# A tricky question involving inequalities

I've tried to solve this but I don't seem to get anywhere.

The question states:

Tom's home is $1800$ m from his school. One morning he walked part of the way to school and then ran the rest. If it took him $20$ mins or less to get to school, and he walks at $70$ m/min and runs at $210$ m/min, how far did he run?

My attempts to solve it got me stuck here:

$$\frac{x}{70}+\frac{y}{210}≥\frac{1800}{20}$$

Any help would be appreciated!

Let the distance he walks be $x$ and the distance he runs be $y$ .
We know that $velocity= \frac{distance}{time}\implies time = \frac{distance}{velocity}$
$\frac x{70}+\frac{y}{210} \le 20$
$\implies \frac{x}{70}+\frac{1800-x}{210}\le 20$$\quad since y = 1800 -x \implies 3x+1800-x\le4200 \implies 2x\le2400 \implies x\le 1200 \therefore y\ge600 Calling$$ r = \mbox{distance covered while running}\\ w = \mbox{distance covered while walking}\\ s_r = \mbox{running speed}\\ s_w = \mbox{walking speed}\\ d= \mbox{total distance}\\ t=\mbox{estimated time to school}\\ \epsilon = \mbox{slack variable} $$we have$$ d = w + r \\ \frac{w}{s_w}+\frac{r}{s_r}=t + \epsilon^2 $$Now solving for w,r we have$$ w=\frac{s_r s_w t}{s_r-s_w}+\frac{s_r s_w \epsilon^2}{s_r-s_w}-\frac{d s_w}{s_r-s_w}\\ r= \frac{d s_r}{s_r-s_w}-\frac{s_r s_w \epsilon^2}{s_r-s_w}-\frac{s_r s_w t}{s_r-s_w} $$or$$ w \le \frac{s_r s_w t-d s_w}{s_r-s_w} = 1200\\ r \ge \frac{s_r (d-s_w t)}{s_r-s_w} = 600$$NOTE We have assumed that$s_r > s_w\$