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3 years ago

2x - 3y = -4 if x + 3y = 7

2 answers:

4 0

(2x-3y=-4)-2(x+3y=7)

2x-3y-2x-6y=-4-14

-9y=-18

y=2, making x+3y=7 become:

x+6=7

x=1

So the solution to the system of equations is the point (1,2)

Serggg [28]3 years ago

3 0

Leave x alone on one side.

x = 7 - 3y

Substitute into equation

2(7 - 3y) - 3y = -4

Distribute

14 - 6y - 3y = -4

Solve

14 - 9y = -4

14 + 4 = 9y

18 = 9y

y = 18/9

y = 2
x = 1

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

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Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

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For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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3 years ago

Deepak randomly chooses two marbles from the bag,

Alika [10]

Green 5/15
red 2/15
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3 0

3 years ago

Read 2 more answers

You plan to borrow $36,500 at a 7.7% annual interest rate. The terms require you to amortize the loan with 7 equal end-of-year p

erma4kov [3.2K]

The amount of interest you would be paying in Year 2 is: $2,492.62.

<h3>Interest</h3>

First step is to calculate the Equal Monthly Payment

Equal Monthly Payment=P×r×(1+r)^t/(1+r)^t-1

Where:

P=Principal=$36,500

r=Rate=7.7%

t=Time=7 years

Equal Monthly Payment=36,500×0.077×(1+0.077)^7÷(1+0.077)^7-1

Equal Monthly Payment=36,500×0.077×(1.077)^7÷(1.077)^7-1

Equal Monthly Payment=36,500×0.077×1.6807763÷1.6807763-1

Equal Monthly Payment=4,723.82/0.6807763

Equal Monthly Payment=$6,938.875

Second step is to calculate Year 1 Closing balance

Year 1 Closing balance = Beginning balance + Interest - EMI Payment

Year 1 Closing balance= $36,500 +($36,500×7.7%) - $6,938.875

Year 1 Closing balance= $36,500 + $2,810.5 -$6,938.875

Year 1 Closing balance = $32,371.625

Third step is to calculate year 2 interest

Year 2 Interest= $32,371.625×7.7%

Year 2 Interest=$2,492.62

Therefore the amount of interest you would be paying in Year 2 is: $2,492.62.

Learn more about interest here:brainly.com/question/15259578

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8 0

2 years ago

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GuDViN [60]

9514 1404 393

Answer:

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Step-by-step explanation:

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The account will have 5450 in it at the end of the year.

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3 years ago

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pishuonlain [190]

Answer:

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3 years ago