To solve the system of linear equations
[tex]\[
\begin{cases}
2e + 3p = 190 \\
5e + 2p = 290
\end{cases}
\][/tex]
we can use the method of elimination or substitution. Here, let's use the elimination method for clarity.
### Step-by-Step Solution
1. Write down the system of equations:
[tex]\[
\begin{cases}
2e + 3p = 190 \quad \text{(Equation 1)} \\
5e + 2p = 290 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
2. Eliminate one variable (let's eliminate [tex]\( p \)[/tex]):
We'll use a technique to make the coefficients of [tex]\( p \)[/tex] in both equations the same, thereby allowing us to eliminate [tex]\( p \)[/tex].
Multiply Equation 1 by 2:
[tex]\[
2(2e + 3p) = 2(190) \implies 4e + 6p = 380 \quad \text{(Equation 3)}
\][/tex]
Multiply Equation 2 by 3:
[tex]\[
3(5e + 2p) = 3(290) \implies 15e + 6p = 870 \quad \text{(Equation 4)}
\][/tex]
3. Subtract Equation 3 from Equation 4:
[tex]\[
(15e + 6p) - (4e + 6p) = 870 - 380
\][/tex]
Simplify:
[tex]\[
15e + 6p - 4e - 6p = 490
\][/tex]
[tex]\[
11e = 490
\][/tex]
[tex]\[
e = \frac{490}{11} = 44.54545454545455
\][/tex]
4. Substitute [tex]\( e \)[/tex] back into one of the original equations to find [tex]\( p \)[/tex]:
Use Equation 1:
[tex]\[
2e + 3p = 190
\][/tex]
Substitute [tex]\( e = 44.54545454545455 \)[/tex]:
[tex]\[
2(44.54545454545455) + 3p = 190
\][/tex]
[tex]\[
89.0909090909091 + 3p = 190
\][/tex]
Subtract [tex]\( 89.0909090909091 \)[/tex] from both sides:
[tex]\[
3p = 190 - 89.0909090909091
\][/tex]
[tex]\[
3p = 100.9090909090909
\][/tex]
[tex]\[
p = \frac{100.9090909090909}{3} = 33.63636363636363
\][/tex]
So, the solution to the system of equations is:
[tex]\[
e = 44.54545454545455, \quad p = 33.63636363636363
\][/tex]
