Question

3. Rezolvatii in numere reale urmatoarele inecuatii si 4 va rog Urgent dau coroana

Answer 1

Răspuns:

3.

a) x < -2

b) y > -2

c) z ≤ -4

d) t ≥ -7/2

4.

a) x > 1

b) x < -30

c) x ≥ 1/7

d) x ≥ -17/7

Explicație pas cu pas:

3.

a) 2x+3 < x+1 ⇔ 2x-x<-3+1  ⇔ x<-2

b) -3y+7 > -5y+3  ⇔ -3y+5y > -7+3  ⇔ 2y > -4 ⇔ y > -4/2   ⇔  y > -2

c) -10 ≥ 2(z+1)+z  ⇔ -10 ≥ 2z+2+z ⇔ -10 ≥ 3z+2  ⇔ -3z ≥ 10+2 ⇔ -3z ≥ 12 ⇔ 3z ≤ -12  ⇔ z ≤ -4

d) $0,5t + \frac{1}{2} \leq 1,5t +4$   ⇔ $0,5t-1,5t \leq -\frac{1}{2} +4$  ⇔ $-t\leq \frac{7}{2}$  ⇔ $t\geq -\frac{7}{2}$

4.

a)   $\frac{2}{3} x - \frac{1}{5} > \frac{7}{15}$   ⇔ $\frac{2}{3} x > \frac{3}{15} + \frac{7}{15}$   ⇔  $\frac{10}{15} x > \frac{10}{15}$  ⇔ x > 1

b) 1,3x - 13 < 2x +8  ⇔ -0,7x < 13+8 ⇔ -0,7x < 21 ⇔ $x < -\frac{21}{0,7}$  ⇔ x < -30

c) $\frac{-2x+1}{5} \geq \frac{-5x+1}{2}$  ⇔ -4x+2 ≥ -25x+5  ⇔ -4x+25x ≥ -2+5 ⇔ 21x ≥ 3 ⇔ $x\geq \frac{1}{7}$

d) $\frac{4(x+3)}{-3} \leq \frac{2x-2}{9}$  ⇔ $\frac{-12x - 36}{9} \leq \frac{2x-2}{9}$ - am trecut semnul - la numarator si am adus la numitor comun cu membrul drept

 -12x - 36 ≤ 2x-2 (fiind numitor comun, inegalitatea se transfera asupra numaratorilor)

-12x-2x ≤ 36-2  ⇔ -14x ≤ 34 ⇔ 7x ≥ -17  ⇔ $x\geq \frac{-17}{7}$