Download Document

3.2
Solve each system graphically.
Then classify it as consistent, inconsistent, dependent or independent
1. 4x + 2y = -8
-2x - y = 6
2. 3x + 2y = 12
x – y = -1
3. 2x + y = 3
4x + 2y = 6
y = -2x – 4
y = -2x - 6
y = -3/2x + 6
y=x+1
y = -2x + 3
y = -2x + 3
same slope
no solution
inconsistent
(2,3)
independent
consistent
same line
infinite solutions
consistent
dependent
3.2
Here are 2 more ways that you can solve a linear system besides graphically.
Substitution
• Solve an equation for one of the variables
• Substitute the new equation into the other equation
and solve for the variable
• Substitute again to find the remaining variable
Elimination
• Multiply one or both equations by a constant so that your
coefficients are the same, but have different signs
• Combine the two equations to eliminate one variable
• Substitute that variable into any equation to find your remaining variable
Solve the system using the substitution method.
EXAMPLE 1
2x + 5y = – 5
x + 3y = 3
Pick the easiest variable to solve for.
(Solve the second equation for x.)
x = – 3y + 3
Substitute x into the first equation and solve for y.
2x +5y = – 5
2(– 3y + 3) + 5y = – 5
-6y + 6 + 5y = – 5
-y + 6 = – 5
-y = – 11
y = 11
Substitute y into any equation and solve for x.
x = – 3y + 3
x = – 3(11) + 3
x = – 33 + 3
x = – 30
The solution is (– 30, 11).
Solve the system using the elimination method.
EXAMPLE 2
3x – 7y = 10
6x – 8y = 8
(Try to only change one equation if possible
and try to pick the smallest number to change)
Multiply the first equation by – 2 so that the coefficients of x differ only in sign.
3x – 7y = 10
6x – 8y = 8
Multiply each # by -2
No change
– 6x + 14y = -20
6x – 8y = 8
6y = – 12
y=–2
Substitute y into one of the original equations. Solve for x.
3x – 7y = 10
3x – 7( – 2) = 10
3x + 14 = 10
3x = -4
4
x =– 3
The solution is ( –
4
, – 2)
3
Solve the systems using the substitution or the elimination method.
1. 4x + 3y = – 2
x + 5y = – 9
(Use substitution because the x has no coefficient)
Solve Equation 2 for x.
x = – 5y – 9
Substitute x into Equation 1 and solve for y.
4x + 3y = – 2
2. 3x + 3y = – 15
5x – 9y = 3
(Multiply Equation 1 by 3 so that the
coefficients of y differ only in sign.)
3x + 3y = – 15
x3
5x – 9y = 3
4(– 5y – 9) + 3y = – 2
-20y - 36 + 3y = – 2
-17y - 36 = – 2
-17y = 34
y=–2
Substitute y into revised Equation 2 and solve for x.
x = – 5y – 9
x = – 5(-2)– 9
x = 10 – 9
x=1
The solution is (1,– 2).
9x + 9y = – 45
5x – 9y = 3
14x
= – 42
x=–3
Substitute the value of x into one of the original
equations. Solve for y.
3x + 3y = – 15
3 ( – 3) + 3y = – 15
– 9 + 3y = – 15
3y = – 6
y= –2
The solution is (-3,– 2).
Solve the system using the substitution or the elimination method.
3. 3x – 6y = 9
– 4x + 7y = – 16
x4
12x – 24y = 36
x3
– 12x + 21y = – 48
– 3y = – 12
y=4
Substitute the value of y into one of
the original equations. Solve for y.
3x – 6y = 9
3x – 6 (4) = 9
3x – 24 = 9
3x = 33
x = 11
The solution is (11 , 4)
Solve
linear systems
with many or no solutions
EXAMPLE
4
a. x – 2y = 4
b.
4x – 10y = 8
3x – 6y = 8
– 14x + 35y = – 28
(use substitution)
(use elimination)
28x – 70y = 56
– 28x + 70y = – 56
0 = 0
Solve the first equation for x.
x = 2y + 4
Substitute
3x – 6y = 8
3(2y + 4) – 6y = 8
6y + 12 – 6y = 8
12 = 8
Because the statement 12 = 8 is never true,
there is no solution.
Because the equation 0 = 0 is always true,
there are infinitely many solutions.
Solve the linear system using any algebraic method.
c. 12x – 3y = – 9
– 4x + y = 3
(Substitution)
– 4x + y = 3
d.
6x + 15y = – 12
– 2x – 5y = 9
6x + 15y = – 12
x3
– 6x – 15y = 27
0 = 15
y = 4x + 3
not true - there are no solutions.
12x – 3y = – 9
12x – 3(4x + 3) = – 9
12x – 12x - 9 = – 9
–9=–9
True - infinitely many solutions
Solve the linear system using any algebraic method.
e. 5x + 3y = 20
–x– 3 y=–4
5
(use substitution)
Solve the 2nd equation for x.
–x– 3 y=–4
5
x=– 3 y+4
5
5x + 3y = 20
5 ( – 3 y + 4 ) + 3y = 20
5
-3y + 20 + 3y = 20
20 = 20
infinitely many solutions
HOMEWORK 3.2
P. 164 #3-40(EOP)47, 48