Solve using elimination.
$\begin{array}{l}
6x + 3y = -18 \\
6x + 7y = -10
\end{array}$
( □ , □ )
We are given a system of two linear equations in two variables \(x\) and \(y\):
1) \(6x + 3y = -18\)
2) \(6x + 7y = -10\)
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously using the elimination method. The elimination method involves eliminating one of the variables by adding or subtracting the equations from each other. In this case, since both equations contain the term \(6x\), subtraction is a natural choice.
To eliminate the variable \(x\), we subtract the first equation from the second equation:
\((6x + 7y) - (6x + 3y) = -10 - (-18)\)
Simplify the left side:
\(6x - 6x + 7y - 3y = 4y\)
Simplify the right side:
\(-10 + 18 = 8\)
So, we have the simplified equation:
\(4y = 8\)
Dividing both sides by 4, we find the value of \(y\):
\(y = 2\)
Now that we have the value \(y = 2\), we can substitute it into the first equation to solve for \(x\):
\(6x + 3(2) = -18\)
Simplify the term \(3(2)\):
\(6x + 6 = -18\)
Subtract 6 from both sides:
\(6x = -18 - 6\)
\(6x = -24\)
Divide by 6:
\(x = -4\)
Checking the values in the second equation:
\(6(-4) + 7(2) = -24 + 14 = -10\), which confirms our result is correct.
The solution to the system is \(x = -4\) and \(y = 2\). The ordered pair is \((-4, 2)\).
Systems of linear equations are fundamental for modeling real-world constraints. For instance, imagine a company produces two types of goods, and you have specific constraints on the total cost and total quantity of production. By representing these constraints as equations, you can solve the system to determine exactly how many of each good to produce to meet your goals within budget. Similarly, this approach is used in physics to find the point where two trajectories intersect or in finance to balance investment portfolios. Understanding how to use the elimination method helps in efficiently solving for variables in such complex situations.