Express y in terms of x using the first equation: y = 1 − x .
Substitute this expression into the second equation: 1 − x = x 2 − 11 .
Rearrange into a quadratic equation: x 2 + x − 12 = 0 .
Solve for x by factoring: ( x + 4 ) ( x − 3 ) = 0 , so x = − 4 or x = 3 . Then find the corresponding y values. The solution set is {( − 4 , 5 ) , ( 3 , − 2 )} .
Explanation
Problem Setup We are given a system of two equations:
Equation 1: x + y = 1 Equation 2: y = x 2 − 11
We will solve this system using the substitution method.
Express y in terms of x From Equation 1, we can express y in terms of x :
y = 1 − x
Substitution Substitute this expression for y into Equation 2:
1 − x = x 2 − 11
Quadratic Form Rearrange the equation into a standard quadratic form:
x 2 + x − 12 = 0
Solve for x Solve the quadratic equation by factoring:
( x + 4 ) ( x − 3 ) = 0
So, x = − 4 or x = 3 .
Solve for y For each value of x , substitute it back into the equation y = 1 − x to find the corresponding value of y .
If x = − 4 , then y = 1 − ( − 4 ) = 5 . So one solution is ( − 4 , 5 ) .
If x = 3 , then y = 1 − 3 = − 2 . So another solution is ( 3 , − 2 ) .
Solution Set The solution set is {( − 4 , 5 ) , ( 3 , − 2 )} .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business, modeling supply and demand in economics, or calculating the trajectory of a projectile in physics. In this case, we found the intersection points of a line and a parabola, which can be visualized graphically. Understanding how to solve systems of equations is crucial for making informed decisions and predictions in many fields.