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Aleksandr-060686 [28]
3 years ago
8

Is (-1,7) a solution for the system of linear equations below? x + 2y = 13 and 3x - y

Mathematics
1 answer:
juin [17]3 years ago
6 0

Answer:

x = -9/7, y = 50/7.

Step-by-step explanation:

x + 2y = 13 ---------(eqn 1)

3x - y = -11 -----------(eqn 2)

Let's solve simultaneously using elimination method by multiplying eqn 2 by 2 so as to eliminate y.

3x - y = - 11 -------( × 2)

It gives

6x - 2y = - 22 ---------- ( new eqn 2)

x + 2y = 13 -----------( eqn 1)

Now add eqn 1 and new eqn 2

6x + x -2y + 2y = -22 + 13

7x = -9

Divide both sides by 7

7x/7  = -9/7

x = -9/7

Put x = -9/7 into eqn 1

x + 2y = 13

-9/7 + 2y = 13

Add 9/7 to both sides

-9/7+9/7 + 2y = 13+9/7

2y = 100/7

Divide both sides by 2

2y/2 = 100/7 ÷ 2

y = 100/7 × 1/2

y = (100÷2)/7

y = 50/7

Therefore

x = -9/7, y = 50/7

(-9/7, 50/7)

Therefore (-1,7) is not the solution for both equations , it is not a solution for NEITHER equation.

I hope this was helpful, please rate as brainliest

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<h3><em>Good</em><em> </em><em>Luck</em><em>!</em></h3>

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B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =
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I suppose you mean

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Differentiate one term at a time.

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Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

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\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

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Then by the chain rule,

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