MA 265 HOMEWORK ASSIGNMENT #1 SOLUTIONS
#1.
Page 8; Exercise 1.
In Exercises 1 through 14, solve each given linear system by the method of
elimination.
x
+
2
y
=
8
3
x

4
y
=
4
Solution:
First we eliminate
x
. Multiply the first equation by (

3), then add to the second:

3
x

6
y
=

24
3
x

4
y
=
4

10
y
=

20
=
⇒
y
= 2
.
Substituting this value into the equation
x
+ 2
y
= 8 yields
x
+ 4 = 8, so that
x
= 4. Hence
x
= 4,
y
= 2.
#2.
Page 8; Exercise 2.
In Exercises 1 through 14, solve each given linear system by the method of
elimination.
2
x

3
y
+
4
z
=

12
x

2
y
+
z
=

5
3
x
+
y
+
2
z
=
1
Solution:
First we eliminate
x
. Multiply the second equation by (

2), then add to the first; and multiply
the second equation by (

3) then add to the third:
2
x

3
y
+
4
z
=

12

2
x
+
4
y

2
z
=
10
y
+
2
z
=

2

3
x
+
6
y

3
z
=
15
3
x
+
y
+
2
z
=
1
7
y

z
=
16
This gives the system of equations
y
+
2
z
=

2
7
y

z
=
16
Now we eliminate
y
. Multiple the first equation by (

7) and add to the second:

7
y

14
z
=
14
7
y

z
=
16

15
z
=
30
=
⇒
z
=

2
.
Substituting this value into the equation
y
+ 2
z
=

2 yields
y

4 =

2, so that
y
= 2.
Finally, sub
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 Spring '10
 ...
 Equations, Elementary algebra

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