Home
Class 11
MATHS
2(X - 1) lt X + 5, 3 (X + 2) gt 2 - X...

`2(X - 1) lt X + 5, 3 (X + 2) gt 2 - X `

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given inequalities step by step, we will break down each part of the problem. ### Step 1: Write down the inequalities We have two inequalities to solve: 1. \( 2(X - 1) < X + 5 \) 2. \( 3(X + 2) > 2 - X \) ### Step 2: Expand the first inequality For the first inequality, we will distribute the 2: \[ 2(X - 1) = 2X - 2 \] So, the inequality becomes: \[ 2X - 2 < X + 5 \] ### Step 3: Rearrange the first inequality Now, we will move all terms involving \(X\) to one side and constant terms to the other side: \[ 2X - X < 5 + 2 \] This simplifies to: \[ X < 7 \] ### Step 4: Expand the second inequality Now, let's expand the second inequality: \[ 3(X + 2) = 3X + 6 \] So, the inequality becomes: \[ 3X + 6 > 2 - X \] ### Step 5: Rearrange the second inequality Now, we will move all terms involving \(X\) to one side and constant terms to the other side: \[ 3X + X > 2 - 6 \] This simplifies to: \[ 4X > -4 \] ### Step 6: Solve for \(X\) in the second inequality Now, divide both sides by 4: \[ X > -1 \] ### Step 7: Combine the results Now we have two results: 1. \(X < 7\) 2. \(X > -1\) ### Step 8: Write the solution in interval notation The solution can be expressed in interval notation as: \[ X \in (-1, 7) \] ### Step 9: Draw the number line (optional) You can represent this solution on a number line, indicating that the interval does not include the endpoints -1 and 7.

To solve the given inequalities step by step, we will break down each part of the problem. ### Step 1: Write down the inequalities We have two inequalities to solve: 1. \( 2(X - 1) < X + 5 \) 2. \( 3(X + 2) > 2 - X \) ### Step 2: Expand the first inequality ...
Promotional Banner

Topper's Solved these Questions

  • LINEAR INEQUALITIES

    NAGEEN PRAKASHAN ENGLISH|Exercise NCERT QUESTION|51 Videos
  • LIMITS AND DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|30 Videos
  • MATHEMATICAL REASONING

    NAGEEN PRAKASHAN ENGLISH|Exercise Misellaneous exercise|7 Videos

Similar Questions

Explore conceptually related problems

3x - 2 lt 2x + 1

The solution set of system of linear inequalities 2(x+1) le x+5, 3(x+2), gt 2 - x , x in R is

5x + 1 gt - 24, 5x - 1 lt 24

The function f(x) ={{:( 5x-4, " for " 0 lt x le 1) ,( 4x^(2) - 3x, " for" 1 lt x lt 2 ),( 3x + 4 , "for" x ge2):}is

If f(x) {:{(x^(2)+4," for "x lt 2 ),(x^(3)," for " x gt 2 ):} , find Lim_(x to 2) f(x)

Let f(x)={{:(x^(2) sin ((pix)/(2)),-1 lt x lt 1, x ne 0),(x|x|, x gt 1 or x le -1):} . Then ,

If the median of the scores 1,2,x,4,5("where " 1 lt 2 lt x lt 4 lt 5) is 3, then the mean of the scores is

Statement - 1 : If x gt 1 then log_(10)x lt log_(3)x lt log_(e )x lt log_(2)x . Statement - 2 : If 0 lt x lt 1 , then log_(x)a gt log_(x)b implies a lt b .

the area of region for which 0 lt y lt 3 - 2x-x^2 and x gt 0 is

Find the value of a ,b if f(x) = {((a e^(1//|x+2|) - 1)/(2-e^(1//|x+2|)),; -3 lt x lt -2), (b,; x = -2), (sin((x^4-16)/(x^5+32)),; -2 lt x lt 0):} is continuous at x = -2.