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2 years ago

Simultaneously x-2y=0 40-y=14

Mathematics

1 answer:

harkovskaia [24]2 years ago

7 0

Answer:

x = 52, y = 26

Step-by-step explanation:

x - 2y = 0

40 - y = 14

(40 - 40) - y = 14 - 40

- y = -26

y = 26

x - 2(26) = 0

x - 52 = 0

x (- 52 + 52) = 0 + 52

x = 52

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$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

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When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

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$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

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$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

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Isolate the equation:

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We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

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$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

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$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

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Thus, our two intersection points are:

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2 years ago

Simultaneously x-2y=0 40-y=14​

Simultaneously x-2y=0 40-y=14