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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!

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Let the distance he walks be $x$ and the distance he runs be $y$ .

We know that $velocity= \frac{distance}{time}\implies time = \frac{distance}{velocity}$

So your equations comes to ;

$\frac x{70}+\frac{y}{210} \le 20 $

$\implies \frac{x}{70}+\frac{1800-x}{210}\le 20$$\quad $ since $y = 1800 -x$

$\implies 3x+1800-x\le4200$

$\implies 2x\le2400$

$\implies x\le 1200$

$\therefore y\ge600$

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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 $s_r > s_w$

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