c. \( 5 x-100=69-8 x \) d. \( 6 x-18+2 x=3 x+17 \cdot 6 \) e. \( 3 \cdot(x+5)-2 x+1=48: 3 \) f. \( (x-2) \cdot 4+36=45 \cdot 2+x+4 \)

To solve these equations step by step can be quite rewarding! First, let's look at c. \( 5x - 100 = 69 - 8x \). To isolate \( x \), you’ll want to add \( 8x \) to both sides, resulting in \( 13x - 100 = 69 \). Then add \( 100 \) to both sides to get \( 13x = 169 \) and finally divide by \( 13 \) to find \( x = 13 \). Now, for d. \( 6x - 18 + 2x = 3x + 17 \cdot 6 \), combine your \( x \) terms on the left side. That gives you \( 8x - 18 = 3x + 102 \). Move \( 3x \) to the left to get \( 5x - 18 = 102 \), then add \( 18 \) to both sides, leading to \( 5x = 120 \). Divide by \( 5 \) for \( x = 24 \). Revving up for e. \( 3 \cdot(x + 5) - 2x + 1 = 48 : 3 \) is also exciting! Start by solving the right side; \( 48 : 3 = 16 \). Now, distribute \( 3 \) to get \( 3x + 15 - 2x + 1 = 16 \), which simplifies to \( x + 16 = 16 \). Finally, subtract \( 16 \) from both sides, giving \( x = 0 \). Last but not least, for f. \( (x - 2) \cdot 4 + 36 = 45 \cdot 2 + x + 4 \), compute \( 45 \cdot 2 = 90 \), leading to \( 4x - 8 + 36 = 90 + x + 4 \). Simplifying yields \( 4x + 28 = 94 + x \). Bring \( x \) to the left: \( 3x + 28 = 94 \), then \( 3x = 66 \), thus \( x = 22 \). Voilà! You’ve just tackled some intriguing algebra problems!