Verified Answer :

calculista

we will proceed to solve all cases to determine the solution of the problem

case A) \(6(x+2)> x-3\)

Eliminate the parenthesis of the left side

\(6x+12> x-3\)

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

\(6x-x> -3-12\)

Combine like terms

\(5x> -15\)

\(x> -3\)

The solution is the interval--------> (-3,∞)

All real numbers greater than \(-3\)

therefore

The inequality A has solution

case B) \(3+4x <2(1+2x)\)

Eliminate the parenthesis of the right side

\(3+4x <2+4x\)

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

\(4x-4x <2-3\)

Combine like terms

\(0 <-1\) ---------> is not true

therefore

The inequality B has no solution

case C) \(-2(x+6)< x-20\)

Eliminate the parenthesis of the left side

\(-2x-12< x-20\)

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

\(-2x-x< -20+12\)

Combine like terms

\(-3x< -8\)

Multiply by \(-1\) both sides

\(3x> 8\)

\(x> (8/3)\)

The solution is the interval--------> (8/3,∞)

All real numbers greater than \(8/3\)

therefore

The inequality C has solution

case D) \(x-9<3(x-3)\)

Eliminate the parenthesis of the right side

\(x-9<3x-9\)

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

\(x-3x< -9+9\)

Combine like terms

\(-2x< 0\)

Multiply by \(-1\) both sides

\(2x> 0\)

\(x>0\)

The solution is the interval--------> (0,∞)

All real numbers greater than \(0\)

therefore

The inequality D has solution

therefore

the answer is

\(3+4x <2(1+2x)\)

\(\boxed{3 + 4x < 2\left( {1 + 2x} \right)}\) inequality has no solution. Option (b) is correct.

Further explanation:

The linear equation with slope m and intercept c is given as follows.

\(\boxed{y = mx + c}\)

The formula for slope of line with points \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right)\) can be expressed as,

\(\boxed{m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}\)

Given:

The inequalities are as follows,

(a). \(6\left( {x + 2} \right) > x - 3\)

(b). \(3 + 4x < 2\left( {1 + 2x} \right)\)

(c). \(- 2\left( {x + 6} \right) < x - 20\)

(d). \(x - 9 < 3\left( {x - 3} \right)\)

Explanation:

Solve the inequality \(6\left( {x + 2} \right) > x - 3\) to check whether the inequality has solution.

\(\begin{aligned}6\left( {x + 2} \right) >& x - 3\\6x + 12 >& x - 3\\6x - x >&  - 3 - 12\\ 5x >&- 15\\x >& \frac{{ - 15}}{5}\\x >&- 3\\\end{aligned}\)

Option (a) has solution.

Solve the inequality \(3 + 4x < 2\left( {1 + 2x} \right)\) to check whether the inequality has solution.

\(\begin{aligned}3+ 4x &< 2\left( {1 + 2x} \right)\\3 + 4x &< 2 + 4x\\3&< 2\\\end{aligned}\)

Option (b) has no solution.

Solve the inequality \(- 2\left( {x + 6} \right) < x - 20\) to check whether the inequality has solution.

\(\begin{aligned}- 2\left( {x + 6}\right) &< x - 20\\- 2x - 12 &< x - 20\\- 12 + 20 &< x + 2x\\8&< 3x\\\frac{8}{3} &< x\\\end{aligned}\)

Option (c) has solution.

Solve the inequality \(x - 9 < 3\left( {x - 3} \right)\) to check whether the inequality has solution.

\(\begin{aligned}x - 9&< 3\left( {x - 3} \right)\\x - 9&< 3x - 9\\x&< 3x \\1&< 3\\\end{aligned}\)

Option (d) has solution.

\(\boxed{3 + 4x < 2\left( {1 + 2x} \right)}\) inequality has no solution. Option (b) is correct.

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear equation

Keywords: numbers,slope, slope intercept, inequality, equation, linear inequality, shaded region, y-intercept, graph, representation, origin.

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