QuestionSolve the compound inequality: $3(x-4)<15$ or $x+10>14$ . What are the solution options?

Studdy Solution

STEP 1

Assumptions1. The given compound inequality is $3(x-4)<15$ or $x+10>14$ . . We need to solve each inequality separately and then combine the solutions according to the 'or' operator.
3. The inequalities are linear, so the solution will be a range of values for $x$ .

STEP 2

First, let's solve the inequality $(x-4)<15$ . We can start by distributing the on the left side.
$x -12 <15$

STEP 3

Next, we add12 to both sides of the inequality to isolate $3x$ on the left side.
$3x -12 +12 <15 +12$ $3x <27$

STEP 4

Then, we divide both sides of the inequality by3 to solve for $x$ .
$\frac{3x}{3} < \frac{27}{3}$ $x <9$

STEP 5

Now, let's solve the second inequality $x+10>14$ . We start by subtracting10 from both sides to isolate $x$ .
$x +10 -10 >14 -10$ $x >4$

STEP 6

Since the original compound inequality is connected by 'or', the solution is the union of the solutions of the two inequalities. This means that $x$ can be any value that satisfies either $x <9$ or $x >4$ .
So, the solution is all real numbers less than9 or greater than4.
Looking at the given choices, the correct choice is not listed. The correct solution should be $x <9$ or $x >4$ .

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QuestionSolve the compound inequality: $3(x-4)<15$ or $x+10>14$ . What are the solution options?

Studdy Solution

STEP 1

Assumptions1. The given compound inequality is $3(x-4)<15$ or $x+10>14$ . . We need to solve each inequality separately and then combine the solutions according to the 'or' operator.
3. The inequalities are linear, so the solution will be a range of values for $x$ .

STEP 2

First, let's solve the inequality $(x-4)<15$ . We can start by distributing the on the left side.
$x -12 <15$

STEP 3

Next, we add12 to both sides of the inequality to isolate $3x$ on the left side.
$3x -12 +12 <15 +12$ $3x <27$

STEP 4

Then, we divide both sides of the inequality by3 to solve for $x$ .
$\frac{3x}{3} < \frac{27}{3}$ $x <9$

STEP 5

Now, let's solve the second inequality $x+10>14$ . We start by subtracting10 from both sides to isolate $x$ .
$x +10 -10 >14 -10$ $x >4$

STEP 6

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QuestionSolve the compound inequality: 3(x−4)<153(x-4)<153(x−4)<15 or x+10>14x+10>14x+10>14. What are the solution options?

Studdy Solution

STEP 1

Assumptions1. The given compound inequality is 3(x−4)<153(x-4)<153(x−4)<15 or x+10>14x+10>14x+10>14. . We need to solve each inequality separately and then combine the solutions according to the 'or' operator.3. The inequalities are linear, so the solution will be a range of values for xxx.

STEP 2

First, let's solve the inequality (x−4)<15(x-4)<15(x−4)<15. We can start by distributing the on the left side.x−12<15x -12 <15x−12<15

STEP 3

Next, we add12 to both sides of the inequality to isolate 3x3x3x on the left side.3x−12+12<15+123x -12 +12 <15 +123x−12+12<15+123x<273x <273x<27

STEP 4

Then, we divide both sides of the inequality by3 to solve for xxx.3x3<273\frac{3x}{3} < \frac{27}{3}33x​<327​x<9x <9x<9

STEP 5

Now, let's solve the second inequality x+10>14x+10>14x+10>14. We start by subtracting10 from both sides to isolate xxx.x+10−10>14−10x +10 -10 >14 -10x+10−10>14−10x>4x >4x>4

STEP 6

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QuestionSolve the compound inequality: $3(x-4)<15$ or $x+10>14$ . What are the solution options?

Assumptions1. The given compound inequality is $3(x-4)<15$ or $x+10>14$ . . We need to solve each inequality separately and then combine the solutions according to the 'or' operator.
3. The inequalities are linear, so the solution will be a range of values for $x$ .

First, let's solve the inequality $(x-4)<15$ . We can start by distributing the on the left side.
$x -12 <15$

Next, we add12 to both sides of the inequality to isolate $3x$ on the left side.
$3x -12 +12 <15 +12$ $3x <27$

Then, we divide both sides of the inequality by3 to solve for $x$ .
$\frac{3x}{3} < \frac{27}{3}$ $x <9$

Now, let's solve the second inequality $x+10>14$ . We start by subtracting10 from both sides to isolate $x$ .
$x +10 -10 >14 -10$ $x >4$