Question

if $tu = 8y - 78$, $sv = y + 58$, and $rw = 2y + 66$, what is the value of $y$?
$y = \square$

Explanation:

Step1: Identify the midline property

In a trapezoid (or a quadrilateral with a midline), the length of the midline (here \( SV \)) is the average of the lengths of the two bases (\( TU \) and \( RW \)). So, the formula is \( SV=\frac{TU + RW}{2} \).

Step2: Substitute the given expressions

Substitute \( TU = 8y - 78 \), \( SV = y + 58 \), and \( RW = 2y + 66 \) into the formula:
\[
y + 58=\frac{(8y - 78)+(2y + 66)}{2}
\]

Step3: Simplify the right - hand side

First, simplify the numerator of the right - hand side:
\((8y - 78)+(2y + 66)=8y+2y-78 + 66=10y-12\)
Then the equation becomes:
\[
y + 58=\frac{10y-12}{2}
\]
Simplify the right - hand side further: \(\frac{10y - 12}{2}=5y-6\)
So the equation is \(y + 58=5y-6\)

Step4: Solve for y

Subtract \( y \) from both sides:
\(58 = 5y-y-6\)
\(58 = 4y-6\)
Add 6 to both sides:
\(58 + 6=4y\)
\(64 = 4y\)
Divide both sides by 4:
\(y=\frac{64}{4}=16\)

Answer:

\(16\)