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Chapter 9: Problem 22

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} y=\frac{10}{x+3} \\ y=-x+8 \end{array}\right. $$

Short Answer

Expert verified
Solutions are \((x, y) = (-2, 10)\) and \((7, 1)\).

Step by step solution

01

Set Equations Equal

The system of equations given is \( y = \frac{10}{x+3} \) and \( y = -x + 8 \). Since both expressions equal \( y \), we can set them equal to each other to solve for \( x \): \( \frac{10}{x+3} = -x + 8 \).
02

Clear the Fraction

To eliminate the fraction, multiply both sides of the equation by \( x+3 \), resulting in \( 10 = (-x + 8)(x + 3) \).
03

Expand the Equation

Now expand the right side: \( 10 = -x^2 - 3x + 8x + 24 \), which simplifies to \( 10 = -x^2 + 5x + 24 \).
04

Rearrange into Quadratic Form

Bring all terms to one side to get a standard quadratic equation: \( 0 = -x^2 + 5x + 24 - 10 \), which simplifies to \( 0 = -x^2 + 5x + 14 \).
05

Solve the Quadratic Equation

To solve \( -x^2 + 5x + 14 = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 5 \), and \( c = 14 \).
06

Calculate Discriminant

The discriminant \( b^2 - 4ac \) is \( 5^2 - 4(-1)(14) = 25 + 56 = 81 \).
07

Find the Roots

Since the discriminant is positive, there are two real roots. Solve \( x = \frac{-5 \pm \sqrt{81}}{-2} = \frac{-5 \pm 9}{-2} \). The solutions are \( x = -2 \) and \( x = 7 \).
08

Solve for y

Substitute \( x = -2 \) into \( y = -x + 8 \): \( y = -(-2) + 8 = 10 \). Substitute \( x = 7 \) into \( y = -x + 8 \): \( y = -(7) + 8 = 1 \).
09

Verify Solutions in Original Equation

Check \( x = -2, y = 10 \) satisfies \( y = \frac{10}{x+3} \) by evaluating \( \frac{10}{-2+3} = 10 \). Check \( x = 7, y = 1 \) satisfies \( y = \frac{10}{x+3} \) by evaluating \( \frac{10}{7+3} = 1 \). Both solutions satisfy both equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a technique used to solve systems of equations, where one is solved for a single variable and substituted into the other. This helps simplify the system into one equation in terms of one variable.
  • Start with both equations in the system. Let's consider the system from the exercise: \( y = \frac{10}{x+3} \) and \( y = -x + 8 \).
  • Notice that both expressions equal \( y \). By setting them equal to each other, we derive a new equation: \( \frac{10}{x+3} = -x + 8 \).
  • This process transforms the system into a manageable single equation, making it easier to find a solution for \( x \).
By eliminating one variable through substitution, we significantly reduce the complexity of solving systems with multiple equations.
Quadratic Formula
The quadratic formula is a powerful algebraic tool used to solve quadratic equations, equations that take the form \( ax^2 + bx + c = 0 \). This formula gives solutions for any quadratic equation across its realm.
  • The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • It applies to equations where only \( a \), \( b \), and \( c \) are constants with known values. In our case, the solution involves substituting these into the formula: \( a = -1 \), \( b = 5 \), \( c = 14 \).
  • Use this structure to solve for \( x \), accounting for both potential solutions due to the \( \pm \) sign, indicating the possibility of two different roots.
This formula is essential for finding precise solutions to quadratic equations, assisting significantly when factoring is impractical.
Discriminant
The discriminant is a crucial part of the quadratic formula, found under the square root sign within the equation. It provides insight into the nature of the roots of a quadratic equation.
  • The discriminant is given by the part \( b^2 - 4ac \) from the quadratic formula.
  • Depending on its value, the discriminant tells us the type of roots:
    • If it is positive, the equation has two distinct real roots.
    • If it is zero, there is exactly one real root, also called a repeated root.
    • If negative, the equation has no real roots but two complex ones.
  • In our example, the discriminant calculated as \( 81 \) (i.e., \( 25 + 56 \)) signals two real roots, guiding us to explore both solutions for \( x \).
Understanding the discriminant ensures we make accurate conclusions about the nature of the solutions.
Real Roots
Real roots of a quadratic equation are simply the solutions that are actual numbers, as opposed to complex or imaginary numbers. In real-world applications, real roots are particularly significant.
  • When the discriminant is positive, we expect two different real roots.
  • These roots can be found by completing the calculation associated with the quadratic formula. In our system, these roots were determined as \( x = -2 \) and \( x = 7 \).
  • After finding the roots, it's essential to verify them by substituting back into the original equations to ensure they satisfy both parts of the system.
Finding real roots helps us apply these solutions in various practical scenarios, highlighting their usability beyond theoretical environments.

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Most popular questions from this chapter

Exer. 1-28: Find the partial fraction decomposition. $$ \frac{8 x-1}{(x-2)(x+3)} $$

Exer. 17-20: Solve the system using the inverse method. Refer to Exercises 3-4 and 9-10. $$ \left\\{\begin{aligned} x+2 y+3 z &=c \\ -2 x+y &=d \\ 3 x-y+z &=e \end{aligned}\right. $$ (a) \(\left[\begin{array}{l}c \\ d \\\ e\end{array}\right]=\left[\begin{array}{r}-1 \\ 4 \\ 2\end{array}\right]\) (b) \(\left[\begin{array}{l}c \\ d \\\ e\end{array}\right]=\left[\begin{array}{r}-3 \\ -2 \\ 1\end{array}\right]\)

Exer. 1-28: Find the partial fraction decomposition. $$ \frac{2 x^{4}-2 x^{3}+6 x^{2}-5 x+1}{x^{3}-x^{2}+x-1} $$

Exer. 33-42: Use Cramer's rule, whenever applicable, to solve the system. $$ \left\\{\begin{array}{rr} x+3 y-z= & -3 \\ 3 x-y+2 z= & 1 \\ 2 x-y+z= & -1 \end{array}\right. $$

Let \(A=\left[\begin{array}{rr}1 & 2 \\ 0 & -3\end{array}\right], \quad B=\left[\begin{array}{rr}2 & -1 \\ 3 & 1\end{array}\right], \quad C=\left[\begin{array}{rr}3 & 1 \\ -2 & 0\end{array}\right]\) $$ (A+B)(A+B) \neq A^{2}+2 A B+B^{2} $$
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