Create an equivalent system of equations using the system and the first equation.
[tex]
\begin{array}{l}
x+4 y=8 \\
4 x+y=2
\end{array}
[/tex]
[tex]
\begin{array}{l}
x+4 y=13 \\
5 x+y=2
\end{array}
[/tex]
[tex]x+4 y=8[/tex]
[tex]5 x+5 y=2[/tex]
[tex]
\begin{array}{l}
x+4 y=8 \\
4 x+5 y=10
\end{array}
[/tex]
[tex]
\begin{array}{l}
x+4 y=8 \\
5 x+5 y=10
\end{array}
[/tex]
Answer (1)
Check each system to see if it is equivalent to the original system.
System 1 is not equivalent because it has x + 4 y = 13 instead of x + 4 y = 8 .
Systems 2, 3, and 4 have x + 4 y = 8 , so we need to check if the second equation is a linear combination of the original equations.
Systems 3 and 4 are equivalent to the original system. x + 4 y = 8 4 x + 5 y = 10 and x + 4 y = 8 5 x + 5 y = 10
Explanation
Understanding the Problem We are given a system of two linear equations:
x + 4 y = 8 4 x + y = 2
Our goal is to identify which of the provided systems is equivalent to the original system, meaning they have the same solution set, and one equation in the new system must be x + 4 y = 8 .
Analyzing System 1 and System 2 Let's analyze each of the provided systems:
System 1:
x + 4 y = 13 5 x + y = 2
This system has x + 4 y = 13 which is different from x + 4 y = 8 , so it is not equivalent.
System 2:
x + 4 y = 8 5 x + 5 y = 2
Let's check if the second equation is a linear combination of the original equations. We want to see if we can obtain 5 x + 5 y by multiplying the first equation by a and the second equation by b and adding them:
a ( x + 4 y ) + b ( 4 x + y ) = ( a + 4 b ) x + ( 4 a + b ) y
We want a + 4 b = 5 and 4 a + b = 5 . Solving this system gives a = 1 and b = 1 . So,
1 ( x + 4 y ) + 1 ( 4 x + y ) = 5 x + 5 y = 1 ( 8 ) + 1 ( 2 ) = 10
Thus, the second equation should be 5 x + 5 y = 10 , but it is given as 5 x + 5 y = 2 , so this system is not equivalent.
Analyzing System 3 and System 4 System 3:
x + 4 y = 8 4 x + 5 y = 10
Let's check if the second equation is a linear combination of the original equations. We want to obtain 4 x + 5 y by multiplying the first equation by a and the second equation by b and adding them:
a ( x + 4 y ) + b ( 4 x + y ) = ( a + 4 b ) x + ( 4 a + b ) y
We want a + 4 b = 4 and 4 a + b = 5 . Solving this system gives 16 b − b = 16 − 5 , so 15 b = 11 , and b = 15 11 . Then a = 4 − 4 ( 15 11 ) = 15 60 − 44 = 15 16 . So,
( 15 16 ) ( x + 4 y ) + ( 15 11 ) ( 4 x + y ) = ( 15 16 ) x + ( 15 64 ) y + ( 15 44 ) x + ( 15 11 ) y = ( 15 60 ) x + ( 15 75 ) y = 4 x + 5 y
And
( 15 16 ) ( 8 ) + ( 15 11 ) ( 2 ) = ( 15 128 + 22 ) = 15 150 = 10
Thus, the second equation is 4 x + 5 y = 10 , so this system is equivalent.
System 4:
x + 4 y = 8 5 x + 5 y = 10
As shown in the analysis of System 2, 5 x + 5 y = 1 ( x + 4 y ) + 1 ( 4 x + y ) = 1 ( 8 ) + 1 ( 2 ) = 10 . Thus, the second equation is 5 x + 5 y = 10 , so this system is equivalent.
Final Answer Therefore, the equivalent systems are:
x + 4 y = 8 4 x + 5 y = 10
and
x + 4 y = 8 5 x + 5 y = 10
Examples
Understanding equivalent systems of equations is crucial in various fields, such as electrical engineering, where you might need to simplify a circuit's equations to find unknown currents and voltages. By manipulating the equations while preserving their solutions, engineers can analyze complex circuits more efficiently. This technique is also used in economics to model market equilibrium, where multiple equations represent supply and demand relationships. Finding an equivalent, simpler system can help economists predict market behavior and make informed policy recommendations.
