[tex]\frac{(x-1)}{2}[/tex] + [tex]\frac{(x+1)}{3}[/tex] ≥ 7

Предыдущий
вопрос Следующий
вопрос

To solve this inequality, we need to find the values of x that satisfy the inequality.

First, let's find a common denominator for the fractions on the left side:

[tex]\frac{3(x-1)}{6}[/tex] + [tex]\frac{2(x+1)}{6}[/tex] ≥ 7

Now combine the fractions:

[tex]\frac{3(x-1) + 2(x+1)}{6}[/tex] ≥ 7

Expand the numerators:

[tex]\frac{3x - 3 + 2x + 2}{6}[/tex] ≥ 7

Combine like terms:

[tex]\frac{5x - 1}{6}[/tex] ≥ 7

Multiply both sides by 6 to get rid of the denominator:

5x - 1 ≥ 42

Add 1 to both sides:

5x ≥ 43

Divide by 5 to isolate x:

x ≥ 8.6

Therefore, the inequality [tex]\frac{(x-1)}{2}[/tex] + [tex]\frac{(x+1)}{3}[/tex] ≥ 7 is satisfied when x is greater than or equal to 8.6.