Solve the system of equations.
[tex]
\begin{array}{l}
y=7 x+43 \\
y=2 x+13
\end{array}
[/tex]
A. (-6,1)
B. (6,-1)
C. (6,1)
D. No solution
Answer (1)
Set the two equations equal to each other: 7 x + 43 = 2 x + 13 .
Solve for x : 5 x = − 30 , so x = − 6 .
Substitute x = − 6 into one of the equations to solve for y : y = 2 ( − 6 ) + 13 = 1 .
The solution to the system of equations is ( − 6 , 1 ) .
Explanation
Analyze the problem We are given a system of two linear equations:
y = 7 x + 43 y = 2 x + 13
Our goal is to find the values of x and y that satisfy both equations.
Set the equations equal Since both equations are equal to y , we can set them equal to each other:
7 x + 43 = 2 x + 13
Solve for x Now, we solve for x . Subtract 2 x from both sides:
7 x − 2 x + 43 = 2 x − 2 x + 13 5 x + 43 = 13
Subtract 43 from both sides:
5 x + 43 − 43 = 13 − 43 5 x = − 30
Divide by 5:
x = 5 − 30 x = − 6
Solve for y Now that we have the value of x , we can substitute it into either equation to find the value of y . Let's use the second equation:
y = 2 x + 13 y = 2 ( − 6 ) + 13 y = − 12 + 13 y = 1
State the solution So the solution to the system of equations is x = − 6 and y = 1 . Therefore, the solution is ( − 6 , 1 ) .
Examples
Systems of equations are used in many real-world applications, such as determining the break-even point for a business. For example, a company might use a system of equations to model its costs and revenues, and then solve the system to find the level of sales needed to cover all costs. Another example is in electrical engineering, where systems of equations are used to analyze circuits and determine the currents and voltages at different points in the circuit. These models help engineers design efficient and reliable electrical systems.
