- The system of linear equations is solved by the method of elimination.
- Multiplying the second equation by 3 and adding it to the first equation eliminates \(y\), resulting in \(13x = 30\).
- Solving for \(x\) gives \(x = \frac{30}{13}\), and substituting this into the second equation yields \(y = \frac{70}{13}\).
- The correct choice is Option A, and the solution set is \(\boxed{\left\{\left(\frac{30}{13}, \frac{70}{13}\right)\right\}}\).
Explanation
- Analysis of the System of Equations
We are given a system of two linear equations in two variables:
\[7x - 3y = 0\]
\[2x + y = 10\]
Our goal is to find the ordered pair \((x, y)\) that satisfies both equations. Since we have a system of two linear equations with two variables, we expect a unique solution unless the lines are parallel or coincident. We can use the method of elimination to solve this system efficiently.
- Eliminating the Variable \(y\)
To eliminate \(y\), we can make the coefficients of \(y\) in both equations opposites. The second equation is \(2x + y = 10\). If we multiply this entire equation by \(3\), we get:
\[3(2x + y) = 3(10)\]
\[6x + 3y = 30\]
Now our system is:
\[7x - 3y = 0\]
\[6x + 3y = 30\]
Adding these two equations together eliminates \(y\):
\[(7x - 3y) + (6x + 3y) = 0 + 30\]
\[13x = 30\]
Dividing by \(13\), we find the value of \(x\):
\[x = \frac{30}{13}\]
- Solving for \(x\) and \(y\)
Now that we have \(x = \frac{30}{13}\), we can substitute this value back into one of the original equations to find \(y\). Using the second equation, \(2x + y = 10\):
\[2\left(\frac{30}{13}\right) + y = 10\]
\[\frac{60}{13} + y = 10\]
To solve for \(y\), subtract \(\frac{60}{13}\) from both sides:
\[y = 10 - \frac{60}{13}\]
Since \(10 = \frac{130}{13}\), we have:
\[y = \frac{130}{13} - \frac{60}{13} = \frac{70}{13}\]
Thus, the solution is the ordered pair \(\left(\frac{30}{13}, \frac{70}{13}\right)\). This confirms that Option A is the correct choice.
Examples
Solving systems of linear equations is a fundamental skill used in fields like economics and engineering to find equilibrium points. For instance, if you have a supply function and a demand function for a product, the intersection point (the solution to the system) tells you the equilibrium price and quantity where supply meets demand. In this case, finding the intersection of two linear paths is essential for resource planning, logistics, and balancing various constraints in real-world projects.