Section 4.5 Systems of Linear Equations in Three Variables
4.5 Check Points
1. Test the ordered triple in each equation.
2 3 22
( 1) 2 ( 4) 3 (5) 22
22 22, true
xyz−+=
−− −+ =
=
23 5
2( 1) 3( 4) (5) 5
5 5, true
xyz−−=
−−− − =
=
3532
3( 1) ( 4) 5(5) 32
32 32, true
xy z+− =−
−+−− =−
−=−
The ordered triple (–1, –4, 5) makes all three equations true, so it is a solution to the system.
2.
420
32 8
232 16
xyz
xyz
xyz
+−=
++=
−+=−
Add the first two equations to eliminate z.
45 24
xy
+=
Multiply the second equation by –1 and add the equations.
46 28
45 24
4
xy
xy
y
+=
−− =−
=
Back-substitute 4 for y to find x.
46 28
46(4)28
xy
x
+=
+=
3.
27
217
232 1
yz
xyz
xyz
−=
++=
−+=−
Chapter 4 Systems of Linear Equations
Back-substitute 5 for y to find z.
27
2(5) 7
10 7
3
3
yz
z
z
z
z
−=
−=
−=
−=−
=
4. Use each ordered pair to write an equation.
() ()
2
2
(, ) (1,4)
41 1
4
xy
yax bxc
abc
abc
=
=++
=++
=++
The system of three equations in three variables is:
4
42 1
abc
abc
++=
++=
Section 4.5 Systems of Linear Equations in Three Variables
Multiply the first equation by –2 and add to the second equation.
12
b
=−
Back-substitute into an original equation to find c.
4
(3) ( 12) 4
94
13
abc
c
c
c
++=
+− + =
−+ =
=
The equation is
2
31213yx x
=−+
.
4.5 Concept and Vocabulary Check
1. triple; all
4.5 Exercise Set
1. Test the ordered triple in each equation.
4
xyz
++=
2 1
xyz
−−=
21
xyz
−−=−
2. Test the ordered triple in each equation.
02353421
xyz x y z x y z
++= + − = + + =−
3. Test the ordered triple in each equation.
22 2311 4 7
xy xy yz
−= += −=−
Chapter 4 Systems of Linear Equations
4. Test the ordered triple in each equation.
25 33 2 4
xz yz xz
−=− −=− −=−
5.
211
314
2 5
xyz
xyz
xyz
++=
++=
+−=
Multiply the second equation by –1 and add to the first equation.
Back-substitute 3 for z in the first and third equations.
()
211 25
23 11 235
28
611
5
xy z xyz
xy xy
xy
xy
xy
++ = +−=
++ = +−=
+=
++=
+=
We now have two equations in two variables.
6. Multiply the first equation by 3 and add to the second equation.
636 3
33 5
9 7 2
xyz
xyz
xz
+−=−
−−=
−=
Multiply the first equation by 2 and add to the third equation.
Section 4.5 Systems of Linear Equations in Three Variables
Multiply the second equation by –7 and eliminate y.
Back-substitute 1 for x into one of the equations in
two variables.
()
51 4
54
1
z
z
z
−=
−=
−=−
7.
4 211
2 1
2 23 1
xyz
xyz
xyz
−+=
+−=−
+−=−
Multiply the second equation by –4 and add to the
first equation.
We now have two equations in two variables.
9615
2 1
yz
yz
−+ =
−−=
Multiply the second equation by 6 and solve by
addition.
Back-substitute –1 for y in one of the equations in
1
1
z
z
−=−
=
Back-substitute –1 for y and 1 for z in one of the
original equations in three variables.
()
21
211 1
xyz
x
+−=−
+−−=−
3 8
3 2 2
4 6
xyz
xyz
xz
−+=
+−=−
+=
Multiply the second equation by –4 and add it to the
third equation.
12 4 8 8
xyz
−−+=
46 46
1
x
x
−=−
=
Back-substitute 1 for x in one of the equations in
two variables.
