Rozwiąż równanie...- Zadanie 1: Matematyka z kluczem 7

Rozwiązanie

POZIOM A

a) $3 x - 8 = - 7 ∣ + 8$

$3 x = - 7 + 8$

$3 x = 1 ∣ : 3$

$x = \frac{1}{3}$

b) $4 x - 2 = - 2 ∣ + 2$

$4 x = - 2 + 2$

$4 x = 0 ∣ : 4$

$x = 0$

c) $15 x + 19 = 10 ∣ - 19$

$15 x = 10 - 19$

$15 x = - 9 ∣ : 15$

$x = - \frac{9}{15}$

$x = - \frac{3}{5}$

d) $2 x + 9 = 1 ∣ - 9$

$2 x = 1 - 9$

$2 x = - 8 ∣ : 2$

$x = - 4$

e) $7 x - 9 = 5 ∣ + 9$

$7 x = 5 + 9$

$7 x = 14 ∣ : 7$

$x = 2$

f) $8 x + 5 = - 7 ∣ - 5$

$8 x = - 7 - 5$

$8 x = - 12 ∣ : 8$

$x = - \frac{12}{8}$

$x = - \frac{3}{2}$

g) $9 x - 40 = 14 ∣ + 40$

$9 x = 14 + 40$

$9 x = 54 ∣ : 9$

$x = 6$

h) $20 x + 7 = 31 ∣ - 7$

$20 x = 31 - 7$

$20 x = 24 ∣ : 20$

$x = \frac{24}{20}$

$x = \frac{6}{5}$

i) $6 x - 25 = - 25 ∣ + 25$

$6 x = - 25 + 25$

$6 x = 0 ∣ : 6$

$x = 0$

POZIOM B

a) $3 x - 8 = 2 x + 4 ∣ + 8$

$3 x = 2 x + 4 + 8$

$3 x = 2 x + 12 ∣ - 2 x$

$3 x - 2 x = 12$

$x = 12$

b) $5 x - 3 = 4 x + 6 ∣ + 3$

$5 x = 4 x + 6 + 3$

$5 x = 4 x + 9 ∣ - 4 x$

$5 x - 4 x = 9$

$x = 9$

c) $7 x - 9 = 6 x - 8 ∣ + 9$

$7 x = 6 x - 8 + 9$

$7 x = 6 x + 1 ∣ - 6 x$

$7 x - 6 x = 1$

$x = 1$

d) $7 x - 5 = 6 x + 8 ∣ + 5$

$7 x = 6 x + 8 + 5$

$7 x = 6 x + 13 ∣ - 6 x$

$7 x - 6 x = 13$

$x = 13$

e) $5 x - 8 = 4 x + 9 ∣ + 8$

$5 x = 4 x + 9 + 8$

$5 x = 4 x + 17 ∣ - 4 x$

$5 x - 4 x = 17$

$x = 17$

f) $9 x - 4 = 8 x + 5 ∣ + 4$

$9 x = 8 x + 5 + 4$

$9 x = 8 x + 9 ∣ - 8 x$

$9 x - 8 x = 9$

$x = 9$

g) $12 x + 4 = 11 x + 15 ∣ - 4$

$12 x = 11 x + 15 - 4$

$12 x = 11 x + 11 ∣ - 11 x$

$12 x - 11 x = 11$

$x = 11$

h) $3 x - 1 = 2 x + 1 ∣ + 1$

$3 x = 2 x + 1 + 1$

$3 x = 2 x + 2 ∣ - 2 x$

$3 x - 2 x = 2$

$x = 2$

i) $2 x + 7 = x - 1 ∣ - 7$

$2 x = x - 1 - 7$

$2 x = x - 8 ∣ - x$

$x = - 8$

POZIOM C

a) $4 x - 3 (3 x - 5) = 6 x + 4$

$4 x - 9 x + 15 = 6 x + 4$

$- 5 x + 15 = 6 x + 4 ∣ - 15$

$- 5 x = 6 x + 4 - 15$

$- 5 x = 6 x - 11 ∣ - 6 x$

$- 5 x - 6 x = - 11$

$- 11 x = - 11 ∣ : (- 11)$

$x = 1$

b) $8 - (4 x - 12) = 8 x - (3 x - 2)$

