Help solve these equations:

1) \( 3y + 1 = 16 \)

2) \( 18 = 4k + 6 \)

3y+1=16 Subtract 1 from both sides (3y+1)-1=(16)-1 3y=15 Divide both sides by 3 (3y)/3=(15)/3 Y=5
18=4k+6 Subtract 6 from both sides (18)-6=(4k+6)-6 12=4k Divide both sides by 4 (12)/4=(4k)/4 3=k
Final answer:

y=5
k=3

Answered by FirstSineOfMadness | 2024-06-10

1 ) 3 y + 1 = 16 ∣ s u b t r a c t 1 3 y = 15 ∣ d i v i d e b y 3 y = 5 2 ) 18 = 4 k + 6 ∣ s u b t r a c t 6 12 = 4 k ∣ d i v i d e b y 4 k = 3

Answered by luana | 2024-06-10

The solution for the first equation is y = 5 , and for the second equation, k = 3 . By isolating the variables through subtraction and division, we successfully found the values for y and k .
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Answered by FirstSineOfMadness | 2024-12-26