Chapter 4 Systems of Linear Equations
The solution is
()
1, 1, 2− and the solution set is
()
{
}
1, 1, 2 .−
9.
323 2
252 2
43410
xyz
xyz
xyz
+−=−
−+=−
−+=
Multiply the second equation by –2 and add to the third
equation.
()
323 2
3223 2
343 2
xyz
xz
xz
+−=−
+−=−
+− =−
Multiply the first equation by –4 and the second
equation by 3.
12 12 24
12 12 48
24 72
3
xz
xz
z
z
−+ =
+=
=
=
15 10 25 110
15 21 9 86
11 16 26
xyz
xyz
yz
−− + =
+−=−
+=
The system of two equations in two variables
becomes:
531 83
11 16 26
yz
yz
−− =−
+=
Multiply the first equation by 11 and the second
two variables.
()
531 83
5313 83
yz
y
−− =−
−− =−
21513
22
1
x
x
x
+=
=−
=−
The solution is
()
1, 2 , 3−−
and the solution set is
()
{
}
1, 2, 3 .−−
Section 4.5 Systems of Linear Equations in Three Variables
11.
24317
2 0
4 6
xyz
xyz
xyz
−+=
+−=
−−=
Multiply the second equation by –1 and add it to the
third equation.
The system in two variables becomes:
33 6
5217
xy
xy
−=
+=
Multiply the first equation by 2 and the second
equation by 3 and solve by addition.
Back-substitute 3 for x and 1 for y in one of the
original equations in three variables.
()
20
321 0
xyz
z
+−=
+−=
12. Multiply the third equation by 2 and add it to the
second equation.
2 1
4226
5 7
xyz
xyz
xz
+−=
−+=
+=
1
x
=
Back-substitute 1 for x in one of the equations in
two variables.
3
13
xz
z
+=
+=
13.
2 2
4
32 0
xy
xyz
xyz
+=
+−=
++=
Add the second and third equations together to
obtain an equation in two variables.
Chapter 4 Systems of Linear Equations
Back-substitute 0 for y in one of the equations in
two unknowns.
22
202
22
xy
x
x
+=
+=
=
14. Multiply the first equation by –3 and add to the
third equation.
3 9 15 60
32 9 36
11 6 24
xy z
xy z
yz
−− − =−
−+ =
−−=−
The system of two equations in two variables
becomes:
Back-substitute 4 for z in one of the equations in
two unknowns.
()
416
44 16
yz
y
−=−
−=−
16 16
0
y
y
−=−
=
15.
4
1
2321
xy
yz
xy z
+=−
−=
++ =−
Multiply the first equation by –1 and add to the
5
24 22
xz
xz
−− =
+=−
Multiply the first equation by 2 and add to the
second equation.
22 10
24 22
2 12
6
xz
xz
z
z
−− =
+=−
=−
=−
original system.
4
14
5
xy
y
y
+=−
+=−
=−
The solution is
()
1, 5, 6−−
and the solution set is
()
{
}
1, 5, 6 .−−
Section 4.5 Systems of Linear Equations in Three Variables
16. Multiply the first equation by –1 and add to the
second equation.
4
4
0
xy
xz
yz
−− =−
+=
−+=
The system of two equations in two variables is as
follows.
Back-substitute 2 for y to find x.
4
24
2
xy
x
x
+=
+=
=
The solution is
()
2, 2, 2
and the solution set is
()
{
}
2, 2, 2 .
17.
2 21
3 2
2 0
xyz
xyz
xyz
++=
−+=
−−=
Add the first and second equations to eliminate y.
We obtain two equations in two variables.
53 3
53 4
xz
xz
+=
−− =−
Adding the two equations, we obtain:
18. Multiply the second equation by 2 and add to the
first equation.
3 4 5 8
24 6 12
5 11 4
xyz
xyz
xz
++=
−+=−
+=−
Multiply the second equation by –2 and add to the
third equation.
15 6 15 3
xy z
−− =
Multiply the first equation by –2 and add to the
second equation.
