What value of $y$ makes the equation true?
$3(2 y-5)+15=12$
$y=$
To solve the equation \(3(2y - 5) + 15 = 12\) for \(y\), we first expand the expression by distributing the constant 3 across the terms inside the parentheses. This gives us:
\[3 \times 2y - 3 \times 5 + 15 = 12\]
\[6y - 15 + 15 = 12\]
Next, we observe that the terms \(-15\) and \(+15\) on the left side of the equation cancel each other out, simplifying the equation to:
\[6y = 12\]
To isolate the variable \(y\), we divide both sides of the equation by 6:
\[y = \frac{12}{6}\]
\[y = 2\]
Finally, we verify the solution by substituting \(y = 2\) back into the original equation:
\[3(2(2) - 5) + 15 = 3(4 - 5) + 15 = 3(-1) + 15 = -3 + 15 = 12\]
Since this matches the original value, the solution is correct. The value of \(y\) is \(\boxed{2}\).
Linear equations like this are fundamental in many real-life situations, such as budgeting. Suppose you have a budget of \$12, and you buy 3 packs of pens. Each pack costs a base price, but you have a \$5 discount applied to each pack's price before multiplying by the quantity, and you receive an additional \$15 rebate. If we represent the original price of a pack of pens as \(y\), the total cost is modeled by the equation \(3(y-5) + 15 = 12\). Solving such equations allows you to determine the unknown variables in financial planning, engineering, and everyday problem-solving.