Which system of equations has a solution of approximately $(-0.3,1.4)$?
A. $2 x-y=-2$ and $22 x+10 y=7$
B. $3 x+2 y=5$ and $4 x-y=2$
C. $4 x-y=2$ and $22 x+10 y=7$
Answer (2)
The system of equations that has a solution of approximately (-0.3, 1.4) is System A, which consists of the equations 2 x − y = − 2 and 22 x + 10 y = 7 . This was determined by substituting the values into each system and checking the results. Only the first system produced results that were approximately valid for both equations.
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Substitute x = − 0.3 and y = 1.4 into the first system of equations: 2 x − y = − 2 and 22 x + 10 y = 7 . The result is approximately satisfied.
Substitute x = − 0.3 and y = 1.4 into the second system of equations: 3 x + 2 y = 5 and 4 x − y = 2 . The result is not approximately satisfied.
Substitute x = − 0.3 and y = 1.4 into the third system of equations: 4 x − y = 2 and 22 x + 10 y = 7 . The result is not approximately satisfied.
Therefore, the first system of equations has a solution of approximately ( − 0.3 , 1.4 ) . 2 x − y = − 2 and 22 x + 10 y = 7
Explanation
Problem Analysis We are given three systems of equations and a solution of approximately ( − 0.3 , 1.4 ) . Our task is to determine which system of equations has the given solution.
Substitution Method Let's substitute x = − 0.3 and y = 1.4 into each system of equations to see which one is approximately satisfied.
Checking System 1 System 1:
The equations are 2 x − y = − 2 and 22 x + 10 y = 7 .
Substituting the values, we get:
2 ( − 0.3 ) − 1.4 = − 0.6 − 1.4 = − 2 . This equation is satisfied.
22 ( − 0.3 ) + 10 ( 1.4 ) = − 6.6 + 14 = 7.4 . This is approximately equal to 7 , so this equation is approximately satisfied.
Checking System 2 System 2:
The equations are 3 x + 2 y = 5 and 4 x − y = 2 .
Substituting the values, we get:
3 ( − 0.3 ) + 2 ( 1.4 ) = − 0.9 + 2.8 = 1.9 . This is not approximately equal to 5 .
4 ( − 0.3 ) − 1.4 = − 1.2 − 1.4 = − 2.6 . This is not approximately equal to 2 .
Checking System 3 System 3:
The equations are 4 x − y = 2 and 22 x + 10 y = 7 .
Substituting the values, we get:
4 ( − 0.3 ) − 1.4 = − 1.2 − 1.4 = − 2.6 . This is not approximately equal to 2 .
22 ( − 0.3 ) + 10 ( 1.4 ) = − 6.6 + 14 = 7.4 . This is approximately equal to 7 , so this equation is approximately satisfied.
Conclusion From the above analysis, only System 1 has both equations approximately satisfied by the given solution ( − 0.3 , 1.4 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the optimal mix of products to maximize profit, calculating the forces acting on a structure, or modeling the flow of traffic in a city. In economics, systems of equations can be used to model the supply and demand of goods and services. By finding the solution to the system, economists can determine the equilibrium price and quantity of a product.
