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: enter image description here

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?

  • $\begingroup$ If you subtract 6 from LHS and 5 from RHS, since 0 > 0 it is not going to be the opposite? $\endgroup$ Sep 19 '17 at 14:30
  • $\begingroup$ What do you mean "the opposite" ? @kimiTanaka $\endgroup$ Sep 19 '17 at 14:30
  • $\begingroup$ the opposite sign of inequality. I mean that L1 - 6 = L2 - 5 makes 2y + 3x = x - 5 $\endgroup$ Sep 19 '17 at 14:32
  • $\begingroup$ The inequality sign only changes direction when both sides are divided by a negative value, as far as I'm concerned $\endgroup$ Sep 19 '17 at 14:36
  • $\begingroup$ Sorry, I might be confused with a different thing after subtracting 6 from L1 and subtracting from 5. Anyway, since L1: 3x+2y - 6 = 0, L2: x-y-5 = 0 then L1 = L2. In your attempt 3x + 2y - 6 > x- y- 5. You need to compare in one variable like the answer you have just confirmed. $\endgroup$ Sep 19 '17 at 14:50

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. $3-1.5x>x-5$, $2.5x<8$, $\large{x<\frac{16}{5}}$

  • $\begingroup$ Excellent, thank you so much. $\endgroup$ Sep 19 '17 at 14:45
  • $\begingroup$ But the inequality stated differently? Wtf is going on. $\endgroup$ Sep 19 '17 at 14:47
  • $\begingroup$ @neonpokharkar it was stated in 2 variables x,y. Def L1: 3x+2y-6 = 0 L2: x+y-5 = 0, L1 = L2. $\endgroup$ Sep 19 '17 at 14:54
  • $\begingroup$ So, it is not actually, $ 2y+3x\gt x-y$? $\endgroup$ Sep 19 '17 at 14:55
  • $\begingroup$ @neonpokharkar Yes, it is in initial attempt. In 2nd one, subtract 6 from LHS and 5 from RHS. then it should be 2x+3y - 6 = x+y -5 since 2y + 3x = 6 and x-y = 5. If you put x, y in those equations it always is L1 = L2. $\endgroup$ Sep 19 '17 at 14:59

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