Solve the system by substitution.
[tex]\begin{array}{l}
y=4 x \\
y=-6 x-30\end{array}[/tex]
Answer (2)
Set the two equations equal to each other: 4 x = − 6 x − 30 .
Solve for x : 10 x = − 30 , so x = − 3 .
Substitute x = − 3 into the first equation to find y : y = 4 ( − 3 ) = − 12 .
The solution to the system of equations is ( − 3 , − 12 ) .
Explanation
Analyze the problem We are given a system of two equations:
y = 4 x y = − 6 x − 30
Our goal is to find the values of x and y that satisfy both equations simultaneously. We will use the substitution method to solve this system.
Set the equations equal Since both equations are already solved for y , we can set them equal to each other:
4 x = − 6 x − 30
Now, we solve for x .
Solve for x Add 6 x to both sides of the equation:
4 x + 6 x = − 6 x − 30 + 6 x 10 x = − 30
Divide both sides by 10:
x = 10 − 30 x = − 3
Solve for y Now that we have the value of x , we can substitute it back into either of the original equations to find the value of y . Let's use the first equation:
y = 4 x y = 4 ( − 3 ) y = − 12
State the solution So, the solution to the system of equations is x = − 3 and y = − 12 . We can write this as an ordered pair: ( − 3 , − 12 ) .
Examples
Systems of equations are used in various real-world applications, 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, suppose a company wants to know how many units of a product they need to sell to cover their costs. They can set up a system of equations representing their revenue and costs and solve for the quantity at which revenue equals costs. This helps them make informed decisions about pricing and production levels.
To solve the system of equations, we set them equal to find x , yielding x = − 3 . Substituting x = − 3 back into the first equation gives y = − 12 . Therefore, the solution is the ordered pair ( − 3 , − 12 ) .
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