Elimination Using Multiplication
Objective Learn how to use the Multiplication
Property of Equality to modify systems of equations and then to
solve them by using elimination by addition or subtraction.
In this lesson, you will learn a method that can be used to
solve all systems of linear equations. The methods introduced so
far only work in special cases, such as when the coefficient of
one variable is 1, or when the coefficient of a variable in
one equation is equal to or opposite of the coefficient in the
other.
Elimination Using Multiplication
Recall the three methods for solving systems of equations that
students have learned so far. Point out that each method has
advantages and disadvantages.
(A) Solving by Graphing This method can be
used for all systems, but produces only approximate solutions.
(B) Solving by Substitution This method can
be used when the coefficient of one of the variables in one of
the equations is 1.
(C) Elimination by Addition or Subtraction
This method works when the coefficients of one variable are equal
or opposite.
Now let's develop an additional method, which will allow us to
solve all systems of linear equations. Look at the following
system of equations.
3x + 5y = 11
6x + 4y = 16
This system of equations cannot easily be solved by
substitution, since none of the coefficients is 1 or 1. Also, it
cannot be solved using elimination by addition or subtraction. In
this case, use the Multiplication Property of Equality to
multiply the first equation by 2.
The reason for multiplying by 2 is that this makes the
coefficient of x in the first equation equal to 6, which is the
coefficient of x in the second equation. Then the x values can be
eliminated by subtracting the second equation from the first.
Another way to solve this system would be to multiply the first
equation by 2 and then add the equations.
After multiplying by 2, the first equation becomes
6x + 10y = 22.
Now subtract the second equation from this modified first
equation.
6x + 10y = 22 

(  )6x + 4y = 16 
Subtract the equations. 
0 + 6y = 6 

6y = 6 

y = 1 
Divide each side by 6. 
Now substitute 1 for y in, say, the first equation.
3x + 5(1) = 11 
Substitute 1 for y. 
3x + 5 = 11 

3x = 6 
Subtract 5 from each side.

x = 2 
Divide each side by 3. 
The solution is (2, 1).
Key Idea
Any equation in a system of equations can be multiplied by a
nonzero number to obtain an equivalent equation.
Use this fact to complete the following exercises.
Exercises
Use elimination to solve each system of equations.
1. 
3x + 6y = 15 
2. 
4x  3y = 14 
3. 
1.5x + 2.5y = 4 

2x + 7y = 13 

7x + 2y = 39 

0.5x  3.5y = 3 

(3, 1) 

(5, 2) 

(1, 1) 
