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2 years ago

Tell whether each equation has one, zero, or infinitely many solutions.

Mathematics

2 answers:

amid [387]2 years ago

7 0

Answer:

Exactly One Solution

Step-by-step explanation:

4x + 2 = x + 8

3x = 6 Simplify.

x = 2 Divide by 2.

raketka [301]2 years ago

3 0

Answer:

One solution, x = 2

Step-by-step explanation:

4x + 2 = x + 8 (Given)

4x = x + 6

(The 2 is subtracted from the right side, therefore the left side is also subtracted. 8 becomes 6.)

3x = 6

(The x is subtracted from the right side to the left side. Thus turning the 4x into 3x)

x = 2

(The 3 is divided out to get x alone. So, the 6 has to be divided by 3 as well, turning it into 2)

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If the point (a, b) is equidistant from the points (-a, 2) and (2, -b), then prove that 3(a+b)+4=0

erastova [34]

Answer: see proof below

Step-by-step explanation:

Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).

$M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}$

3(a + b) - 4 = 0

$3\bigg(\dfrac{2}{3}+\dfrac{2}{3}\bigg)-4=0\\\\\\3\bigg(\dfrac{4}{3}\bigg)-4=0\\\\\\4-4=0\\\\0=0\qquad \text{TRUE!}$

Notice that I changed the equation to "negative 4" because the equation you provided did not make a true statement.

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