Solve the system of equations.
[tex]
\begin{array}{l}
y=6 x+13 \\
y=2 x+1
\end{array}
[/tex]
A. (5,3)
B. (-3,-5)
C. (3,-5)
D. No solution
Answer (1)
Set the two equations equal to each other: 6 x + 13 = 2 x + 1 .
Solve for x : 4 x = − 12 , so x = − 3 .
Substitute x = − 3 into one of the equations to solve for y : y = 2 ( − 3 ) + 1 = − 5 .
The solution to the system of equations is ( − 3 , − 5 ) .
Explanation
Analyze the problem We are given a system of two linear equations:
Equation 1: y = 6 x + 13
Equation 2: y = 2 x + 1
Our goal is to find the values of x and y that satisfy both equations simultaneously.
Set equations equal Since both equations are already solved for y , we can set them equal to each other:
6 x + 13 = 2 x + 1
Solve for x Now, let's solve for x . Subtract 2 x from both sides:
6 x − 2 x + 13 = 2 x − 2 x + 1 4 x + 13 = 1
Subtract 13 from both sides:
4 x + 13 − 13 = 1 − 13 4 x = − 12
Divide by 4:
x = 4 − 12 x = − 3
Solve for y Now that we have the value of x , we can substitute it back into either equation to find the value of y . Let's use the second equation:
y = 2 x + 1 y = 2 ( − 3 ) + 1 y = − 6 + 1 y = − 5
State the solution So, the solution to the system of equations is x = − 3 and y = − 5 . We can write this as an ordered pair: ( − 3 , − 5 ) .
Comparing this to the given options, we see that the correct answer is option b.
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling supply and demand in economics. For example, a company might use a system of equations to find the number of units they need to sell to cover their costs and start making a profit. By solving the system, they can determine the exact point where their revenue equals their expenses, providing valuable insights for decision-making.
