Solve the simultaneous equations:
$\begin{aligned}
6x + y &= -8 \\
10x + y &= -20
\end{aligned}$
$x = \quad y =$
We are presented with a system of two linear equations with two variables, \(x\) and \(y\):
Our objective is to determine the values of \(x\) and \(y\) that satisfy both equations simultaneously. This is a classic system of linear equations, and we can solve it effectively using the method of elimination, as the coefficient of \(y\) is the same in both equations.
To eliminate the variable \(y\), we can subtract the first equation from the second equation. This will leave us with a single equation containing only one variable, \(x\).
Subtracting the equations:
\((10x + y) - (6x + y) = -20 - (-8)\)
Simplifying the left side:
\(10x - 6x + y - y = 4x\)
Simplifying the right side:
\(-20 + 8 = -12\)
This gives us the simple linear equation \(4x = -12\).
Now, we solve for \(x\) by dividing both sides of the equation by 4:
\(x = \frac{-12}{4}\)
\(x = -3\)
With the value of \(x = -3\) known, we can substitute it back into the first equation to solve for \(y\):
\(6(-3) + y = -8\)
\(-18 + y = -8\)
Adding 18 to both sides:
\(y = -8 + 18\)
\(y = 10\)
Thus, the solution to the system is \(x = -3\) and \(y = 10\). We can check this in the second equation as well: \(10(-3) + 10 = -30 + 10 = -20\), which confirms our result is correct.
Systems of linear equations are frequently used to solve real-world problems involving balancing two related quantities. For example, if you are analyzing the cost of two different service plans, one might be represented as \(6x + y = -8\) and another as \(10x + y = -20\), where \(x\) could represent the number of months and \(y\) a fixed discount or fee. By solving the system, you can identify the exact point where both plans yield the same result, helping in making informed financial decisions.