Area of triangle!

Solving two equations at a time, we get these three points of intersection: $(2,0), (4,0), (\frac{32}{13}, \frac{15}{13})$

\[\begin{align} \text{Area }&= \frac{1}{2} \cdot b \cdot h \\ &= \frac{1}{2} \cdot(4-2)\cdot \bigg(\frac{15}{13} \bigg) \\ &= \frac{1}{\cancel{2}} \cdot(\cancel{2})\cdot \bigg(\frac{15}{13} \bigg) \\ &\approx \boxed{1.15}

\end{align}\]

To find the area you must know where the 3 points of the triangle lie:

• One is at the intersection of x-axis and $5x-2y=10$
• One is at the intersection of x-axis and $3x+4y=12$
• One is at the intersection of $5x-2y=10$ and $3x+4y=12$

Any line $ax+by=c$ intersects x-axis at $\frac{c}{a},0$

Therefore 2 of the lines are $(2,0)$ and $(4,0)$

The third line lies at the intersection of $5x-2y=10$ and $3x+4y=12$ , that is it is the solution of these 2 equations

Solution of the equations: $3x+4y=12$ $5x-2y=10$ Multiplying our last equation by 2 $10x-4y=20$ Adding this and our first equation $13x=32$ $\boxed{x=\frac{32}{13}}$ Put this in our first equation $3(\frac{32}{13})+4y=12$ $\frac{96}{13}+4y=12$ $4y=12-\frac{96}{13}$ $y=3-\frac{24}{13}=\frac{39-24}{13}=\boxed{\frac{15}{13}}$

Now consider the line from $(2,0)$ to $(4,0)$ as the base of the triangle. Then the height will be the distance of $(\frac{32}{13},\frac{15}{13})$ from the base and as the base is the x-axis the height is the ordinate of the point.

Therefore area= $\frac{1}{2}(4-2)(\frac{15}{13})=\frac{15}{13}=\boxed{1.15}384615...)$