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Arada [10]
2 years ago
7

Solve the system of equations {x−4y=3

Mathematics
2 answers:
lilavasa [31]2 years ago
8 0

Answer:

-1/3, 5/3

Step-by-step explanation:

Isolate x for x: -4y=3  x=3+4y

Substitute: x=3+4y

(2(3+4y)+y=3)

Isolate y for: 2(3+4y)+y=3    y=-1/3

Substitute: y=-1/3

3+4(-1/3)

3+4(-1/3)=5/3

x=5/3

-1/3,5/3


Arte-miy333 [17]2 years ago
5 0

Answer:

<h2>(-1, -1)</h2>

Step-by-step explanation:

Substitution\ method:\\\\\left\{\begin{array}{ccc}x-4y=3&\text{add 4y to both sides}\\2x+y=-3\end{array}\right\\\\\left\{\begin{array}{ccc}x=4y+3\\2x+y=-3\end{array}\right\\\\\text{Substitute from the first equation to the second equation:}\\\\2(4y+3)+y=-3\qquad\text{use distributive property}\\\\(2)(4y)+(2)(3)+y=-3\\\\8y+6+y=-3\qquad\text{subtract 6 from both sides}\\\\9y=-9\qquad\text{divide both sides by 9}\\\\\boxed{y=-1}\\\\\text{Put the value of y to the first equation:}\\\\x=4(-1)+3\\\\x=-4+3\\\\\boxed{x=-1}\\\\\boxed{\boxed{x=-1,\ y=-1\to(-1,\ -1)}}


Elimination\ method:\\\\\left\{\begin{array}{ccc}x-4y=3\\2x+y=-3&\text{multiply both sides by 4}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}x-4y=3\\8x+4y=-12\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad9x=-9\qquad\text{divide both sides by 9}\\.\qquad\qquad\boxed{x=-1}\\\\\text{Put the value of x to the first equation:}\\\\-1-4y=3\qquad\text{add 1 to both sides}\\\\-4y=4\qquad\text{divide both sides by (-4)}\\\\\boxed{y=-1}\\\\\boxed{\boxed{x=-1,\ y=-1\to(-1,\ -1)}}

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Answer:

a. See Attachment 1

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c. HT = 31.1\ m

d. OH = 28.4\ m

Step-by-step explanation:

Calculating PT

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tan50 = \frac{OT}{20}

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The relationship between these parameters is;

tan30 = \frac{OP}{20}

Multiply both sides by 20

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--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

HT^2 = OT^2 + OH^2

Substitute 20 for OH and OT = 23.836

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Take Square Root of both sides

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--------------------------------------------------------

Calculating the position of H

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See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

tan50 = \frac{OH}{OT}

Multiply both sides by OT

OT * tan50 = \frac{OH}{OT} * OT

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Substitute OT = 23.836

23.836 * 1.1918 = OH

28.4= OH

OH = 28.4\ m<em> (Approximated)</em>

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