Solve the system of equations and enter the solution as an ordered pair in the box below.
[tex]
\begin{array}{l}
3 y=2 x+1 \\
5 y=2 x+3
\end{array}
[/tex]
Answer (1)
Multiply the first equation by 5 and the second equation by 3 to prepare for elimination.
Subtract the modified equations to eliminate y and solve for x , resulting in x = 1 .
Substitute x = 1 into one of the original equations to solve for y , resulting in y = 1 .
Express the solution as an ordered pair: ( 1 , 1 ) .
Explanation
Analyze the problem We are given a system of two linear equations:
{ 3 y = 2 x + 1 5 y = 2 x + 3
Our goal is to find the values of x and y that satisfy both equations. We will use the elimination method to solve this system.
Prepare for elimination To eliminate x , we can multiply the first equation by 5 and the second equation by 3:
{ 5 ( 3 y ) = 5 ( 2 x + 1 ) 3 ( 5 y ) = 3 ( 2 x + 3 )
This simplifies to:
{ 15 y = 10 x + 5 15 y = 6 x + 9
Eliminate y Now, subtract the second equation from the first equation to eliminate y :
( 15 y − 15 y ) = ( 10 x + 5 ) − ( 6 x + 9 )
This simplifies to:
0 = 4 x − 4
Solve for x Solve for x :
4 x = 4
x = 1
Solve for y Substitute the value of x into either of the original equations to solve for y . Let's use the first equation:
3 y = 2 ( 1 ) + 1
3 y = 2 + 1
3 y = 3
y = 1
State the solution The solution to the system of equations is x = 1 and y = 1 . Therefore, the ordered pair is ( 1 , 1 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, a company might model its costs and revenues as linear equations and solve the system to find the production level at which costs equal revenues, indicating profitability. Understanding how to solve systems of equations is crucial for making informed business decisions and optimizing resource allocation.
