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4 years ago

Solve the following System of Equations 5x+4y-z=1 2x-2y+z=1 -x-y+z=2

Mathematics

1 answer:

podryga [215]4 years ago

3 0

Answer:

x = 0 , y = 1 , z = 3

Step-by-step explanation:

Solve the following system:

{5 x + 4 y - z = 1 | (equation 1)

2 x - 2 y + z = 1 | (equation 2)

-x - y + z = 2 | (equation 3)

Subtract 2/5 × (equation 1) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - (18 y)/5 + (7 z)/5 = 3/5 | (equation 2)

-x - y + z = 2 | (equation 3)

Multiply equation 2 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

-x - y + z = 2 | (equation 3)

Add 1/5 × (equation 1) to equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y/5 + (4 z)/5 = 11/5 | (equation 3)

Multiply equation 3 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y + 4 z = 11 | (equation 3)

Subtract 1/18 × (equation 2) from equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+(65 z)/18 = 65/6 | (equation 3)

Multiply equation 3 by 18/65:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 7 × (equation 3) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y+0 z = -18 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 2 by -18:

{5 x + 4 y - z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 4 × (equation 2) from equation 1:

{5 x + 0 y - z = -3 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Add equation 3 to equation 1:

{5 x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 1 by 5:

{x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Collect results:

Answer: {x = 0 , y = 1 , z = 3

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3 years ago

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A. [4 marks]

storchak [24]

The value of d is -2

Step-by-step explanation:

The nth term of the arithmetic sequence is $u_{n}=u_{1}+(n-1)d$ , where

$u_{1}$ is the first term
d is the common difference between each 2 consecutive terms

The nth term of the geometric sequence is $u_{n}=u_{1}r^{n-1}$ , where

$u_{1}$ is the first term
r is the common ratio between each two consecutive terms $r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}$

∵ $u_{1}$ of an arithmetic sequence is 1

∵ The common difference is d, where d ≠ 0

∵ $u_{2}$ , $u_{3}$ , $u_{6}$ are the first 3 terms of a geometric sequence

∵ $u_{2}$ = 1 + (2 - 1)d

∴ $u_{2}$ = 1 + d

∵ $u_{3}$ = 1 + (3 - 1)d

∴ $u_{2}$ = 1 + 2d

∵ $u_{6}$ = 1 + (6 - 1)d

∴ $u_{2}$ = 1 + 5d

∴ The first 3 terms of the geometric sequence are (1 + d) , (1 + 2d) ,

(1 + 5d)

∵ The common ratio in the geometric sequence is $r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}$

∴ $r=\frac{1+2d}{1+d}=\frac{1+5d}{1+2d}$

- Use cross multiplication with the equal fractions

∵ $\frac{1+2d}{1+d}=\frac{1+5d}{1+2d}$

∴ (1 + 2d)(1 + 2d) = (1 + d)(1 + 5d)

∴ 1 + 2d + 2d + 4d² = 1 + 5d + d + 5d²

- Add like terms in each side

∴ 1 + 4d + 4d² = 1 + 6d + 5d²

- Subtract 1 from both sides

∴ 4d + 4d² = 6d + 5d²

- Subtract 4d from both sides

∴ 4d² = 2d + 5d²

- Subtract 4d² from each side

∴ 0 = 2d + d²

- Take d as a common factor

∴ 0 = d(2 + d)

- Equate each factor by 0

∴ d = 0 but we will reject it because d ≠ 0

∴ 2 + d = 0

- Subtract 2 from both sides

∴ d = -2

The value of d is -2

Learn more:

You can learn more about the sequence in brainly.com/question/1522572

#LearnwithBrainly

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