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Katarina [22]
3 years ago
8

Q8 Q17.) Find the numbers.

Mathematics
2 answers:
Katarina [22]3 years ago
7 0
Let x = the first number
      y = the second number
-------------------------------------
First: set up the equations.

Four times a first number: 4x
Decreased by a second number: - y
Is 4: = 4

4x - y = 4.

The first number increased by: x + 
three times the second number: 3y
is -25: = -25

x + 3y = -25.
------------------------
You can make your system of equations now.

4x - y = 4
x + 3y = -25

I would use substitution to solve these.

To do substitution, isolate a variable first, in one of the equations. I would probably isolate x in the second equation.

x + 3y = -25
x = -25 - 3y

Now, you can substitute -25 - 3y into 4x - y = 4, as x.

4x - y = 4
4(-25 - 3y) - y = 4

Now just solve for y.

-100 - 12y - y = 4
-13y = 104
y = -8

Now, you can put -8 as y into one of the original equations, I'll just use 4x - y = 4.

4x -  y = 4
4x - (-8) = 4
4x + 8 = 4
4x = -4
x = -1

So the first number is -1, and the second -2.
Scorpion4ik [409]3 years ago
7 0

let x be 1st no. and y be the 2nd no.

so 4x - y = 4

and x + 3y = -25

re-arrange first eqn y = 4x - 4

substitute y into second eqn

x + 3(4x - 4) = -25

13x - 12 = -25

13x = -13

x = -1

y = 4(-1 - 1) = -8



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Answer:

Each term is a sum of the two previous terms like the fibonacci sequence, but the sum of the two starting terms of the lucas sequence is followed by 2 then 1, rather than 0 then 1 for the fibannaci sequence.

These are integer sequences.

The sum of the two previous terms can be represented by:

fn = fn-1 + fn-2.

These numbers can be obtained by adding the previous corresponding term of the fibonacci sequence by the next corresponding term to get the corresponding number of the lucas sequence.

Ln = { 2 if n = 0

1 if n = 1

Ln-1 + Ln-2 if n > 1

________________

Fn+1 + Fn-1 = Ln

___________________

f(0) = -1

f(1) = 2, f(2) = 1, f(3) = 3, f(4) = 4, f(5) = 7, f(6) = 11

________________________________

f(12) = 199

________

f(12) = f(11) + f(10) = (f(10) + f(9)) + (f(9) + f(8)) = .....

given f(n) = f(n-1) + f(n-2)

____________________

f(12) = f(1) + f(2) = f(2) + f(3) = f(3) + f(4) =

f(4) + f(5) = f(5) + f(6) = f(6) + f(7) =

f(7) + f(8) = f(8) + f(9) = f(9) + f(10) =

f(10) + f(11) = f(12) →

f(12) = 2 + 1 = 1 + [3] = 3 + [4] = 4 + [7] =

7 + [11] = 11 + [18] = 18 + [29] = 29 + [47] =

47 + [76] = 76 + [123] = <u>199</u>

_____________________

Explicit formula

for n is an integer.

Nth term of the lucas sequence.

As you can see, the golden ratio

( 1 + √5 / 2) appears in here.

L(n) = ((1 + √5) / 2)ⁿ⁻¹ + ((1 – √5) / 2)ⁿ⁻¹

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