$8 - 4 x + 12 = 8 x - 3 x + 2$

$20 - 4 x = 5 x + 2 ∣ - 20$

$- 4 x = 5 x + 2 - 20$

$- 4 x = 5 x - 18 ∣ - 5 x$

$- 4 x - 5 x = - 18$

$- 9 x = - 18 ∣ : (- 9)$

$x = 2$

c) $\frac{5}{4} (1 - x) - \frac{1}{4} = \frac{3}{4} (x - 4) + 3$

$\frac{5}{4} - \frac{5}{4} x - \frac{1}{4} = \frac{3}{4} x - 3 + 3$

$1 - \frac{5}{4} x = \frac{3}{4} x ∣ - \frac{3}{4} x$

$1 - \frac{5}{4} x - \frac{3}{4} x = 0$

$1 - \frac{8}{4} x = 0$

$1 - 2 x = 0 ∣ - 1$

$- 2 x = 0 - 1$

$- 2 x = - 1 ∣ : (- 2)$

$x = \frac{1}{2}$

d) $7 (x + 2) - x = 5 (x - 2)$

$7 x + 14 - x = 5 x - 10$

$6 x + 14 = 5 x - 10 ∣ - 14$

$6 x = 5 x - 10 - 14$

$6 x = 5 x - 24 ∣ - 5 x$

$6 x - 5 x = - 24$

$x = - 24$

e) $0, 6 (x + 2) - (0, 4 x - 0, 6) = 0, 2 (1 - x)$

$0, 6 x + 1, 2 - 0, 4 x + 0, 6 = 0, 2 - 0, 2 x$

$0, 2 x + 1, 8 = 0, 2 - 0, 2 x ∣ - 1, 8$

$0, 2 x = 0, 2 - 0, 2 x - 1, 8$

$0, 2 x = - 1, 6 - 0, 2 x ∣ + 0, 2 x$

$0, 2 x + 0, 2 x = - 1, 6$

$0, 4 x = - 1, 6 ∣ \cdot 10$

$4 x = - 16 ∣ : 4$

$x = - 4$

f) $2 (x + \frac{1}{2}) - \frac{2}{3} (3 - 6 x) = x + 2$

$2 x + 1 - 2 + 4 x = x + 2$

$6 x - 1 = x + 2 ∣ + 1$

$6 x = x + 2 + 1$

$6 x = x + 3 ∣ - x$

$6 x - x = 3$

$5 x = 3 ∣ : 5$

$x = \frac{3}{5}$

g) $5 x - 6 (x + 7) = 8 x - 9 (10 + x)$

$5 x - 6 x - 42 = 8 x - 90 - 9 x$

$- x - 42 = - x - 90 ∣ + x$

$- 42 \neq = - 90$

h) $0, 1 (x - 10) - 0, 6 x = 1, 5 x - 4$

$0, 1 x - 1 - 0, 6 x = 1, 5 x - 4$

$- 0, 5 x - 1 = 1, 5 x - 4 ∣ + 1$

$- 0, 5 x = 1, 5 x - 4 + 1$

$- 0, 5 x = 1, 5 x - 3 ∣ - 1, 5 x$

$- 0, 5 x - 1, 5 x = - 3$

$- 2 x = - 3 ∣ : (- 2)$

$x = \frac{3}{2}$

POZIOM D

a) $4 x - 3 (x - 3) = x - 4$

$4 x - 3 x + 9 = x - 4$

$x + 9 = x - 4 ∣ - x$

$9 \neq = - 4$

b) $2 x - (3 - 2 x) = 5 + x - x$

$2 x - 3 + 2 x = 5$

$4 x - 3 = 5 ∣ + 3$

$4 x = 5 + 3$

$4 x = 8 ∣ : 4$

$x = 2$

c) $5 (x - 2) + (3 + x) = 6 x - 7$

$5 x - 10 + 3 + x = 6 x - 7$

$6 x - 7 = 6 x - 7$

$0 = 0$

d) $- (x + 1) + 8 x = 5 x - 9$

$- x - 1 + 8 x = 5 x - 9$

$7 x - 1 = 5 x - 9 ∣ + 1$

$7 x = 5 x - 9 + 1$

$7 x = 5 x - 8 ∣ - 5 x$

$7 x - 5 x = - 8$

