What is the solution to this system of equations?
$\begin{array}{l}
3 x+2 y=10 \\
2 x+4 y=4
\end{array}$
A. $(4,-1)$
B. $(4,1)$
C. $(1,2)$
D. $(-4,3)$
Answer (1)
Substitute each given point into the system of equations.
Check if both equations are satisfied for each point.
The point (4, -1) satisfies both equations: 3 ( 4 ) + 2 ( − 1 ) = 10 and 2 ( 4 ) + 4 ( − 1 ) = 4 .
Therefore, the solution to the system of equations is ( 4 , − 1 ) .
Explanation
Problem Analysis We are given a system of two linear equations with two variables, x and y :
3 x + 2 y = 10 2 x + 4 y = 4
We are also given four possible solutions: ( 4 , − 1 ) , ( 4 , 1 ) , ( 1 , 2 ) , and ( − 4 , 3 ) . Our goal is to determine which, if any, of these pairs ( x , y ) satisfies both equations.
Checking the Solutions To check if a given point ( x , y ) is a solution to the system of equations, we substitute the values of x and y into both equations. If both equations are true, then the point is a solution.
Testing (4, -1) Let's test the first solution, ( 4 , − 1 ) .
For the first equation: 3 x + 2 y = 3 ( 4 ) + 2 ( − 1 ) = 12 − 2 = 10 The first equation is satisfied.
For the second equation: 2 x + 4 y = 2 ( 4 ) + 4 ( − 1 ) = 8 − 4 = 4 The second equation is also satisfied. Therefore, ( 4 , − 1 ) is a solution to the system of equations.
Checking Other Options Since we found a solution, we can stop here. However, for completeness, let's check the other options as well.
Testing ( 4 , 1 ) :
For the first equation: 3 x + 2 y = 3 ( 4 ) + 2 ( 1 ) = 12 + 2 = 14 = 10 Since the first equation is not satisfied, ( 4 , 1 ) is not a solution.
Testing ( 1 , 2 ) :
For the first equation: 3 x + 2 y = 3 ( 1 ) + 2 ( 2 ) = 3 + 4 = 7 = 10 Since the first equation is not satisfied, ( 1 , 2 ) is not a solution.
Testing ( − 4 , 3 ) :
For the first equation: 3 x + 2 y = 3 ( − 4 ) + 2 ( 3 ) = − 12 + 6 = − 6 = 10 Since the first equation is not satisfied, ( − 4 , 3 ) is not a solution.
Final Answer Therefore, the only solution among the given options is ( 4 , − 1 ) .
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business. For example, suppose a company produces and sells a product. The cost to produce the product is represented by a linear equation, and the revenue from selling the product is also represented by a linear equation. The point where the cost and revenue equations intersect (the solution to the system of equations) is the break-even point, where the company neither makes a profit nor incurs a loss. Finding this point helps the company make informed decisions about pricing and production levels. Another example is calculating mixtures of different concentrations to achieve a desired concentration.