10 4 10 2
10 4 10 2
0 0
xy z
xy z
−++ =−
−− =
=
The system is dependent and has infinitely many
solutions.
20. Multiply the first equation by 2 and add to the
second equation.
24 2 8
xyz
++=
becomes:
5312
10 6 24
xz
xz
+=
+=
Multiply the first equation by –2 and add to the
21.
(
)
()
() ( )
32 5 1
234 9
41 3 3
xy z
xyz
xzy
++=−
−+ =−
+=− −
Rewrite each equation and obtain the system of
three equations in three variables.
Multiply the second equation by –2 and add to the
third equation.
41216 18
4 9 3 4
3 13 14
xyz
xy z
yz
−+ − =
−+ =−
−=
The system of two variables in two equations is:
21 19 26
31314
yz
yz
−=
−=
Multiply the second equation by –7 and add to the
Back-substitute –1 for z and
1
3
for y in one of the
original equations in three variables.
635 1
615 1
64 1
xyz
x
x
++=−
+− =−
−=−
22. After rewriting the equations, the system becomes
267 3
453 7
635 4
xyz
xyz
xyz
−+ + =
−+ + =
−+ + =−
Multiply the first equation by –2 and add to the
15 16 13
yz
−−=−
The system of two variables in two equations is as
follows.
711 1
15 16 13
yz
yz
−− =
−−=−
Multiply the first equation by –15 and the second
equation by 7 to eliminate y.
105 165 15
105 112 91
yz
yz
+=−
−−=−
() ( )
263723
218143
243
21
1
x
x
x
x
−+ +−=
−+− =
−+=
−=−
Section 4.5 Systems of Linear Equations in Three Variables
23. Use each ordered pair to write an equation.
2
(, ) (1,6)
xy
yax bxc
=−
=++
() ()
()
2
2
(, ) (2,9)
92 2
942
94 2
xy
yax bxc
abc
abc
abc
=
=++
=++
=++
=++
The system of three equations in three variables is:
The system of two equations in two variables
becomes:
2210
6321
ac
ac
+=
+=
Multiply the first equation by –3 and add to the
second equation.
()
2210
22310
2610
ac
a
a
+=
+=
+=
The equation is
2
23.yxx=−+
24. Use each ordered pair to write an equation.
42 7
2
42 3
abc
abc
abc
−+=
++=−
++=
Multiply the first equation by –1 and add to the
third equation.
4 5
1
ac
ac
+=
+=−
Multiply the first equation by –1 and add to the
second equation.
45
ac
−−=−
25. Use each ordered pair to write an equation.
() ()
2
2
(, ) (1, 4)
41 1
xy
yax bxc
abc
=− −
=++
−= − + − +
The system of three equations in three variables is:
4
2
42 5
abc
abc
abc
−+=−
++=−
++=
Multiply the second equation by –1 and add to the
first equation.
4
abc
−+=−
The system of two equations in two variables
becomes:
3
4 3
ac
ac
+=−
+=
Multiply the first equation by –1 and add to the
Back-substitute 2 for a and 1for b in one of the
equations in three variables.
4
21 4
abc
c
−+=−
−+ =−
third equation.
3
16 4 0
15 3 3
abc
abc
ab
−− − =−
++=
+=−
The system of two equations in two variables
becomes:
82 4
ab
+=−
()
81 2 4
82 4
212
6
b
b
b
b
+=−
+=−
=−
=−
Back-substitute 1 for a and –6 for b in one of the
equations in three variables.
Section 4.5 Systems of Linear Equations in Three Variables
27. Let x = the first number.
Let y = the second number.
Let z = the third number.
16
234 46
5 31
xyz
xyz
xy
++=
++=
−=
Multiply the first equation by –1 and add to the
second equation.
531
218
7 49
7
xy
xy
x
x
−+=−
−−=−
−=−
=
Back-substitute 7 for x in one of the equations in
two variables.
28. Let x = the first number.
Let y = the second number.
Let z = the third number.
