s check our knowledge: 

1. Solve the system by elimination method  3c + 6d = 18 and 6c – 4d = 4  

2. Solve the system by elimination method  2x – 2y = -4 and  2x + y = 11 

3. Solve the system by elimination method 3x + y = 8 and x – y = 12 

4. Solve the system by elimination method 2r + 3s = 14 and 6r – 3s = 6 

5. Solve the system by elimination method x + y = 56 and x + 2y = 94 

Answers: 

1. Solve the system by elimination method 3c+ 6d = 18 and 6c – 4d = 4  

Solution: 

Step 1: 

Write the given equations  

3c + 6d = 18  

6c – 4d =4 

Step 2: 

Eliminate one variable. The coefficients of c and d  are not the same or opposites. Multiply one or both of the equations so that the variables are the same or opposites. 

                               6c + 12d = 36 

                            -( 6c – 4d  )  = -4 

                            ____________________ 

                               0 + 16d  = 32 

                                            d  = 2  

Step 3 : 

             Solve for the other variable. 

                  6c – 4d = 4  

              6c – 4(2) = 4      (∵ d =2) 

                   6c – 8  = 4 

                          6c = 4+8 

                          6c = 12 

                             c = 2 

The values of c and d are 2 and 2 . 

2. Solve the system by elimination method  

             2x – 2y = -4 

             2x + y  = 11  

Solution: 

Step 1: Write the given system of equations. 

                    2x – 2y = -4 

                    2x + y  = 11  

Step 2: Eliminate one variable. The difference of the coefficients of x is 0. Apply the Subtraction Property of equality to subtract the equations to eliminate x. 

                           2 x – 2y = -4 

                      -( 2x + y  ) = -11    

                        _________________ 

                           0 – 3y    =  -15  

                               – 3y   = -15 

                                     y  = 5 

Step 3: Solve for the other variable. 

                         2x + y = 11  

                         2x + 5 = 11   (∵ y=5) 

                                 2x = 11 – 5 

                                 2x = 6 

                                   x = 3 

∴The values of x and y are 3 and 5 

3. Solve the system of equations by elimination method.  

               3x + y = 8 and x – y = 12  

Solution: 

Step 1:    Eliminate one variable. 

                     3x + y = 8 

                  +  x  –  y = 12 

                  _________________ 

                      4x + 0 = 20 

                              4x = 20 

                                x  = 5 

Step 2:   Solve for the other variable. 

                         x – y = 12 

                         5 – y = 12  

                              y  = 5 – 12 

                              y  = -7 

The solution is (5, -7), so x = 5 and y = -7.  

4. Solve the system of equations by elimination method.  

       2r + 3s = 14 and 6r – 3s = 6 

Solution : 

Step 1: Write the given system of equations. 

                                  2r + 3s = 14 

                                  6r – 3s   = 6 

 Step 2:  add both equations  

Step 3:  Solve for the other variable. 

5. Solve the system of equations by elimination method.  

       x + y = 56 and x + 2y = 94 

Solution: 

Step 1: Write the given system of equations. 

                                       x + y = 56 

                                     x + 2y = 94 

Step 2: Eliminate one variable. The difference of the coefficients of x is 0. Apply the Subtraction Property of equality to subtract the equations to eliminate x. 

                                        x + 2y = 94 

                                   – (x + y )= -56 

                                 ___________________ 

                                     0 + y     = 38 

                                             y     = 38 

Step 3: Solve for the other variable. 

                                x + y = 56 

                             x + 38 = 56 

                                      x   = 56 – 38 

                                     x    = 18    

The Solution of (x, y) is ( 18, 38).                                

Key concept  

You can apply the properties of equality to solve systems of linear equations algebraically by eliminating a variable. 

Elimination is an efficient method when: 

• Like terms have the same or opposite coefficients. 

• One or both equations can be multiplied so that like variable terms have the same or opposite coefficients.  

Exercise:

1. A father’s age is three times the sum of the ages of his children. After five years, his
   age will be two times the sum of their ages. Find the present age of the father.
2. The ratio of the incomes of the two persons is 9:7 and the ratio of their expenditure
   is 4:3. If each of them saves $200 per month. Find their monthly income?
3. Seven times a 2-digit number is equal to four times the number obtained by
   reversing the order of the digits. If the sum of both the digits is 9. Find the number.

Concept map: