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macylen3900
16.06.2021 •
Mathematics
Solve the system of equations.
3x + 4y + z = 3
4x + 3y + 32 - 4
5x+67+72-5
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Ответ:
First, we write the augmented matrix.
⎡
⎢
⎣
1
−
1
1
2
3
−
1
3
−
2
−
9
|
8
−
2
9
⎤
⎥
⎦
[1−1123−13−2−9 | 8−29]
Next, we perform row operations to obtain row-echelon form.
−
2
R
1
+
R
2
=
R
2
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
3
−
2
−
9
|
8
−
18
9
⎤
⎥
⎦
−
3
R
1
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
5
−
3
0
1
−
12
|
8
−
18
−
15
⎤
⎥
⎦
−2R1+R2=R2→[1−1105−33−2−9|8−189]−3R1+R3=R3→[1−1105−301−12|8−18−15]
The easiest way to obtain a 1 in row 2 of column 1 is to interchange
\displaystyle {R}_{2}
R2
and
\displaystyle {R}_{3}
R3
.
Interchange
R
2
and
R
3
→
⎡
⎢
⎣
1
−
1
1
8
0
1
−
12
−
15
0
5
−
3
−
18
⎤
⎥
⎦
InterchangeR2andR3→[1−11801−12−1505−3−18]
Then
−
5
R
2
+
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
57
|
8
−
15
57
⎤
⎥
⎦
−
1
57
R
3
=
R
3
→
⎡
⎢
⎣
1
−
1
1
0
1
−
12
0
0
1
|
8
−
15
1
⎤
⎥
⎦
−5R2+R3=R3→[1−1101−120057|8−1557]−157R3=R3→[1−1101−12001|8−151]
The last matrix represents the equivalent system.
x
−
y
+
z
=
8
y
−
12
z
=
−
15
z
=
1
x−y+z=8 y−12z=−15 z=1
Using back-substitution, we obtain the solution as
\displaystyle \left(4,-3,1\right)
(4,−3,1)
Ответ:
16
Step-by-step explanation:
so in this equation we have 2 variables, and we must figure out what those variables mean:
y: "y" is the total amount of money joe has after subtracting his debt from it
x: "x" is the # of months as the cost of each month(30) is placed just before this variable
as y is the total amount of money joe has we need to substitute it for 230 to figure out how long it took for joe to recieve that much money
230=30x-250
480=30x
x=16
it took joe 16 months to get $230