The solution to the system of equations below is (-2, -1):
[tex]\begin{array}{l}
2 x-3 y=-1 \\
11 x-9 y=-13
\end{array}[/tex]
When the first equation is multiplied by -3, the sum of the two equations is equivalent to [tex]$5 x=-10$[/tex].
Which system of equations will also have a solution of (-2, -1)?
A. [tex]\begin{array}{r}
5 x=-10 \\
11 x-9 y=-13
\end{array}[/tex]
B. [tex]\begin{array}{c}
2 x+9 y=-1 \\
11 x-9 y=-13
\end{array}[/tex]
C. [tex]\begin{array}{c}
-6 x+9 y=-1 \\
11 x-9 y=-13
\end{array}[/tex]
Answer (1)
Substitute the solution ( − 2 , − 1 ) into System 1: 5 x = − 10 and 11 x − 9 y = − 13 . Both equations are satisfied.
Substitute the solution ( − 2 , − 1 ) into System 2: 2 x + 9 y = − 1 and 11 x − 9 y = − 13 . The first equation is not satisfied.
Substitute the solution ( − 2 , − 1 ) into System 3: − 6 x + 9 y = − 1 and 11 x − 9 y = − 13 . The first equation is not satisfied.
Therefore, the system of equations that also has a solution of ( − 2 , − 1 ) is System 1: 5 x = − 10 11 x − 9 y = − 13
Explanation
Understanding the Problem We are given that the solution to the system of equations
2 x − 3 y = − 1 11 x − 9 y = − 13
is ( − 2 , − 1 ) . We need to determine which of the given systems of equations also has the solution ( − 2 , − 1 ) .
Analyzing System 1 Let's analyze each system of equations:
System 1:
5 x = − 10 11 x − 9 y = − 13
Substitute x = − 2 into the first equation: 5 ( − 2 ) = − 10 , which is true. Substitute x = − 2 and y = − 1 into the second equation: 11 ( − 2 ) − 9 ( − 1 ) = − 22 + 9 = − 13 , which is also true. Therefore, ( − 2 , − 1 ) is a solution to System 1.
Analyzing System 2 System 2:
2 x + 9 y = − 1 11 x − 9 y = − 13
Substitute x = − 2 and y = − 1 into the first equation: 2 ( − 2 ) + 9 ( − 1 ) = − 4 − 9 = − 13 , which is not equal to − 1 . Therefore, ( − 2 , − 1 ) is not a solution to System 2.
Analyzing System 3 System 3:
− 6 x + 9 y = − 1 11 x − 9 y = − 13
Substitute x = − 2 and y = − 1 into the first equation: − 6 ( − 2 ) + 9 ( − 1 ) = 12 − 9 = 3 , which is not equal to − 1 . Therefore, ( − 2 , − 1 ) is not a solution to System 3.
Conclusion Since only System 1 has ( − 2 , − 1 ) as a solution, the answer is System 1.
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, if a company has fixed costs and variable costs, and they sell a product at a certain price, a system of equations can be set up to find the number of units that need to be sold to cover all costs and start making a profit. Similarly, systems of equations are used in physics to solve problems involving multiple forces or velocities.
