Which point is a solution to the system of linear equations?
[tex]
\begin{array}{l}
y=-x+3 \
2 x-y=6
\end{array}
[/tex]
A. (3,0)
B. (3,-1)
C. (0,3)
D. (-1,3)
Answer (2)
Substitute each point into both equations.
Check if both equations are satisfied.
Point ( 3 , 0 ) : 0 = − 3 + 3 and 2 ( 3 ) − 0 = 6 , so it is a solution.
The solution to the system of linear equations is ( 3 , 0 ) .
Explanation
Understanding the Problem We are given a system of two linear equations:
y = − x + 3
2 x − y = 6
We need to determine which of the given points, ( 3 , 0 ) , ( 3 , − 1 ) , ( 0 , 3 ) , and ( − 1 , 3 ) , is a solution to this system. A point is a solution if it satisfies both equations simultaneously. We will test each point by substituting its x and y coordinates into both equations.
Testing Point (3, 0) Let's test the point ( 3 , 0 ) .
For the first equation, y = − x + 3 , we substitute x = 3 and y = 0 :
0 = − 3 + 3
0 = 0 . This equation is satisfied.
For the second equation, 2 x − y = 6 , we substitute x = 3 and y = 0 :
2 ( 3 ) − 0 = 6
6 = 6 . This equation is also satisfied.
Since ( 3 , 0 ) satisfies both equations, it is a solution to the system.
Testing Point (3, -1) Let's test the point ( 3 , − 1 ) .
For the first equation, y = − x + 3 , we substitute x = 3 and y = − 1 :
− 1 = − 3 + 3
− 1 = 0 . This equation is not satisfied.
Since ( 3 , − 1 ) does not satisfy the first equation, it is not a solution to the system.
Testing Point (0, 3) Let's test the point ( 0 , 3 ) .
For the first equation, y = − x + 3 , we substitute x = 0 and y = 3 :
3 = − 0 + 3
3 = 3 . This equation is satisfied.
For the second equation, 2 x − y = 6 , we substitute x = 0 and y = 3 :
2 ( 0 ) − 3 = 6
− 3 = 6 . This equation is not satisfied.
Since ( 0 , 3 ) does not satisfy the second equation, it is not a solution to the system.
Testing Point (-1, 3) Let's test the point ( − 1 , 3 ) .
For the first equation, y = − x + 3 , we substitute x = − 1 and y = 3 :
3 = − ( − 1 ) + 3
3 = 1 + 3
3 = 4 . This equation is not satisfied.
Since ( − 1 , 3 ) does not satisfy the first equation, it is not a solution to the system.
Conclusion Only the point ( 3 , 0 ) satisfies both equations in the system. Therefore, ( 3 , 0 ) is the solution to the system of equations.
Examples
Systems of linear equations are used in various real-life applications, such as determining the break-even point for a business, calculating the optimal mix of products to maximize profit, or even in network analysis to determine traffic flow. For instance, imagine a small business that sells two types of products. By setting up a system of equations that represents the cost and revenue for each product, the business owner can find the exact number of units they need to sell to break even, ensuring they don't operate at a loss. This blend of algebra and practical application highlights the importance of understanding linear systems.
The point ( 3 , 0 ) is the only one that satisfies both equations in the system. It fulfills the requirements for both equations, making it the correct solution. Thus, the chosen option is A. (3, 0).
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