3.2 Solve Linear Systems Algebraically
Substitution method
Substitute an expression into one of the equations to solve for
the variable
Elimination method
Eliminate one of the variables by adding equations
THE SUBSTITUTION METHOD
Step 1 Solve one of the equations
for one of its variables.
Step 2 Substitute the expression from _Step 1_
into the other equation and solve for the other variable.
Step 3 Substitute the value from _Step 2_ into
the revised equation from Step 1 and solve.
Example 1
Use the substitution method
Solve the system using the substitution method.
|
x + 2y = -2 |
Equation 1 |
|
3x + 4y = 6 |
Equation 2 |
1.
Solve Equation 1
for x.
|
x =_ - 2y - 2_ |
Revised Equation 1. |
2.
Substitute into
Equation 2 and solve for y.
|
3(- 2y — 2
) + 4y = 6 |
Substitute for x. |
|
y =_ -6_ |
Solve for y. |
3.
Substitute into
revised Equation 1 and solve for x.
|
x =_ -2y— 2_ |
Write revised Equation 1. |
|
x =_ -2(-6) - 2 _ |
Substitute -6 for y. |
|
x =_ 10 _ |
Simplify. |
|
The solution is (_ 10 _,_ -6 _). |
|
THE ELIMINATION METHOD
Step 1 Multiply one or both
of the equations by a _constant_ to obtain coefficients that differ only
in _sign_ for one of its variables.
Step 2 Add the revised equations from _Step
1 _.Combining like terms will _eliminate one
of the variables. Solve for the remaining variable.
Step 3 Substitute
the value obtained in _Step 2 _into either of the original equations and
solve for the other variable.
Example 2
Use the elimination method
Solve the system using the elimination method.
|
2x + 5y = 14 |
Equation 1 |
|
4x + 2y = -4 |
Equation 2 |
1.
Multiply Equation 1
by _-2_ so that
the coefficients of x differ only in sign.
|
2x + 5y = 14 x
_ -2_ |
_-4x - 10y
= -28_ |
|
4x + 2y = -4 |
|
|
2.
Add the
revised equations |
_-8y
= -32_ |
|
and solve for y. |
y = _4_ |
3.
Substitute the value
of y into one of the original equations. Solve for x.
|
2x + 5y = 14 |
Write Equation 1. |
|
2x + 5(_4_) = 14 |
Substitute for y. |
|
2x + _20_ = 14 |
Simplify. |
|
x = _-3_ |
Solve for x. |
|
The solution is (_-3_, _4_) |
|
CHECK You can check the solution
algebraically using the method shown in Example 1. You can also use a graphing
calculator to check the solution.
Example 3
Use the elimination method
Solve the linear system using the elimination method.
|
3x - 4y = -37 |
Equation 1 |
|
-5x + 3y = 14 |
Equation 2 |
Solution
Multiply Equation 1 by _5_ and Equation 2 by _3_ so
that the coefficients of x only differ in sign.
|
3x - 4y
= -37 ´ _5_ |
_15x - 20y
= -185_ |
|
-5x + 3y = 14 ´_3_ |
|
|
Add the
revised equations and |
_-11y
= -143_ |
|
Solve for y. |
y = _13_ |
Substitute the value of y into one of the original
equations and solve for x.
|
3x - 4y
= -37 |
Write equation 1. |
|
3x - 4(_13_) = -37 |
Substitute for y. |
|
x = _5_ |
Solve for x. |
The solution is (_5_, _13_). Check the solution
Algebraically or graphically.
Example 4
Solve linear systems with many or no solutions
Solve the linear system.
|
a. x -3y = 7 |
b. 2x - 6y
= 12 |
|
2x - 6y
= 12 |
-5x + 15y = -30 |
Solution
a. Because
the coefficient of x in the first equation is _1_,use the substitution method. Solve the first equation for x.
x - 3y =
7
x = _3y+ 7_
Substitute the expression forx into the second equation.
|
2x - 6y
= 12 |
Write second equation. |
|
2( 3y+ 7 ) - 6y = 12 |
Substitute for x. |
|
_14_ = _12_ |
Simplify |
Because the statement _14_ = _12_ is _never true_
there _is no solution_.
b. Because
no coefficient is _1 or -1_ , use the elimination method. Multiply the first equation by _5_
and the second equation by _2_.
|
2x - 6y
= 12 ´ _5_ |
_10x-30y
= 60_ |
|
-5x + 15y = -30 ´_2_ |
|
|
Add the revised equations |
_0 = 0_ |
Because the equation _0_ = _0_
is _always true_ ,there _are infinitely many
solutions_.