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Nezavi [6.7K]
3 years ago
13

I don't know how do do these problems​

Mathematics
1 answer:
zubka84 [21]3 years ago
7 0

\bf \begin{array}{ll} term&\stackrel{a(n-1)-3}{value}\\ \cline{1-2} a(1)&-6\\ a(2)&a(2-1)-3\\ &a(1)-3\\ &-6-3\\ &-9 \blacktriangleleft\\ a(3)&a(3-1)-3\\ &a(2)-3\\ &-9-3\\ &-12 \blacktriangleleft\\ a(4)&a(4-1)-3\\ &a(3)-3\\ &-15 \blacktriangleleft\\ a(5)&a(5-1)-3\\ &a(4)-3\\ &-18 \blacktriangleleft \end{array}

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The two lines in this system of equations are parallel

Step-by-step explanation:

Let us revise the relation between 2 lines

  • If the system of linear equations has one solution, then the two line are intersected
  • If the system of linear equations has no solution, then the two line are parallel
  • If the system of linear equations has many solutions, then the two line are coincide (over each other)

∵ The system of equation is

3x - 6y = -12 ⇒ (1)

x - 2y = 10 ⇒ (2)

To solve the system using the substitution method, find x in terms of y in equation (2)

∵ x - 2y = 10

- Add 2y to both sides

∴ x = 2y + 10 ⇒ (3)

Substitute x in equation (1) by equation (3)

∵ 3(2y + 10) - 6y = -12

- Simplify the left hand side

∴ 6y + 30 - 6y = -12

- Add like terms in the left hand side

∴ 30 = -12

∴ The left hand side ≠ the right hand side

∴ There is no solution for the system of equations

∴ The system of equations represents two parallel lines

The two lines in this system of equations are parallel

Learn more:

You can learn more about the equations of parallel lines in brainly.com/question/8628615

#LearnwithBrainly

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