Solve the simultaneous equations:
$\begin{aligned}
5 x+2 y & =21 \\
x+3 y & =12
\end{aligned}$
We are given the following system of linear equations:
(1) \(5x + 2y = 21\)
(2) \(x + 3y = 12\)
These two equations represent two lines in a 2D Cartesian plane. Our objective is to find the point \((x, y)\) where these two lines intersect, which represents the solution satisfying both equations simultaneously. To solve this, we can use the elimination method, which involves manipulating the equations to cancel out one of the variables.
To eliminate \(x\), we can express equation (2) in a form that makes it easy to subtract from equation (1). Let's multiply the entire second equation by 5:
\(5 \times (x + 3y) = 5 \times 12\)
\(5x + 15y = 60\)
Now we have a new version of the second equation. We can subtract equation (1) from this new equation:
\((5x + 15y) - (5x + 2y) = 60 - 21\)
Simplifying the left side:
\(5x - 5x + 15y - 2y = 13y\)
Simplifying the right side:
\(60 - 21 = 39\)
Thus, we are left with \(13y = 39\). Dividing both sides by 13 gives \(y = 3\).
Now that we have found \(y = 3\), we substitute this value back into one of the original equations to find \(x\). Using equation (2):
\(x + 3(3) = 12\)
\(x + 9 = 12\)
Subtract 9 from both sides:
\(x = 12 - 9\)
\(x = 3\)
We have found that \(x = 3\) and \(y = 3\). We can verify this with equation (1): \(5(3) + 2(3) = 15 + 6 = 21\), which matches the original equation. Thus, the solution is \(\boxed{x=3, y=3}\).
In real-life scenarios, systems of equations are used to balance budgets or optimize resources. For example, if you buy 5 notebooks and 2 pens for a total of \$21, and a friend buys 1 notebook and 3 pens for \$12, you can use these equations to determine the price of a single notebook and a single pen. By solving the system \(5n + 2p = 21\) and \(n + 3p = 12\), you find that both items cost \(\$3\) each.