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4 years ago

Substitute each equation by multiplication. 6x-2y=-6 -2x-3y=2

Mathematics

1 answer:

Georgia [21]4 years ago

7 0

Answer:

x = -1

y = 0

Step-by-step explanation:

to solve the equation by substitution in order to evaluate for x and y

let

6x-2y=-6 ................................................ equation 1

-2x-3y=2 .................................................. equation2

from equation 2

-2x-3y=2 .................................................. equation2

-2x -2 = 3y

divide both sides by 3

(-2x -2)/3 = y

y = (-2x -2)/3 ................................... equation3

substitute y = (-2x -2)/3 into equation 1

6x-2y=-6 ................................................ equation 1

6x - 2[ (-2x -2)/3] = -6

6x + (4x + 4)/ 3 = -6

multiply through by 3

18x + 4x + 4 = -18

22x = -18-4

22x = -22

divide both sides by the coefficient of x which is 22

22x/22 = -22/22

x = -1

substitute for x in equation 3

y = (-2x -2)/3 ................................... equation3

y = -2(-1) - 2/3

y = 2 -2/3

y = 0/3

y = 0

therefore the value of x and y is -1 and 0 respectively

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riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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