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3 years ago

The system of equations shown below is graphed on a coordinate grid:

Mathematics

2 answers:

Vlad1618 [11]3 years ago

8 0

Answer:

um idek

Step-by-step explanation:

DaniilM [7]3 years ago

5 0

4y + x = 4 . . . . . . . (1)
2y - x = 8 . . . . . . . (2)

(1) + (2) => 6y = 12
y = 12/6 = 2

From (2), 2(2) - x = 8
4 - x = 8
x = 4 - 8 = -4

Therefore solution is (-4, 2) and lies on both lines.

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<img src="https://tex.z-dn.net/?f=%20x%20-%202%5Csqrt%7Bx%20-%203%20%3D%200%7D%20" id="TexFormula1" title=" x - 2\sqrt{x - 3 = 0

GrogVix [38]

Answer:

No Real Solutions

Step-by-step explanation:

So we have:

$x-2\sqrt{x-3}=0$

First, determine the domain restrictions. The expression under the radical cannot be less than 0. Therefore:

$x-3\geq 0\\x\geq 3$

Therefore, our final answers must be greater than or equal to 3:

Now, going back to the original equation, subtract x from both sides:

$-2\sqrt{x-3}=-x$

Now, square both sides:

$4(x-3)=x^2$

Distribute:

$4x-12=x^2$

Subtract 4x and add 12 to both sides:

$0=x^2-4x+12$

This isn't factor-able. Let's use the quadratic formula. a is 1, b is -4 and c is 12:

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

Substitute:

$x=\frac{4\pm\sqrt{(-4)^2-4(1)(12)}}{2(1)}$

Simplify the radical:

$x=\frac{4\pm\sqrt{-32}}{2(1)}$

The number under the radical is negative. In other words, there are no real solutions.

7 0

3 years ago

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