Solved on Sep 28, 2023

Solve $|2x-2| + |x-1| = 6$ graphically. Find the solution set. A. $\{-1,3\}$ B. The solution set is the set of real numbers. C. The solution set is the empty set.

STEP 1

Assumptions1. The given equation is $|x-|+|x-1|=6$ . We are asked to solve this equation graphically3. The absolute value function $|x|$ is defined as $x$ if $x \geq0$ and $-x$ if $x <0$
4. The solution set is the set of all $x$ values that satisfy the given equation

STEP 2

The absolute value function splits the equation into different cases depending on the sign of the expressions inside the absolute value. We will split the equation into four cases1. $2x-2 \geq0$ and $x-1 \geq0$
2. $2x-2 <0$ and $x-1 \geq0$ . $2x-2 \geq0$ and $x-1 <0$
4. $2x-2 <0$ and $x-1 <0$

STEP 3

Let's solve the first case where both $2x-2 \geq0$ and $x-1 \geq0$ . This implies that $x \geq1$ . The equation becomes $2x-2 + x-1 =6$

STEP 4

implify the equation from3 $3x -3 =6$

STEP 5

olve the equation from4 for $x$ $3x =9$ $x =3$ Since $x =3$ is in the range $x \geq1$ , it is a valid solution for this case.

STEP 6

Now, let's solve the second case where $2x-2 <0$ and $x-1 \geq0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

STEP 7

Now, let's solve the third case where $2x-2 \geq0$ and $x-1 <0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

STEP 8

Finally, let's solve the fourth case where $2x-2 <0$ and $x-1 <0$ . This implies that $x <1$ . The equation becomes $-(2x-2) - (x-1) =6$

STEP 9

implify the equation from8 $-2x +2 - x + =6$ $-3x +3 =6$

STEP 10

olve the equation from9 for $x$ $-3x =3$ $x = -$ Since $x = -$ is in the range $x <$ , it is a valid solution for this case.

STEP 11

The solutions to the given equation are the solutions from the first and fourth cases, which are $x =3$ and $x = -$ . Therefore, the solution set is $\{-,3\}$ .
The correct answer is A. $\{-,3\}$ .

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Solve $|2x-2| + |x-1| = 6$ graphically. Find the solution set. A. $\{-1,3\}$ B. The solution set is the set of real numbers. C. The solution set is the empty set.

STEP 1

Assumptions1. The given equation is $|x-|+|x-1|=6$ . We are asked to solve this equation graphically3. The absolute value function $|x|$ is defined as $x$ if $x \geq0$ and $-x$ if $x <0$
4. The solution set is the set of all $x$ values that satisfy the given equation

STEP 2

The absolute value function splits the equation into different cases depending on the sign of the expressions inside the absolute value. We will split the equation into four cases1. $2x-2 \geq0$ and $x-1 \geq0$
2. $2x-2 <0$ and $x-1 \geq0$ . $2x-2 \geq0$ and $x-1 <0$
4. $2x-2 <0$ and $x-1 <0$

STEP 3

Let's solve the first case where both $2x-2 \geq0$ and $x-1 \geq0$ . This implies that $x \geq1$ . The equation becomes $2x-2 + x-1 =6$

STEP 4

implify the equation from3 $3x -3 =6$

STEP 5

olve the equation from4 for $x$ $3x =9$ $x =3$ Since $x =3$ is in the range $x \geq1$ , it is a valid solution for this case.

STEP 6

Now, let's solve the second case where $2x-2 <0$ and $x-1 \geq0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

STEP 7

Now, let's solve the third case where $2x-2 \geq0$ and $x-1 <0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

STEP 8

Finally, let's solve the fourth case where $2x-2 <0$ and $x-1 <0$ . This implies that $x <1$ . The equation becomes $-(2x-2) - (x-1) =6$

STEP 9

implify the equation from8 $-2x +2 - x + =6$ $-3x +3 =6$

STEP 10

olve the equation from9 for $x$ $-3x =3$ $x = -$ Since $x = -$ is in the range $x <$ , it is a valid solution for this case.