()
()
325
332
23 1
xy z
xz y
xyz
++ =
+−=
+−=
Multiply the first equation by 3 and add to the
second equation.
93615
33 2
10 9 17
xyz
xyz
xz
++=
−+=
+=
equation by 10 to eliminate x.
30 27 51
30 20 30
7 21
3
xz
xz
z
z
−− =−
+=
−=−
=
Back-substitute 3 for z in one of the equations in
two variables.
323
xz
+=
The numbers are –1, 2, and 3.
Chapter 4 Systems of Linear Equations
29. Simplify each equation.
()
24 0
632
24
660
632
xyz
xyz
++
−+=
++
−+=
()()()
51219
4324
512 19
12 12
432 4
35416257
3 154 46 12 57
34680
xyz
xyz
xyz
xyz
xyz
−+−
++ =
−+−
++ =
−+ ++ − =
−+ ++ − =
++=
Now solve the equivalent system.
Using the two reduced equations, we solve the
system.
32 24
844
xz
xz
+=
−+ =
Multiply the second equation by 3 and add the
equations.
32 24
3 24 132
26 156
6
xz
xz
z
z
+=
−+ =
=
=
30. Simplify each equation.
()()()
3123
2242
312 3
44
224 2
2321 26
2622 26
22 4
xyz
xyz
xyz
xyz
xyz
+−+
−+ =
+−+
−+ =
+− −++ =
+− +++=
−+=−
()()()
44
422 2
32 12 3 10
32 22 6 10
221
xyz
xyz
xyz
−+ =−
−− ++ −=−
−− −+ − =−
−+=
Now solve the equivalent system.
Section 4.5 Systems of Linear Equations in Three Variables
22 4
643 24
221
xyz
xyz
xyz
−+=−
+−=−
−+=
Multiply the first equation by 2 and add it to the
second equation.
Multiply the second equation by 1− and add it to
the first.
10 32
5
9 27
3
xz
xz
x
x
−=−
−+ =
=−
=−
Back-substitute to solve for z.
The solution is
()
3, 0, 2−
and the solution set is
()
{
}
3, 0, 2 .−
31. Selected points may vary, but the equation will be
the same.
2
yax bxc=++
Use the points
()
2, 2−
,
()
4,1
, and
()
6, 2−
to get
32 4 0
ab
+=
Using the two reduced equations, we get the system
12 2 3
32 4 0
ab
ab
+=
+=
Multiply the first equation by
2−
and add to the
second equation.
24 4 6
ab
−−=−
212
6
b
b
=
=
Back-substitute to solve for c.
()
42 2
3
4262
4
312 2
11
abc
c
c
c
++=−
−+ +=−
−+ + =−
=−
The equation is
2
36 11.
4
yxx=− + −
Chapter 4 Systems of Linear Equations
32. Selected points may vary, but the equation will be
the same.
2
yax bxc=++
Use the points
()
3, 4
,
()
4, 2
, and
()
5, 2
to get the
7 2
ab
+=−
Multiply the first equation by
1−
and add to the
third equation.
93 4
25 5 2
16 2 2
abc
abc
ab
−−−=−
++=
+=−
Use the two reduced equations to get the system
1
a
=
Back-substitute to solve for b.
()
72
71 2
72
9
ab
b
b
b
+=−
+=−
+=−
=−
Back-substitute to solve for c.
33.
221
0
214
ax by cz
ax by cz
ax by cz
−− =
++=
−+=
Add the first two equations.
Use the two reduced equations to get the following
system:
221
37
ax cz
ax cz
−=
+=−
Multiply the second equation by
2−
and add the
equations.
221
ax cz
−=
15 7
8
8
ax
ax
xa
−=−
=
=
Back-substitute to solve for y.
0
ax by cz
++=
Section 4.5 Systems of Linear Equations in Three Variables
34.
24
31
232
ax by cz
ax by cz
ax by cz
−+ =−
+−=
++ =
Multiply the first equation by
1−
and add to the
second equation.
Use the two reduced equations to get the following
system:
435
310
by cz
by cz
−=
−=
Multiply the second equation by
3−
and add to the
first equation.
