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Montano1993 [528]
2 years ago
14

Carina begins to solve the equation -4 - 2/3x = -6 by adding 4 to both sides. Which statements regarding the rest of the solving

process could be true? Select three options.
Mathematics
1 answer:
Andru [333]2 years ago
5 0

Answer:

A.) After adding 4 to both sides, the equation is -\frac{2}{3}x=-2

C.) The equation can be solved for x using exactly one more step by multiplying both sides by -\frac{3}{2}

D.) The equation can be solved for x using exactly one more step by dividing both sides by -\frac{2}{3}

Step-by-step explanation:

The correct questions is as follows:

Carina begins to solve the equation -4-2/3x=-6 by adding 4 to both sides. Which statements regarding the rest of the solving process could be true? Check all that apply.

A.) After adding 4 to both sides, the equation is -2/3x=-2.

B.) After adding 4 to both sides, the equation is -2/3x=-10 .

C.) The equation can be solved for x using exactly one more step by multiplying both sides by -3/2.

D.) The equation can be solved for x using exactly one more step by dividing both sides by -2/3.

E.) The equation can be solved for x using exactly one more step by multiplying both sides by -2/3.

Given equation:

-4-\frac{2}{3}x=-6

To show the steps we will carry out in order to solve for x

Solution:

Solving for x

Step 1:

Adding both sides by 4

4-4-\frac{2}{3}x=-6+4

Thus, we get:

-\frac{2}{3}x=-2

Thus statement A is correct.

Step 2:

Multiplying both sides by -\frac{3}{2}

-\frac{3}{2}\times -\frac{2}{3}x=-2\times -\frac{3}{2}

Thus, we get:

x=3  [Two negatives multiply to give a positive]

This proves that statement C is correct.

Or Step 2:

Dividing both sides by -\frac{2}{3}

\frac{-\frac{2}{3}x}{-\frac{2}{3}}=\frac{-2}{-\frac{2}{3}}

Thus, we get:

x=-2\times -\frac{3}{2} [On dividing with a fractional divisor, we take reciprocal and multiply it with the dividend.]

x=3  [Two negatives multiply to give a positive]

This prove that statement D is correct.

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Step-by-step explanation:

It is helpful to remember the factoring of the difference of squares:

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