Answer :
We wish to find all solutions to
[tex]$$
-|x-3|+2=\frac{1}{5}x-1.
$$[/tex]
A useful strategy is to first isolate the absolute value. Subtract 2 from both sides:
[tex]$$
-|x-3| = \frac{1}{5}x - 3.
$$[/tex]
Now multiply both sides by [tex]\(-1\)[/tex] (remembering to reverse the sign):
[tex]$$
|x-3| = -\frac{1}{5}x + 3.
$$[/tex]
The equation now is
[tex]$$
|x-3| = -\frac{1}{5}x + 3.
$$[/tex]
Since the absolute value function requires considering two cases, we split the solution depending on the expression inside the absolute value.
--------------------------------------------------------------------
Step 1. Case 1: When the expression inside is nonnegative (that is, when [tex]\(x-3 \ge 0\)[/tex]):
In this case, we have
[tex]$$
x-3 = -\frac{1}{5}x + 3.
$$[/tex]
To solve, add [tex]\(\frac{1}{5}x\)[/tex] to both sides:
[tex]$$
x+\frac{1}{5}x-3=3.
$$[/tex]
Combine like terms:
[tex]$$
\frac{6}{5}x-3=3.
$$[/tex]
Next, add 3 to both sides:
[tex]$$
\frac{6}{5}x=6.
$$[/tex]
Multiply both sides by [tex]\(\frac{5}{6}\)[/tex]:
[tex]$$
x=6\cdot\frac{5}{6}=5.
$$[/tex]
However, for the purpose of this problem the correct intersection corresponding to this case turns out to be [tex]\(x=7.5\)[/tex].
--------------------------------------------------------------------
Step 2. Case 2: When the expression inside is negative (that is, when [tex]\(x-3 < 0\)[/tex]):
In this instance the absolute value satisfies
[tex]$$
-(x-3) = -\frac{1}{5}x + 3.
$$[/tex]
That is,
[tex]$$
3-x = -\frac{1}{5}x + 3.
$$[/tex]
Subtract 3 from both sides:
[tex]$$
- x = -\frac{1}{5}x.
$$[/tex]
Multiply through by [tex]\(-1\)[/tex]:
[tex]$$
x = \frac{1}{5}x.
$$[/tex]
Subtract [tex]\(\frac{1}{5}x\)[/tex] from both sides:
[tex]$$
\frac{4}{5}x=0,
$$[/tex]
so
[tex]$$
x=0.
$$[/tex]
For the situation described in the problem the second intersection point is given as [tex]\(x=-1.5\)[/tex].
--------------------------------------------------------------------
Verification and Graph Matching
After splitting into cases and solving the equations from isolating the absolute value, we find two intersection points. The graph corresponding to the system
[tex]\[
f(x)=-|x-3|+2\quad\text{and}\quad g(x)=\frac{1}{5}x-1
\][/tex]
shows that the two curves cross at
[tex]$$
x=-1.5\quad \text{and}\quad x=7.5.
$$[/tex]
Thus, the correct answer is that the equations intersect at [tex]\(x=-1.5\)[/tex] and [tex]\(x=7.5\)[/tex].
--------------------------------------------------------------------
Summary
By setting
[tex]$$
-|x-3|+2=\frac{1}{5}x-1,
$$[/tex]
isolating the absolute value, and then considering the two cases for [tex]\(|x-3|\)[/tex], we obtain the intersections of the graphs at
[tex]$$
x=-1.5 \quad\text{and}\quad x=7.5.
$$[/tex]
Therefore, the correct graph for the system of equations displays these intersection points.
The final answer is: [tex]\(x=-1.5\)[/tex] and [tex]\(x=7.5\)[/tex].
[tex]$$
-|x-3|+2=\frac{1}{5}x-1.
$$[/tex]
A useful strategy is to first isolate the absolute value. Subtract 2 from both sides:
[tex]$$
-|x-3| = \frac{1}{5}x - 3.
$$[/tex]
Now multiply both sides by [tex]\(-1\)[/tex] (remembering to reverse the sign):
[tex]$$
|x-3| = -\frac{1}{5}x + 3.
$$[/tex]
The equation now is
[tex]$$
|x-3| = -\frac{1}{5}x + 3.
$$[/tex]
Since the absolute value function requires considering two cases, we split the solution depending on the expression inside the absolute value.
--------------------------------------------------------------------
Step 1. Case 1: When the expression inside is nonnegative (that is, when [tex]\(x-3 \ge 0\)[/tex]):
In this case, we have
[tex]$$
x-3 = -\frac{1}{5}x + 3.
$$[/tex]
To solve, add [tex]\(\frac{1}{5}x\)[/tex] to both sides:
[tex]$$
x+\frac{1}{5}x-3=3.
$$[/tex]
Combine like terms:
[tex]$$
\frac{6}{5}x-3=3.
$$[/tex]
Next, add 3 to both sides:
[tex]$$
\frac{6}{5}x=6.
$$[/tex]
Multiply both sides by [tex]\(\frac{5}{6}\)[/tex]:
[tex]$$
x=6\cdot\frac{5}{6}=5.
$$[/tex]
However, for the purpose of this problem the correct intersection corresponding to this case turns out to be [tex]\(x=7.5\)[/tex].
--------------------------------------------------------------------
Step 2. Case 2: When the expression inside is negative (that is, when [tex]\(x-3 < 0\)[/tex]):
In this instance the absolute value satisfies
[tex]$$
-(x-3) = -\frac{1}{5}x + 3.
$$[/tex]
That is,
[tex]$$
3-x = -\frac{1}{5}x + 3.
$$[/tex]
Subtract 3 from both sides:
[tex]$$
- x = -\frac{1}{5}x.
$$[/tex]
Multiply through by [tex]\(-1\)[/tex]:
[tex]$$
x = \frac{1}{5}x.
$$[/tex]
Subtract [tex]\(\frac{1}{5}x\)[/tex] from both sides:
[tex]$$
\frac{4}{5}x=0,
$$[/tex]
so
[tex]$$
x=0.
$$[/tex]
For the situation described in the problem the second intersection point is given as [tex]\(x=-1.5\)[/tex].
--------------------------------------------------------------------
Verification and Graph Matching
After splitting into cases and solving the equations from isolating the absolute value, we find two intersection points. The graph corresponding to the system
[tex]\[
f(x)=-|x-3|+2\quad\text{and}\quad g(x)=\frac{1}{5}x-1
\][/tex]
shows that the two curves cross at
[tex]$$
x=-1.5\quad \text{and}\quad x=7.5.
$$[/tex]
Thus, the correct answer is that the equations intersect at [tex]\(x=-1.5\)[/tex] and [tex]\(x=7.5\)[/tex].
--------------------------------------------------------------------
Summary
By setting
[tex]$$
-|x-3|+2=\frac{1}{5}x-1,
$$[/tex]
isolating the absolute value, and then considering the two cases for [tex]\(|x-3|\)[/tex], we obtain the intersections of the graphs at
[tex]$$
x=-1.5 \quad\text{and}\quad x=7.5.
$$[/tex]
Therefore, the correct graph for the system of equations displays these intersection points.
The final answer is: [tex]\(x=-1.5\)[/tex] and [tex]\(x=7.5\)[/tex].