Let's solve each problem one at a time using the substitution method.

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1. Solve the system:

$2x + 3y = 4$ $-4x - 6y = -8$

Start by solving the first equation for $y$: $3y = 4 - 2x$ $y = \frac{4 - 2x}{3}$

Substitute $y$ in the second equation: $-4x - 6\left(\frac{4 - 2x}{3}\right) = -8$

Simplify this expression, and you'll find that both equations describe the same line, indicating infinite solutions.
Answer: A. Infinite solutions

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2. Solve the system:

$x = y - 4$ $x + 8y = 2$

Substitute $x$ from the first equation into the second: $(y - 4) + 8y = 2$ $9y - 4 = 2$ $9y = 6$ $y = \frac{6}{9} = \frac{2}{3}$

Plug $y = \frac{2}{3}$ back into $x = y - 4$: $x = \frac{2}{3} - 4 = \frac{2 - 12}{3} = -\frac{10}{3}$

This gives a unique solution: $\left(-\frac{10}{3}, \frac{2}{3}\right)$ Answer: The correct option isn’t listed.

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3. Solve the system:

$2.5x + y = -2$ $3x + 2y = 0$

From the first equation, solve for $y$: $y = -2.5x - 2$

Substitute $y$ in the second equation: $3x + 2(-2.5x - 2) = 0$ $3x - 5x - 4 = 0$ $-2x = 4$ $x = -2$

Substitute $x = -2$ back into $y = -2.5x - 2$: $y = -2.5(-2) - 2 = 5 - 2 = 3$

Answer: B. (-2, 3)

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4. Solve the linear-quadratic system:

$y = x^2 - 4x + 8$ $y = 3x - 4$

Substitute $y = 3x - 4$ into the quadratic equation: $3x - 4 = x^2 - 4x + 8$ $0 = x^2 - 7x + 12$ $0 = (x - 3)(x - 4)$

Thus, $x = 3$ or $x = 4$. Substitute back to find $y$:

• For $x = 3$: $y = 3(3) - 4 = 5$
• For $x = 4$: $y = 3(4) - 4 = 8$

Answer: A. (4, 8) and (3, 5)

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5. Solve the linear-quadratic system:

$y = x^2 - x - 40$ $y = -3x - 5$

Substitute $y = -3x - 5$ into the quadratic equation: $-3x - 5 = x^2 - x - 40$ $0 = x^2 + 2x - 35$ $0 = (x - 5)(x + 7)$

Thus, $x = 5$ or $x = -7$. Substitute back to find $y$:

• For $x = 5$: $y = -3(5) - 5 = -15 - 5 = -10$
• For $x = -7$: $y = -3(-7) - 5 = 21 - 5 = 26$

Answer: A. (5, -10) and (-7, 26)

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Would you like further clarification on any steps or concepts?

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Related Practice Questions

1. Use the substitution method to solve $x + 2y = 3$ and $y = x + 1$.
2. Determine if the system $y = x^2 + 2x + 1$ and $y = 4x + 3$ has any solutions using substitution.
3. Solve $3x - y = 7$ and $x + 4y = 5$ by substitution.
4. Solve the quadratic-linear system $y = x^2 + 6x + 9$ and $y = -x + 3$.
5. Using substitution, solve $x = y + 5$ and $2x - y = 4$.

Tip: When using substitution, always isolate one variable first to simplify the calculations in the other equation.