Question

9)
-4p + 12q - 36 = 0
-p + 3q - 9 = 0

10)
-c + 10d = 0
-20d + 2c = 3

Explanation:

Step1: Analyze system 9 equations

Equation 1: $-4p + 12q - 36 = 0$
Equation 2: $-p + 3q - 9 = 0$

Step2: Simplify Equation 1

Divide Equation 1 by 4:
$\frac{-4p}{4} + \frac{12q}{4} - \frac{36}{4} = \frac{0}{4}$
$-p + 3q - 9 = 0$

Step3: Compare simplified equations

The simplified Equation 1 is identical to Equation 2. This means the two equations represent the same line, so there are infinitely many solutions. We express $p$ in terms of $q$:
$-p = -3q + 9$
$p = 3q - 9$

---

Step1: Analyze system 10 equations

Equation A: $-c + 10d = 0$
Equation B: $-20d + 2c = 3$

Step2: Isolate $c$ from Equation A

$c = 10d$

Step3: Substitute $c$ into Equation B

$-20d + 2(10d) = 3$
$-20d + 20d = 3$
$0 = 3$

Step4: Interpret the result

The statement $0=3$ is false, so this system has no solution.

Answer:

1. Infinitely many solutions: $p = 3q - 9$ (where $q$ is any real number)
2. No solution