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3 years ago

Can somebody solve this system equation using addition please?

Mathematics

1 answer:

givi [52]3 years ago

7 0

-4x - 5y = 7

3x + 5y = -14

You can add these two equations together straightaway since the y-terms have opposite coefficients.

-4x - 5y = 7

3x + 5y = -14

+___________

-x - 0 = -7

-x = -7

x = 7

Substitute 7 for x into either of the original equations and solve algebraically to find y.

3x + 5y = -14

3(7) + 5y = -14

21 + 5y = -14

21 = -14 - 5y

35 = -5y

-7 = y

Finally, check work by substituting both x- and y-values into both original equations.

-4x - 5y = 7

-4(7) - 5(-7) = 7

-28 + 35 = 7

7 = 7

3x + 5y = -14

3(7) + 5(-7) = -14

21 - 35 = -14

-14 = -14

Answer:

x = 7 and y = -7; (7, -7).

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$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

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$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

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