14. Given: [tex] \frac{5y-1}{2} = 7 [/tex]; Prove: [tex] y = 3 [/tex]

\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{Statements} & \multicolumn{1}{|c|}{Reasons} \\
\hline
[tex] \frac{5y-1}{2} = 7 [/tex] & Given \\
\hline
[tex] 5y - 1 = 14 [/tex] & Multiply both sides by 2 \\
\hline
[tex] 5y = 15 [/tex] & Add 1 to both sides \\
\hline
[tex] y = 3 [/tex] & Divide both sides by 5 \\
\hline
\end{tabular}



Answer :

Sure! Let's solve the given equation step by step and prove [tex]\( y = 3 \)[/tex].

Given equation:
[tex]\[ \frac{5y - 1}{2} = 7 \][/tex]

Step-by-Step Solution:

1. Multiply both sides by 2:

[tex]\[ \frac{5y - 1}{2} \times 2 = 7 \times 2 \][/tex]

This gives:

[tex]\[ 5y - 1 = 14 \][/tex]

Reason: To eliminate the fraction.

2. Add 1 to both sides:

[tex]\[ 5y - 1 + 1 = 14 + 1 \][/tex]

This simplifies to:

[tex]\[ 5y = 15 \][/tex]

Reason: To isolate the term with the variable [tex]\( y \)[/tex].

3. Divide both sides by 5:

[tex]\[ \frac{5y}{5} = \frac{15}{5} \][/tex]

This simplifies to:

[tex]\[ y = 3 \][/tex]

Reason: To solve for [tex]\( y \)[/tex].

Thus, we have proved that:
[tex]\[ y = 3 \][/tex]

To summarize:

[tex]\[ \frac{5y - 1}{2} = 7 \quad \text{(Given)} \][/tex]
[tex]\[ \implies 5y - 1 = 14 \quad \text{(Multiplying both sides by 2)} \][/tex]
[tex]\[ \implies 5y = 15 \quad \text{(Adding 1 to both sides)} \][/tex]
[tex]\[ \implies y = 3 \quad \text{(Dividing both sides by 5)} \][/tex]

Hence, we have proved that [tex]\( y = 3 \)[/tex].