we will proceed to solve all cases to determine the solution of the problem
case A) [tex]6(x+2)> x-3[/tex]
Eliminate the parenthesis of the left side
[tex]6x+12> x-3[/tex]
Group terms that contain the same variable, and move the constants to the opposite side of the inequality
[tex]6x-x> -3-12[/tex]
Combine like terms
[tex]5x> -15[/tex]
[tex]x> -3[/tex]
The solution is the interval--------> (-3,∞)
All real numbers greater than [tex]-3[/tex]
therefore
The inequality A has solution
case B) [tex]3+4x <2(1+2x)[/tex]
Eliminate the parenthesis of the right side
[tex]3+4x <2+4x[/tex]
Group terms that contain the same variable, and move the constants to the opposite side of the inequality
[tex]4x-4x <2-3[/tex]
Combine like terms
[tex]0 <-1[/tex] ---------> is not true
therefore
The inequality B has no solution
case C) [tex]-2(x+6)< x-20[/tex]
Eliminate the parenthesis of the left side
[tex]-2x-12< x-20[/tex]
Group terms that contain the same variable, and move the constants to the opposite side of the inequality
[tex]-2x-x< -20+12[/tex]
Combine like terms
[tex]-3x< -8[/tex]
Multiply by [tex]-1[/tex] both sides
[tex]3x> 8[/tex]
[tex]x> (8/3)[/tex]
The solution is the interval--------> (8/3,∞)
All real numbers greater than [tex]8/3[/tex]
therefore
The inequality C has solution
case D) [tex]x-9<3(x-3)[/tex]
Eliminate the parenthesis of the right side
[tex]x-9<3x-9[/tex]
Group terms that contain the same variable, and move the constants to the opposite side of the inequality
[tex]x-3x< -9+9[/tex]
Combine like terms
[tex]-2x< 0[/tex]
Multiply by [tex]-1[/tex] both sides
[tex]2x> 0[/tex]
[tex]x>0[/tex]
The solution is the interval--------> (0,∞)
All real numbers greater than [tex]0[/tex]
therefore
The inequality D has solution
therefore
the answer is
[tex]3+4x <2(1+2x)[/tex]
[tex]\boxed{3 + 4x < 2\left( {1 + 2x} \right)}[/tex] inequality has no solution. Option (b) is correct.
Further explanation:
The linear equation with slope m and intercept c is given as follows.
[tex]\boxed{y = mx + c}[/tex]
The formula for slope of line with points [tex]\left( {{x_1},{y_1}} \right)[/tex] and [tex]\left( {{x_2},{y_2}} \right)[/tex] can be expressed as,
[tex]\boxed{m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}[/tex]
Given:
The inequalities are as follows,
(a). [tex]6\left( {x + 2} \right) > x - 3[/tex]
(b). [tex]3 + 4x < 2\left( {1 + 2x} \right)[/tex]
(c). [tex]- 2\left( {x + 6} \right) < x - 20[/tex]
(d). [tex]x - 9 < 3\left( {x - 3} \right)[/tex]
Explanation:
Solve the inequality [tex]6\left( {x + 2} \right) > x - 3[/tex] to check whether the inequality has solution.
[tex]\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}[/tex]
Option (a) has solution.
Solve the inequality [tex]3 + 4x < 2\left( {1 + 2x} \right)[/tex] to check whether the inequality has solution.
[tex]\begin{aligned}3+ 4x &< 2\left( {1 + 2x} \right)\\3 + 4x &< 2 + 4x\\3&< 2\\\end{aligned}[/tex]
Option (b) has no solution.
Solve the inequality [tex]- 2\left( {x + 6} \right) < x - 20[/tex] to check whether the inequality has solution.
[tex]\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}[/tex]
Option (c) has solution.
Solve the inequality [tex]x - 9 < 3\left( {x - 3} \right)[/tex] to check whether the inequality has solution.
[tex]\begin{aligned}x - 9&< 3\left( {x - 3} \right)\\x - 9&< 3x - 9\\x&< 3x \\1&< 3\\\end{aligned}[/tex]
Option (d) has solution.
[tex]\boxed{3 + 4x < 2\left( {1 + 2x} \right)}[/tex] 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.
