18 uždavinys
Dvi sesutės – Irutė ir Birutė – kurį laiką gaudė pokemonus. Irutė kasdien sugaudavo po x pokemonų, o Birutė – trimis pokemonais daugiau. Irutė pokemonus gaudė viena diena ilgiau negu Birutė. Birutė iš viso sugavo 484 pokemonus, o Irutė iš viso sugavo 437 pokemonus.
Apskaičiuokite x reikšmę
Sprendimas:
| Kas dien sugaudavo po | Gaudė dienų | Iš viso sugavo | |
| Irutė | x | y | 437 |
| Birutė | x + 3 | y - 1 | 484 |
Lygčių sistema:
437 = xy (1)
484 = (x + 3)(y-1) (2)
Iš pirmos lygties $$y = \frac{437}{x}$$
Statome y į antrą lygtį:
484 =
(x+3)* (
-1)
(x+3)* (
| 437 |
| / x |
|
484 = (x+3)* (
-1)$$484$$ = $$(x+3)\cdot (\frac{437}{x}-1)$$
| 437 |
| / x |
484 = (
- x* 1+
- 3* 1)$$484$$ = $$(\frac{x\cdot 437}{x}-x\cdot 1+\frac{3\cdot 437}{x}-3\cdot 1)$$
| x* 437 |
| / x |
| 3* 437 |
| / x |
484 = (437- x* 1+
- 3* 1)$$484$$ = $$(437-x\cdot 1+\frac{3\cdot 437}{x}-3\cdot 1)$$
| 3* 437 |
| / x |
484 = (437-x+
- 3* 1)$$484$$ = $$(437-x+\frac{3\cdot 437}{x}-3\cdot 1)$$
| 3* 437 |
| / x |
484 = (437-x+
- 3* 1)$$484$$ = $$(437-x+\frac{1311}{x}-3\cdot 1)$$
| 1311 |
| / x |
484 = (437-x+
-3)$$484$$ = $$(437-x+\frac{1311}{x}-3)$$
| 1311 |
| / x |
484 = 437-x+
-3$$484$$ = $$437-x+\frac{1311}{x}-3$$
| 1311 |
| / x |
x+484 = 437+
-3$$x+484$$ = $$437+\frac{1311}{x}-3$$
| 1311 |
| / x |
x-
+484 = 437-3$$x-\frac{1311}{x}+484$$ = $$437-3$$
| 1311 |
| / x |
x-
+484-437 = -3$$x-\frac{1311}{x}+484-437$$ = $$-3$$
| 1311 |
| / x |
x-
+484-437+3 = 0$$x-\frac{1311}{x}+484-437+3$$ = $$0$$
| 1311 |
| / x |
x-
+47+3 = 0$$x-\frac{1311}{x}+47+3$$ = $$0$$
| 1311 |
| / x |
x-
+50 = 0$$x-\frac{1311}{x}+50$$ = $$0$$
| 1311 |
| / x |
x* x-
+ 50* x = 0$$x\cdot x-\frac{1311\cdot x}{x}+50\cdot x$$ = $$0$$
| 1311* x |
| / x |
x^2-
+ 50* x = 0$$x^{2}-\frac{1311\cdot x}{x}+50\cdot x$$ = $$0$$
| 1311* x |
| / x |
x^2-1311+ 50* x = 0$$x^{2}-1311+50\cdot x$$ = $$0$$
x^2+ 50* x-1311 = 0$$x^{2}+50\cdot x-1311$$ = $$0$$
Paaiškinimas: (x-19)* (x+69) = 0$$(x-19)\cdot (x+69)$$ = $$0$$
x-19 = 0$$x-19$$ = $$0$$
x = 0+19$$x$$ = $$0+19$$
x = 19$$x$$ = $$19$$
$$484$$ = $$(x+3)\cdot (\frac{437}{x}-1)$$
$$484$$ = $$437-x+\frac{1311}{x}-3$$
$$x-\frac{1311}{x}+50$$ = $$0$$
$$x^{2}+50\cdot x-1311$$ = $$0$$
$$(x-19)\cdot (x+69)$$ = $$0$$
$$x-19$$ = $$0$$
$$x$$ = $$19$$
Atsakymas: 19