Find the area of the triangle formed by the lines 5x + 7y = 35 and 4x + 3y = 12 together with the x-axis.

Introduction / Context:
Two nonparallel lines intersect above the x-axis, and each meets the x-axis at a positive x-intercept. Along with the x-axis, these form a triangle. We need the triangle’s area using intercepts and the intersection point.

Given Data / Assumptions:

• Lines: 5x + 7y = 35 and 4x + 3y = 12.
• x-axis: y = 0.
• Area unit: square units.

Concept / Approach:
Find the x-intercepts by setting y = 0. Then find the intersection of the two lines. With a base on the x-axis, the triangle’s area equals (1/2) * base * height, where the base is the distance between the two x-intercepts and the height is the y-coordinate of the intersection point (vertical distance to the x-axis).

Step-by-Step Solution:
For 5x + 7y = 35 with y = 0 ⇒ x = 7. First intercept: (7, 0).For 4x + 3y = 12 with y = 0 ⇒ x = 3. Second intercept: (3, 0).Base length on x-axis = 7 − 3 = 4.Solve the system: 5x + 7y = 35 and 4x + 3y = 12 ⇒ intersection P(−21/13, 80/13).Height (vertical) = y of P = 80/13.Area = (1/2) * base * height = (1/2) * 4 * (80/13) = 160/13.

Verification / Alternative check:
Plotting sketches reveal a base segment from x = 3 to x = 7 on the x-axis, with the third vertex above the axis, matching the computed height and area formula.

Why Other Options Are Wrong:
150/13 and 140/13 arise from arithmetic slips. 10 and 12 are rough guesses that ignore the exact intersection height 80/13.

Common Pitfalls:
Using y-intercepts instead of x-intercepts for the base; sign mistakes when solving the system; or forgetting that area must be positive and uses the vertical height to the base line.

Final Answer:
160/13 sq. unit