Solve using elimination.
$\begin{array}{l}
-7 x+4 y=5 \\
-7 x+7 y=14
\end{array}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
1) \(-7x + 4y = 5\)
2) \(-7x + 7y = 14\)
Our goal is to solve this system using the method of elimination. The elimination method involves manipulating the equations so that one of the variables is removed when we add or subtract the equations from each other. In this system, we observe that both equations contain the term \(-7x\). This makes the system particularly well-suited for elimination by subtraction.
To eliminate the variable \(x\), we subtract the first equation from the second equation:
\((-7x + 7y) - (-7x + 4y) = 14 - 5\)
Distributing the negative sign:
\(-7x + 7y + 7x - 4y = 14 - 5\)
Combining like terms:
\((-7x + 7x) + (7y - 4y) = 9\)
\(3y = 9\)
Dividing by 3:
\(y = \frac{9}{3} = 3\)
We have successfully found the value of \(y\).
Now that we have \(y = 3\), we can substitute this value back into either of the original equations to solve for \(x\). Let's use the first equation, \(-7x + 4y = 5\):
\(-7x + 4(3) = 5\)
\(-7x + 12 = 5\)
Subtract 12 from both sides:
\(-7x = 5 - 12\)
\(-7x = -7\)
Divide by -7:
\(x = \frac{-7}{-7} = 1\)
Thus, the solution to the system is \(x = 1\) and \(y = 3\). We can verify this by checking the second equation:
\(-7(1) + 7(3) = -7 + 21 = 14\).
This confirms that our solution is correct. The final solution is the ordered pair \((1, 3)\).
Systems of equations are used whenever you need to find two unknown values that satisfy two different conditions simultaneously. For example, if you are buying school supplies, a shop might charge a fixed price for a bundle of items and a different price if you add extra units. If you know the total cost for two different quantities, you can use a system of equations to determine the unit price of each item or the base fee for a package. Mastering this technique helps in planning budgets, balancing chemical equations in science, and optimizing logistics in business.