5r +2r= 6 +8 substitute r =2
Answers
Answer:
5(2) + 2(2) = 6+8
10 +4 = 14
14 =14
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Answer:
Write the augmented matrix.
⎡
⎢
⎣
−
1
−
2
1
2
3
0
0
1
−
2
|
−
1
2
0
⎤
⎥
⎦
First, multiply row 1 by \displaystyle -1−1 to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.
−
R
1
→
⎡
⎢
⎣
1
2
−
1
1
2
3
0
2
0
1
−
2
0
⎤
⎥
⎦
R
2
↔
R
3
→
⎡
⎢
⎣
1
2
−
1
0
1
−
2
2
3
0
|
1
0
2
⎤
⎥
⎦
−
2
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
2
−
1
0
1
−
2
0
−
1
2
|
1
0
0
⎤
⎥
⎦
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
2
−
1
0
1
−
2
0
0
0
|
2
1
0
⎤
⎥
⎦
The last matrix represents the following system.
x
+
2
y
−
z
=
1
y
−
2
z
=
0
0
=
0
We see by the identity \displaystyle 0=00=0 that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for \displaystyle yy and substituting it into the first equation we can solve for \displaystyle zz in terms of \displaystyle xx.
x
+
2
y
−
z
=
1
y
=
2
z
x
+
2
(
2
z
)
−
z
=
1
x
+
3
z
=
1
z
=
1
−
x
3
Now we substitute the expression for \displaystyle zz into the second equation to solve for \displaystyle yy in terms of \displaystyle xx.
y
−
2
z
=
0
z
=
1
−
x
3
y
−
2
(
1
−
x
3
)
=
0
y
=
2
−
2
x
3
The generic solution is \displaystyle \left(x,\frac{2 - 2x}{3},\frac{1-x}{3}\right)(x,
3
2−2x
,
3
1−x
).