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Arada [10]
3 years ago
7

Solve the system of equations {x−4y=3

Mathematics
2 answers:
lilavasa [31]3 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]3 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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Determine the equation of the circle graphed below. <br><br> ( please help me )
Brums [2.3K]

Answer:

Equation : \\(x+3)^2 + ( y -2)^2 = 36

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: <em><u>Find the centre</u></em>

If the centre is not given find the end points of the diameter

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Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

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Step 2 : <u>Find radius</u>

<u></u>Radius = \frac{Diameter }{2}<u></u>

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

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Therefore, the equation of the circle with centre ( -3, 2)

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    (x - (-3))^2 +  (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36

5 0
2 years ago
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