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DaniilM [7]
4 years ago
5

For each linear system,tell whether you would multiply the terms in the first or second equation in order to eliminate one of th

e variables.Give the number by which you could multiply.
X+3y=-14, 2x + y =-3
Mathematics
1 answer:
liq [111]4 years ago
7 0

We can multiply first equation x + 3y = -14 by 2 to get rid of varibale "x"

<em><u>Solution:</u></em>

<em><u>Given that, the system of equations are:</u></em>

x + 3y = -14 ----- eqn 1

2x + y = -3 --------- eqn 2

We can solve by elimination method

For doing elimination, we have to make the any one of coefficient of variable same

Here, in eqn 1 we have 1 as coefficient for "x"

In eqn 2, we have 2 as coeficient for "x"

So, we can multiply eqn 1 by 2, to get rid of "x"

2(x + 3y = -14)

2x + 6y = -28 ------- eqn 3

If we subtract eqn 2 from eqn 3, we will get the solution

2x + 6y = -28

2x + y = -3

( - ) ----------------

5y = -25

<h3>y = -5</h3>

Substitute y = -5 in eqn 1

x + 3(-5) = -14

x -15 = -14

<h3>x = 1</h3>

Thus the solution is x = 1 and y = -5. So we have multiplied equation 1 by 2 to get of varibale "x" and solved the equations

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What are axioms in algebra called in geometry
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The compound interest on a sum of money in
Yakvenalex [24]

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

5,460  = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}

P = \dfrac{100}{21} \times 5,460 = 26,000

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

5 0
3 years ago
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