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The worksheet involves solving equivalent systems of equations, specifically parts (2) and (3). Here's the overview:

Part 2: Explain why system A is equivalent to system B.

System A:
$2x + 1.5y = -9$
$y = x - 6$

System B:
$3x + 0.5y = -3$
$y = x - 6$

To verify if the systems are equivalent:

• The second equation is identical in both systems, $y = x - 6$, so it does not change.
• To see if the first equations are equivalent, we manipulate $2x + 1.5y = -9$.
  Multiply the entire equation by $\frac{3}{2}$: $\frac{3}{2}(2x + 1.5y) = \frac{3}{2}(-9) \implies 3x + 0.5y = -3$ Thus, the first equations are equivalent after scaling.

Conclusion: System A and System B are equivalent because the first equations are scalar multiples, and the second equation is the same.

Part 3a: Solve system A using substitution.

System A:
$2x + 1.5y = -9$
$y = x - 6$

Substitute $y = x - 6$ into $2x + 1.5y = -9$: $2x + 1.5(x - 6) = -9$ Simplify: $2x + 1.5x - 9 = -9$ $3.5x - 9 = -9$ Add 9 to both sides: $3.5x = 0$ $x = 0$

Substitute $x = 0$ into $y = x - 6$: $y = 0 - 6 = -6$

Solution to system A: $x = 0, y = -6$.

Part 3b: Solve system B by graphing.

System B:
$3x + 0.5y = -3$
$y = x - 6$

Rewrite $3x + 0.5y = -3$ in slope-intercept form ($y = mx + b$): $0.5y = -3 - 3x$ $y = -6 - 6x$

Graph $y = x - 6$ and $y = -6 - 6x$:

• The lines intersect at $(0, -6)$.

Solution to system B: $x = 0, y = -6$.

Part 3c: Compare solutions to parts a and b.

The solutions to both systems are the same: $(0, -6)$. This confirms that the systems are equivalent, as solving them produces the same result.

Would you like more details on any step?
Here are 5 related questions to explore:

1. How do you determine if two systems are equivalent?
2. What are the methods to solve systems of equations (substitution, elimination, graphing)?
3. How do you convert equations to slope-intercept form for graphing?
4. Can equivalent systems have different numbers of solutions? Why or why not?
5. What are scalar multiples in the context of equations?

Tip: When solving systems by substitution, always simplify the substituted equation as much as possible before solving. This minimizes errors.