SECTION 4.3 Solving Systems of Linear Equations by Elimination^267
NOTE In Example 3,we eliminated the variable x. Alternatively, we could multiply
each equation of the system by a suitable number so that the variable yis eliminated.
Multiply by 2.
(1)
Multiply by
(2)Complete this approach and confirm that the same solution results.
5 x+ 2 y= 1 - 15 x- 6 y=- 3
2 x+ 3 y=- 15 4 x+ 6 y=- 30
NOW TRY
EXERCISE 4
Solve the system.
5 y= 6 - 3 x4 x+ 9 y= 3OBJECTIVE 3 Use an alternative method to find the second value in a
solution. Sometimes it is easier to find the value of the second variable in a solu-
tion by using the elimination method twice.
CAUTION When using the elimination method, remember to multiply both
sidesof an equation by the same nonzero number.
Finding the Second Value by Using an Alternative MethodSolve the system.
(1)
(2)Write equation (1) in standard form by adding 3yto each side.
(3)
(2)One way to proceed is to eliminate yby multiplying each side of equation (3) by 2
and each side of equation (2) by 3 and then adding.
Multiply equation (3) by 2.
Multiply equation (2) by 3.
Add.x= Divide by 23.
42
23
23 x = 42
15 x-^6 y=^24
8 x+ 6 y= 18
5 x- 2 y= 8
4 x+ 3 y= 9
5 x- 2 y= 8
4 x= 9 - 3 y
EXAMPLE 4
Substituting for xin one of the given equations would give y, but the arithmetic
would be messy. Instead, solve for yby starting again with the original equations
written in standard form (equations (3) and (2)) and eliminating x.
Multiply equation (3) by 5.
Multiply equation (2) by.
Add.Divide by 23.Check that the solution set is EA^4223 ,^1323 BF. NOW TRY
y=
13
23
23 y= 13
-^20 x+^8 y=-^32 -^4
20 x+ 15 y= 45
42
23The coefficients of y
are opposites.NOTE When the value of the first variable is a fraction, the method used in Example 4
helps avoid arithmetic errors. This method could be used to solve any system.
NOW TRY ANSWER
- EA^397 , -^157 BF
The coefficients of x
are opposites.