Use substitution to solve the system:
$\begin{array}{l}
y=3.2 x+1.9 \\
2.3 x+2 y=17.72
\end{array}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
\[ \begin{cases} y = 3.2x + 1.9 \quad (1) \ 2.3x + 2y = 17.72 \quad (2) \end{cases} \]
The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. The method of substitution is suitable here because the first equation already expresses \(y\) in terms of \(x\).
To solve the system, we substitute the expression for \(y\) given by equation (1) into equation (2):
\[2.3x + 2(3.2x + 1.9) = 17.72\]
Next, we distribute the \(2\) across the terms inside the parentheses:
\[2.3x + (2 \times 3.2x) + (2 \times 1.9) = 17.72\]
\[2.3x + 6.4x + 3.8 = 17.72\]
Now, combine the like terms involving \(x\):
\[(2.3 + 6.4)x + 3.8 = 17.72\]
\[8.7x + 3.8 = 17.72\]
We isolate \(x\) by subtracting \(3.8\) from both sides of the equation:
\[8.7x = 17.72 - 3.8\]
\[8.7x = 13.92\]
Divide by \(8.7\) to solve for \(x\):
\[x = \frac{13.92}{8.7} = 1.6\]
Now that we have the value of \(x\), we substitute it back into equation (1) to find \(y\):
\[y = 3.2(1.6) + 1.9\]
\[y = 5.12 + 1.9 = 7.02\]
The solution to the system is the ordered pair \((x, y) = (1.6, 7.02)\).
In real-life scenarios, systems of equations are used to compare costs. For example, if you are comparing two internet plans, one might have a fixed monthly fee plus a variable cost per gigabyte, while another has a different fixed fee and a different per-gigabyte rate. By setting these expressions equal to each other (or substituting one into another), you can find the exact usage amount where both plans cost the same, helping you make the best financial decision.