Question: Solve for the value of $k$.
(2k+3)°
(4k-3)°
Graph/shape description: A vertical line with arrows at both ends is intersected by two rays on the right; the upper right ray and lower right ray form a right angle at the vertex, with the angle between the upper right ray and the top part of the vertical line labeled $(2k+3)^{\circ}$ near the top-right, and the angle between the lower right ray and the bottom part of the vertical line labeled $(4k-3)^{\circ}$ near the lower-right.
1. **State the problem:** We need to find the value of $k$ given two angles $(2k+3)^{\circ}$ and $(4k-3)^{\circ}$ that together form a right angle.
2. **Understand the geometry:** The two angles are adjacent and together form a right angle, which means their sum is $90^{\circ}$.
3. **Write the equation:**
$$
(2k+3) + (4k-3) = 90
$$
4. **Simplify the equation:**
$$
2k + 3 + 4k - 3 = 90
$$
$$
(2k + 4k) + (3 - 3) = 90
$$
$$
6k + 0 = 90
$$
$$
6k = 90
$$
5. **Solve for $k$:**
$$
\cancel{6}k = \cancel{90}15
$$
$$
k = 15
$$
6. **Conclusion:** The value of $k$ is $15$.
This means the angles are:
- $(2k+3)^{\circ} = 2(15) + 3 = 33^{\circ}$
- $(4k-3)^{\circ} = 4(15) - 3 = 57^{\circ}$
Their sum is $33^{\circ} + 57^{\circ} = 90^{\circ}$, confirming the solution is correct.