Question: 2) Solve each linear system using the method of SUBSTITUTION [9]
a) l₁: y = -6x
l₂: y = -2x + 8
Solution:
b) l₁: 7x + 10y = 36
l₂: -2x + y = 9
Solution:
c) l₁: x - 3y = 6
l₂: 3x = 18 + 9y
Solution:
1. **Problem a:** Solve the system \(\begin{cases} y = -6x \\ y = -2x + 8 \end{cases}\) using substitution.
2. Since both expressions equal \(y\), set them equal: \(-6x = -2x + 8\).
3. Solve for \(x\):
$$-6x = -2x + 8$$
$$-6x + 2x = 8$$
$$-4x = 8$$
$$x = \frac{8}{-4} = -2$$
4. Substitute \(x = -2\) into \(y = -6x\):
$$y = -6(-2) = 12$$
5. **Solution for a:** \(x = -2, y = 12\).
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6. **Problem b:** Solve the system \(\begin{cases} 7x + 10y = 36 \\ -2x + y = 9 \end{cases}\) using substitution.
7. From the second equation, express \(y\) in terms of \(x\):
$$-2x + y = 9 \implies y = 9 + 2x$$
8. Substitute \(y = 9 + 2x\) into the first equation:
$$7x + 10(9 + 2x) = 36$$
$$7x + 90 + 20x = 36$$
$$27x + 90 = 36$$
$$27x = 36 - 90$$
$$27x = -54$$
$$x = \frac{-54}{27} = -2$$
9. Substitute \(x = -2\) back into \(y = 9 + 2x\):
$$y = 9 + 2(-2) = 9 - 4 = 5$$
10. **Solution for b:** \(x = -2, y = 5\).
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11. **Problem c:** Solve the system \(\begin{cases} x - 3y = 6 \\ 3x = 18 + 9y \end{cases}\) using substitution.
12. From the first equation, express \(x\) in terms of \(y\):
$$x = 6 + 3y$$
13. Substitute \(x = 6 + 3y\) into the second equation:
$$3(6 + 3y) = 18 + 9y$$
$$18 + 9y = 18 + 9y$$
14. Simplify:
$$18 + 9y = 18 + 9y$$
This is an identity, meaning the two equations represent the same line.
15. **Solution for c:** Infinite solutions along the line \(x = 6 + 3y\).
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**Final answers:**
- a) \(x = -2, y = 12\)
- b) \(x = -2, y = 5\)
- c) Infinite solutions along \(x = 6 + 3y\)