Inequalities: Solving Linear Inequalities

Hello ! This is the second note in the series Inequalities.

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LINEAR−INEQUALITIES \mathbf{LINEAR-INEQUALITIES}

  1. a≥b⇒ a \geq b \Rightarrow either a>b a > b or a=b a = b

  2. If a>b a > b and b>c⇒ b > c \Rightarrow a>c a > c [ Transition property ]

  3. For a>b⇔a±c<b±c a > b \Leftrightarrow a \pm c < b \pm c

  4. If a>b a > b , then: ac>bc ifc>0ac<bc ifc<0 \begin{aligned} ac > bc & \text{ if} & c > 0 \\ ac < bc & \text{ if} & c < 0 \end{aligned}

    In other words, the sign of inequality flips if we multiply a negative real number to both side. For example: 3>2⇔−3<−2 3 > 2 \Leftrightarrow -3 < -2 . Which is definitely true.

  5. If ax>b ax > b , then: x>b/a ifa>0 [Sign remains unchanged] x<b/a ifa<0 [Sign changes]  \begin{aligned} x > b/a & \text{ if} & a > 0 & \text{ [Sign remains unchanged] }\\ x < b/a & \text{ if} & a < 0 & \text{ [Sign changes] } \end{aligned}

    In other words, the sign of inequality flips if we divide both sides by a negative real number.

SOLVED−EXAMPLE \mathbf{SOLVED-EXAMPLE} 1. Find the range of values of x x satisfying the inequality: 3≥1−2x>−19 3 \geq 1-2x > -19 Solution: −3≤2x−1<19 [Multiplying both sides by -1] −2≤2x<20−1≤x<10 [Dividing both sides by 2]  \begin{aligned} -3 \leq 2x-1& < 19 & \text{ [Multiplying both sides by -1] } \\ -2 \leq 2x & < 20 & \\ -1 \leq x & < 10 & \text{ [Dividing both sides by 2] } \end{aligned} Thus the answer is: x∈[−1,10) x \in [-1,10)

PRACTICE \mathbf{PRACTICE} 1. Find the range of values of y y in interval form, satisfying the inequality: −4≤y−13≤a -4 \leq \dfrac{y-1}{3} \leq a Where ′a′ 'a' is number of irrational roots of: x2+x2+11+x2−x2+11=4 \sqrt{x^2 + \sqrt{x^2 + 11}} + \sqrt{x^2 - \sqrt{x^2 + 11}} = 4

2. Algebraic mess

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Note by Priyansh Sangule
6 years, 11 months ago

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