Question

when $ac = 3y - 5$ and $hj = 4y + 2$, what is hb ?
$hb = \square$

Explanation:

Step1: Identify midsegment relationship

From the triangle diagram, $AC$ is the midsegment parallel to $HJ$, so $AC = \frac{1}{2}HJ$. Also, $HB = \frac{1}{2}HJ$ (since $B$ is the midpoint of $HJ$).

Step2: Set up equation for $y$

Substitute given expressions:
$$3y - 5 = \frac{1}{2}(4y + 2)$$
Simplify right side:
$$3y - 5 = 2y + 1$$

Step3: Solve for $y$

Subtract $2y$ from both sides, add 5 to both sides:
$$3y - 2y = 1 + 5$$
$$y = 6$$

Step4: Calculate $HJ$

Substitute $y=6$ into $HJ = 4y + 2$:
$$HJ = 4(6) + 2 = 24 + 2 = 26$$

Step5: Calculate $HB$

Since $HB = \frac{1}{2}HJ$:
$$HB = \frac{1}{2}(26) = 13$$

Answer:

13