Problem 24
Question
Solve each system. $$\begin{aligned} &y-x=1\\\ &4 y^{2}-16 x^{2}=64 \end{aligned}$$
Step-by-Step Solution
Verified Answer
The given system of equations
\[
\begin{cases}
y-x=1\\
4y^2-16(x-1)^2=64
\end{cases}
\]
has no real solutions, because after substituting \(x=y-1\) into the second equation, simplifying, and attempting to solve the resulting quadratic equation, we find a negative discriminant, which means there are no real solutions for y.
1Step 1: Solve the first equation for x or y
The first equation is \(y-x=1\). Let's solve for x:
\(x = y-1\)
2Step 2: Substitute the expression for x into the second equation
Now substitute the expression we found for x into the second equation:
\(4 y^2 - 16(y - 1)^2=64\)
3Step 3: Simplify and solve for y
First, expand the second equation and simplify:
\[4y^2 - 16(y^2 - 2y + 1) = 64\]
\[4y^2 - 16y^2 + 32y - 16 = 64\]
Now, collect like terms and solve for y:
\[-12y^2 + 32y - 80 = 0\]
Divide through by -4:
\[3y^2 - 8y + 20 = 0\]
4Step 4: Solve the quadratic equation for y
We can use the quadratic formula to solve for y:
\[y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
For our equation, a=3, b=-8, and c=20.
\[y=\frac{8\pm\sqrt{(-8)^2-4(3)(20)}}{2(3)}\]
Now calculate the discriminant:
\((-8)^2-4(3)(20)=64-240=-176\)
Since the discriminant is negative, the quadratic equation has no real solutions.
5Step 5: Conclusion
Since there are no real solutions for y in the quadratic equation, the system of equations has no real solutions. The given system of equations cannot be solved for real values of x and y.
Key Concepts
Quadratic EquationsDiscriminantSubstitution MethodNo Real Solutions
Quadratic Equations
Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations involve one unknown variable, usually \(x\), and the highest exponent of the variable is 2. The graph of a quadratic equation is a parabola.
- The vertex of the parabola can be a maximum or minimum point.
- Quadratic equations can have zero, one, or two solutions depending on the discriminant (which is related to the roots).
Discriminant
The discriminant is part of the quadratic formula and helps determine the nature of the solutions of a quadratic equation. It is found using the expression \(b^2 - 4ac\) where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
- If the discriminant is positive, there are two distinct real solutions.
- If it is zero, there is exactly one real solution (a repeated root).
- If the discriminant is negative, there are no real solutions, but two complex solutions.
Substitution Method
The substitution method is a technique used for solving systems of equations, particularly when one equation can be easily solved for one of the variables. The steps include:
- Solve one equation for one variable.
- Substitute the expression for the solved variable into the other equation.
- Simplify and solve the resulting equation.
No Real Solutions
When a system of equations, like the one in the problem, results in no real solutions, it means there are no points on a Cartesian plane that satisfy both equations simultaneously. The reasons can include:
- For linear-quadratic systems, the quadratic graph (parabola) and the line may not intersect at all.
- This typically occurs if the solutions involve imaginary or complex numbers, indicated by a negative discriminant in the quadratic equation.
Other exercises in this chapter
Problem 24
Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$1-d^{2}>0$$
View solution Problem 24
Identify the center of each ellipse and graph the equation. $$9 x^{2}+y^{2}=36$$
View solution Problem 24
Find an equation of the circle with the given center and radius. Center \((0,0) ;\) radius \(=\sqrt{6}\)
View solution Problem 24
Solve each system. $$\begin{array}{c} y-x=1 \\ 4 y^{2}-16 x^{2}=64 \end{array}$$
View solution