Use the graphing calculator to locate the solutions of this system of equations:
[tex]
\begin{array}{l}
y=(x-2)^2-15 \\
-5 x+y=-1
\end{array}
[/tex]
One solution of the system of equations is [$\square$]
A second solution of the system of equations is [$\square$]
Answer (2)
Solve the second equation for y , obtaining y = 5 x − 1 .
Substitute this expression into the first equation, resulting in 5 x − 1 = ( x − 2 ) 2 − 15 .
Simplify and solve the quadratic equation x 2 − 9 x − 10 = 0 , which factors to ( x − 10 ) ( x + 1 ) = 0 , giving x = 10 and x = − 1 .
Substitute x values back into y = 5 x − 1 to find corresponding y values, yielding the solutions ( 10 , 49 ) and ( − 1 , − 6 ) .
Explanation
Understanding the Problem We are given a system of two equations:
y = ( x − 2 ) 2 − 15 (a parabola)
− 5 x + y = − 1 (a line)
Our goal is to find the points (x, y) where these two equations intersect. These points are the solutions to the system of equations.
Isolating y in the Linear Equation First, let's solve the second equation for y:
− 5 x + y = − 1
y = 5 x − 1
Now we have an expression for y in terms of x.
Substitution and Simplification Next, we substitute the expression for y from the linear equation into the quadratic equation:
5 x − 1 = ( x − 2 ) 2 − 15
Expand the quadratic term:
5 x − 1 = x 2 − 4 x + 4 − 15
Simplify the equation:
5 x − 1 = x 2 − 4 x − 11
Move all terms to one side to set the equation to zero:
0 = x 2 − 9 x − 10
Solving for x Now we solve the quadratic equation x 2 − 9 x − 10 = 0 . We can factor this equation:
( x − 10 ) ( x + 1 ) = 0
So the solutions for x are:
x = 10 or x = − 1
Finding the Corresponding y Values Now we find the corresponding y values for each x value using the equation y = 5 x − 1 :
For x = 10 :
y = 5 ( 10 ) − 1 = 50 − 1 = 49
So one solution is ( 10 , 49 ) .
For x = − 1 :
y = 5 ( − 1 ) − 1 = − 5 − 1 = − 6
So the other solution is ( − 1 , − 6 ) .
Final Answer Therefore, the solutions to the system of equations are ( 10 , 49 ) and ( − 1 , − 6 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, if a company's cost function is represented by a quadratic equation and its revenue function is linear, solving the system of equations will give the production levels where the company neither makes a profit nor incurs a loss. This helps in making informed business decisions.
The solutions to the system of equations are (10, 49) and (-1, -6). To find these solutions, we substituted the line equation into the parabola equation and solved for x, and then found corresponding y values for each x. Thus, the complete set of solutions is identified as two points of intersection between the parabola and the line.
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