$2 x = - 8 ∣ : 2$

$x = - 4$

e) $2 (2 x + 1) = 4 x + 3$

$4 x + 2 = 4 x + 3 ∣ - 4 x$
$2 \neq = 3$

f) $9 x + 3 = 7 (x - 2) + 17$

$9 x + 3 = 7 x - 14 + 17$

$9 x + 3 = 7 x + 3 ∣ - 3$

$9 x = 7 x + 3 - 3$

$9 x = 7 x ∣ - 7 x$

$9 x - 7 x = 0$

$2 x = 0 ∣ : 2$

$x = 0$

g) $5 x - 9 + x = 3 (2 x - 3)$

$6 x - 9 = 6 x - 9$

$0 = 0$

h) $4 (3 x - 2) + 5 = 2 (6 x - 4) + x$

$12 x - 8 + 5 = 12 x - 8 + x$

$12 x - 3 = 13 x - 8 ∣ + 3$

$12 x = 13 x - 8 + 3$

$12 x = 13 x - 5 ∣ - 13 x$

$12 x - 13 x = - 5$

$- x = - 5 ∣ : (- 1)$

$x = 5$

MISTRZ

a) $\frac{12 - x}{4} = \frac{x + 4}{12} ∣ \cdot 12$

$12 \cdot \frac{12 - x}{4} = 12 \cdot \frac{x + 4}{12}$

$12^{3} \cdot \frac{12 - x}{4 ^{1}} = 12^{1} \cdot \frac{x + 4}{12 ^{1}}$

$3 \cdot (12 - x) = x + 4$

$36 - 3 x = x + 4 ∣ - 36$

$- 3 x = x + 4 - 36$

$- 3 x = x - 32 ∣ - x$

$- 3 x - x = - 32$

$- 4 x = - 32 ∣ : (- 4)$

$x = 8$

b) $\frac{3 x - 2}{4} = \frac{4 x - 5}{3} ∣ \cdot 12$

$12 \cdot \frac{3 x - 2}{4} = 12 \cdot \frac{4 x - 5}{3}$

$12^{3} \cdot \frac{3 x - 2}{4 ^{1}} = 12^{4} \cdot \frac{4 x - 5}{3 ^{1}}$

$3 \cdot (3 x - 2) = 4 \cdot (4 x - 5)$

$9 x - 6 = 16 x - 20 ∣ + 6$

$9 x = 16 x - 20 + 6$

$9 x = 16 x - 14 ∣ - 16 x$

$9 x - 16 x = - 14$

$- 7 x = - 14 ∣ : (- 7)$

$x = 2$

c) $\frac{x - 3}{6} = 2 (3 x + 1) ∣ \cdot 6$

$6 \cdot \frac{x - 3}{6} = 6 \cdot 2 (3 x + 1)$

$6^{1} \cdot \frac{x - 3}{6 ^{1}} = 12 (3 x + 1)$

$x - 3 = 12 (3 x + 1)$

$x - 3 = 36 x + 12 ∣ + 3$

$x = 36 x + 12 + 3$

$x = 36 x + 15 ∣ - 36 x$

$x - 36 x = 15$

$- 35 x = 15 ∣ : (- 35)$

$x = \frac{15}{- 35}$

$x = \frac{3}{- 7}$

$x = - \frac{3}{7}$

d) $3 (1 - \frac{x}{3}) = \frac{5 x - 1}{9} ∣ \cdot 9$

$9 \cdot 3 (1 - \frac{x}{3}) = 9 \cdot \frac{5 x - 1}{9}$

$27 (1 - \frac{x}{3}) = 9^{1} \cdot \frac{5 x - 1}{9 ^{1}}$

$27 - 9 x = 5 x - 1 ∣ - 27$

$- 9 x = 5 x - 1 - 27$

$- 9 x = 5 x - 28 ∣ - 5 x$

$- 9 x - 5 x = - 28$

$- 14 x = - 28 ∣ : (- 14)$

$x = 2$

Magdalena Matusik

Nauczycielka matematyki

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