Back-substitute to solve for x.
24
55
24
ax by cz
ax b c
bc
−+ =−
−+ =−
35. a.
()
0,5
,
()
50,31
,
()
100,15
b. Substituting each ordered pair gives:
2
2
(0) (0) 5
ax bx c y
abc
++=
++=
36. a.
()
10,16
,
()
20,52
,
()
40, 23
b. Substituting each ordered pair gives:
2
2
2
(10) (10) 16
(20) (20) 52
ax bx c y
abc
abc
++=
++=
++=
224
93 176
8 2 48
abc
abc
ab
−−− =−
++=
+=−
Multiply the first equation by
1−
and add to the
Chapter 4 Systems of Linear Equations
24 6 144
30 6 240
6 96
16
ab
ab
a
a
−−=
+=−
=−
=−
Back-substitute to solve for b.
b. When
5x=
, we get
() ()
()
2
16 5 40 5 200
16 25 200 200
400 400
0
y=− + +
=− + +
=− +
=
After 5 seconds, the ball hits the ground.
38. a.
46
42 84
93 114
abc
abc
abc
++=
++=
++=
Multiply the first equation by –1 and add to the
Multiply the first equation by –1 and add to the
second equation.
62 76
82 68
2 8
4
ab
ab
a
a
−− =−
+=
=−
=−
Back-substitute –4 for a in one of the equations
in two variables.
()
338
34 38
12 38
ab
b
b
+=
−+=
−+=
39. Let x = hours Chemical Engineering Majors study
per week
Let y = hours Mathematics Majors study per week.
Let z = hours Psychology Majors study per week.
52
6
8
xyz
xy
xz
++=
−=
−=
Solve the second equation for y.
6
6
6
xy
yx
yx
−=
−=−+
=−
22
x
=
Back-substitute to solve for y and z.
Section 4.5 Systems of Linear Equations in Three Variables
6
22 6
yx
=−
=−
40. Let x = hours Physics Majors study per week
Let y = hours English Majors study per week.
Let z = hours Sociology Majors study per week.
50
xyz
++=
Substitute the expressions for x and z into the first
equation and solve for y.
50
( 4) ( 6) 50
yz
xyz
xx x
++ =
+−+−=
Back-substitute to solve for y and z.
4
20 4
16
yx
=−
=−
=
41. Let x = the amount invested at 8%.
Let y = the amount invested at 10%.
Add the first and third equations to find z.
6700
300
xy z
xy z
++ =
−− + =
The system of two equations in two variables
becomes:
3200
0.08 0.10 296
xy
xy
+=
+=
Multiply the second equation by –10 and add it to
1200
x
=
Back-substitute 1200 for x in one of the equations in
two variables.
3200
xy
+=
Chapter 4 Systems of Linear Equations
42. Let x = the amount invested at 10%.
Let y = the amount invested at 12%.
Let z = the amount invested at 15%.
17000
0.10 0.12 0.15 2110
1000
xyz
xyz
xz y
++=
++=
+=+
8000
y
=
Back-substitute 9000 for y to obtain two equations
in two variables.
17000
8000 17000
9000
xyz
xz
xz
++=
++=
+=
()
0.10 0.12 0.15 2110
0.10 0.12 8000 0.15 2110
xyz
xz
++=
++=
Multiply the second equation by –10 and add it to
the first equation.
9000
1.5 11500
0.5 2500
xz
xz
z
+=
−− =−
−=−
43. Let x = the number of $8 tickets.
Let y = the number of $10 tickets.
Let z = the number of $12 tickets.
400
81012 3700
7
xyz
xyz
xy z
++=
++=
+=
50
z
=
Back-substitute 50 for z in two of the original
equations to obtain two of equations in two
variables.
400
50 400
350
xyz
xy
xy
++=
++ =
+=
81012 3700
xyz
++=
810 3100
xy
+=
Multiply the first equation by –8 and add to the
second equation.
8 8 2800
xy
−− =−