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Pie
3 years ago
9

Use Gauss-Jordan elimination to solve the following linear system.

Mathematics
1 answer:
svp [43]3 years ago
8 0
<span>x – 2z = 9 so x = 2z + 9
6x – 2y – 5z = 29
–5x + 5y + 3z = –14

substitute </span>x = 2z + 9 into 6x – 2y – 5z = 29 and –5x + 5y + 3z = –14

6(2z + 9) – 2y – 5z = 29
12z + 54 - 2y -5z =29
-2y + 7z = - 25 (1st equation)

–5(2z + 9) + 5y + 3z = –14
-10z - 45 + 5y + 3z = -14
5y - 7z = 31 (2nd equation)

multiply (1st) equation by 5 and (2nd) equation by 2
-10y + 35z = - 125
 10y - 14z = 62
-----------------------add
21z = - 63
 z = -3

substitute z = - 3 into x = 2z + 9

x = 2z + 9
x = 2(-3) + 9 
x = 3

substitute x =  3 and z = 3 into <span>6x – 2y – 5z = 29
</span>
<span>6x – 2y – 5z = 29
</span><span>6(3) – 2y – 5(-3) = 29
18 - 2y + 15 = 29
-2y + 33 = 29
-2y = -4
  y  = 2
x = 3, y = 2 and z = -3

answer
</span><span>B. (3, 2, –3)</span>
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Terrance works part time and deposits 134.12 into his checking account every week. he began with 207.43 in the account Identify
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5 0
3 years ago
Y=-2x+5<br> X | Y<br> ____<br> 5<br> -1<br> 2<br> 0
Y_Kistochka [10]

Answers:

<h3>1. -5</h3><h3>2. 7</h3><h3>3. 1</h3><h3>4. 5</h3>

Step-by-step explanation:

Plug in the x into the equation and solve.

<h3>Value 1: X = 5</h3>

Y = -2(5) + 5

Y = -10 + 5

Y = -5

<h3>Value 2: X = -1</h3>

Y = -2(-1) + 5

Y = 2 + 5

Y = 7

<h3>Value 3: X = 2</h3>

Y = -2(2) + 5

Y = -4 + 5

Y = 1

<h3>Value 4: X = 0</h3>

Y = -2(0) + 5

Y = 0 + 5

Y = 5

I'm always happy to help :)

8 0
3 years ago
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