To solve this problem, let's start by setting up our equations based on the given information. We know that the total charge is partly constant and partly varies with the number of units consumed. Let:
- [tex]\( a \)[/tex] be the constant part of the charge.
- [tex]\( b \)[/tex] be the variable charge per unit of electricity.
So for a consumption of [tex]\( r \)[/tex] units, the charge [tex]\( \varepsilon \)[/tex] can be expressed as:
[tex]\[ \varepsilon = a + b \cdot r \][/tex]
We are given two scenarios from which we can create a system of linear equations:
1. For 163 units, the total charge is [tex]$453.80. So:
\[ a + b \cdot 163 = 453.80 \]
2. For 42 units, the total charge is $[/tex]139.20. So:
[tex]\[ a + b \cdot 42 = 139.20 \][/tex]
Now, we have the following system of equations:
[tex]\[ \begin{cases}
a + 163b = 453.80 \\
a + 42b = 139.20
\end{cases} \][/tex]
To find the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we can solve these equations simultaneously.
### Step 1: Eliminate [tex]\( a \)[/tex] by Subtraction
First, subtract the second equation from the first:
[tex]\[ (a + 163b) - (a + 42b) = 453.80 - 139.20 \][/tex]
[tex]\[ a + 163b - a - 42b = 314.60 \][/tex]
[tex]\[ 121b = 314.60 \][/tex]
### Step 2: Solve for [tex]\( b \)[/tex]
[tex]\[ b = \frac{314.60}{121} \][/tex]
[tex]\[ b = 2.60 \][/tex]
So, the charge per unit of electricity is [tex]\( \$2.60 \)[/tex].
### Step 3: Solve for [tex]\( a \)[/tex]
Substitute [tex]\( b = 2.60 \)[/tex] back into one of the original equations. Using the second equation:
[tex]\[ a + 42 \cdot 2.60 = 139.20 \][/tex]
[tex]\[ a + 109.20 = 139.20 \][/tex]
[tex]\[ a = 139.20 - 109.20 \][/tex]
[tex]\[ a = 30.00 \][/tex]
So, the constant part of the charge is [tex]\( \$30.00 \)[/tex].
### Conclusion
Thus, the charge per unit of electricity is [tex]\( \$2.60 \)[/tex].
