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Solutions to the exercises

Exercise 1

$3 y = 2 x + 6$ therefore $y = \frac{2}{3} x + 2$ , gradient = $\frac{2}{3}$

$2 y = 6 x - 24$ therefore $y = 3 x - 12$ , gradient = $3$

$y = 2 x + 1$ , gradient = $2$

$y = -4 x + 10$ , gradient = $- 4$

$x = 2,25$ , no gradient

$y = -2$ , gradient = $0$

Exercise 2

$l$ : $(-1 \frac{1}{2}, 0)$ and $(0, \frac{3}{4})$
$m$ : $(\frac{8}{5}, 0)$ and $(0, 2)$

Enter: Y1=1/2X+3/4 and Y2=-5/4X+2. Window: $-4 \leq x \leq 4$ and $-4 \leq y \leq 4$ .

The calculator gives you coordinates $(0.71; 1.11)$ .

Exercise 3

$a + s = 90$ and $0.90 a + 1.05 s = 100$

Enter: Y1=-X+90 and Y2=-7/6X+100. Window: $0 \leq x \leq 100$ and $0 \leq y \leq 100$ .

The point of intersection of the two graphs is $(30, 60)$ , so $30$ packs of apple juice and $60$ packs of orange juice.

Exercise 4

The donkey is carrying $x$ bags and the mule is carrying $y$ bags.
If the donkey gave one bag to the mule, then he would be left with $x - 1$ and the mule would have $y + 1$ and also $2 (x - 1) = y + 1$ . It follows that $y = 2 x - 3$ .
If the mule gave a bag to the donkey then the donkey would have $x + 1$ and the mule $y - 1$ and also $x + 1 = y - 1$ , therefore $y = x + 2$ .
So $2 x - 3 = x + 2$ , or $x = 5$ and $y = 7$ .
The donkey is carrying $5$ bags and the mule $7$ .

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