STEP 11

The solutions to the given equation are the solutions from the first and fourth cases, which are $x =3$ and $x = -$ . Therefore, the solution set is $\{-,3\}$ .
The correct answer is A. $\{-,3\}$ .

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Solve ∣2x−2∣+∣x−1∣=6|2x-2| + |x-1| = 6∣2x−2∣+∣x−1∣=6 graphically. Find the solution set. A. {−1,3}\{-1,3\}{−1,3} B. The solution set is the set of real numbers. C. The solution set is the empty set.

STEP 1

STEP 2

STEP 3

Let's solve the first case where both 2x−2≥02x-2 \geq02x−2≥0 and x−1≥0x-1 \geq0x−1≥0. This implies that x≥1x \geq1x≥1. The equation becomes2x−2+x−1=62x-2 + x-1 =62x−2+x−1=6

STEP 4

implify the equation from33x−3=63x -3 =63x−3=6

STEP 5

olve the equation from4 for xxx3x=93x =93x=9x=3x =3x=3Since x=3x =3x=3 is in the range x≥1x \geq1x≥1, it is a valid solution for this case.

STEP 6

Now, let's solve the second case where 2x−2<02x-2 <02x−2<0 and x−1≥0x-1 \geq0x−1≥0. This implies that 1≤x<11 \leq x <11≤x<1, which is not possible. Therefore, this case has no solution.

STEP 7

Now, let's solve the third case where 2x−2≥02x-2 \geq02x−2≥0 and x−1<0x-1 <0x−1<0. This implies that 1≤x<11 \leq x <11≤x<1, which is not possible. Therefore, this case has no solution.

STEP 8

Finally, let's solve the fourth case where 2x−2<02x-2 <02x−2<0 and x−1<0x-1 <0x−1<0. This implies that x<1x <1x<1. The equation becomes−(2x−2)−(x−1)=6-(2x-2) - (x-1) =6−(2x−2)−(x−1)=6

STEP 9

implify the equation from8−2x+2−x+=6-2x +2 - x + =6−2x+2−x+=6−3x+3=6-3x +3 =6−3x+3=6

STEP 10

olve the equation from9 for xxx−3x=3-3x =3−3x=3x=−x = -x=−Since x=−x = -x=− is in the range x<x <x<, it is a valid solution for this case.

STEP 11

The solutions to the given equation are the solutions from the first and fourth cases, which are x=3x =3x=3 and x=−x = -x=−. Therefore, the solution set is {−,3}\{-,3\}{−,3}.The correct answer is A. {−,3}\{-,3\}{−,3}.

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Solve $|2x-2| + |x-1| = 6$ graphically. Find the solution set. A. $\{-1,3\}$ B. The solution set is the set of real numbers. C. The solution set is the empty set.

Let's solve the first case where both $2x-2 \geq0$ and $x-1 \geq0$ . This implies that $x \geq1$ . The equation becomes $2x-2 + x-1 =6$

implify the equation from3 $3x -3 =6$

olve the equation from4 for $x$ $3x =9$ $x =3$ Since $x =3$ is in the range $x \geq1$ , it is a valid solution for this case.

Now, let's solve the second case where $2x-2 <0$ and $x-1 \geq0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

Now, let's solve the third case where $2x-2 \geq0$ and $x-1 <0$ . This implies that $1 \leq x <1$ , which is not possible. Therefore, this case has no solution.

Finally, let's solve the fourth case where $2x-2 <0$ and $x-1 <0$ . This implies that $x <1$ . The equation becomes $-(2x-2) - (x-1) =6$

implify the equation from8 $-2x +2 - x + =6$ $-3x +3 =6$

olve the equation from9 for $x$ $-3x =3$ $x = -$ Since $x = -$ is in the range $x <$ , it is a valid solution for this case.

The solutions to the given equation are the solutions from the first and fourth cases, which are $x =3$ and $x = -$ . Therefore, the solution set is $\{-,3\}$ .
The correct answer is A. $\{-,3\}$ .