Question

y = 2x - 2
x\ty
\t-2
\t0
\t2
\t4

Explanation:

Step1: Solve for x when y = -2

We have the equation \( y = 2x - 2 \). Substitute \( y=-2 \) into the equation:

$$ -2 = 2x - 2 $$

Add 2 to both sides:

$$ -2 + 2=2x - 2 + 2 $$

$$ 0 = 2x $$

Divide both sides by 2:

$$ x = 0 $$

Step2: Solve for x when y = 0

Substitute \( y = 0 \) into \( y=2x - 2 \):

$$ 0=2x - 2 $$

Add 2 to both sides:

$$ 0 + 2=2x - 2 + 2 $$

$$ 2 = 2x $$

Divide both sides by 2:

$$ x = 1 $$

Step3: Solve for x when y = 2

Substitute \( y = 2 \) into \( y=2x - 2 \):

$$ 2=2x - 2 $$

Add 2 to both sides:

$$ 2 + 2=2x - 2 + 2 $$

$$ 4 = 2x $$

Divide both sides by 2:

$$ x = 2 $$

Step4: Solve for x when y = 4

Substitute \( y = 4 \) into \( y=2x - 2 \):

$$ 4=2x - 2 $$

Add 2 to both sides:

$$ 4 + 2=2x - 2 + 2 $$

$$ 6 = 2x $$

Divide both sides by 2:

$$ x = 3 $$

Answer:

When \( y=-2 \), \( x = 0 \); when \( y = 0 \), \( x=1 \); when \( y = 2 \), \( x = 2 \); when \( y = 4 \), \( x=3 \)

So the completed table (filling in the \( x \) values) is:

\( x \)	\( y \)
\( 1 \)	\( 0 \)
\( 2 \)	\( 2 \)
\( 3 \)	\( 4 \)