Let $L_1$ be the graph of $6=3x+2y$ and let $L_2$ be the graph of $5=x-y$. Find the coordinates of the intersection, point $P$.
Hence or otherwise, write down the solution to the inequality $$2y+3x>x-y$$ the two graphs are illustrated below:
Now, the point of intersection can be easily calculated by letting $y=x-5$ and then solving for $x$ when $2(x-5)+3x=6$, and the point that gives is at $P(3.2, -1.8)$, so therefore the solution to the inequality (which is basically asking $L_1 > L_2$) is obviously when $x < 3.2$. But I'm considering examination circumstances, where the same question arises with no graph. I've therefore tried to solve the inequality algebraically and I'm slightly stuck.
Here's my attempt:
since when $L_1=0$, $L_1 = 2y+3x-6$ and when $L_2=0$, $L_2=x-y-5$, the inequality can be restated as $2y+3x -6 > x-y-5$
from here: $$3y+2x>1$$ and then, after letting $y=x-5$, $$3(x-5)+2x>1$$ $$5x>16$$ $$x>\frac{16}{5}$$ which is $3.2$ the only issue here is obviously that this is saying $L_1$ is greater than $L_2$ when $x > 3.2$ when the answer is really $x < 3.2$
Can anybody talk me through how to solve this inequality algebraically?
For $L_1$ we have $y=3-1.5x$, for $L_2$ we have $y=x-5$. We need to solve $L_1>L_2$, i.e. $3-1.5x>x-5$, $2.5x<8$, $\large{x<\frac{16}{5}}$