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Chapter 13: Problem 26

Solve. Remember that graphs can be used to confirm all real solutions. $$ \begin{aligned} &x y=4\\\ &x+y=5 \end{aligned} $$

Short Answer

Expert verified
The solutions are (4, 1) and (1, 4).

Step by step solution

01

- Solve for one variable

From the second equation, express one variable in terms of the other. Let's solve for y in terms of x: \[ y = 5 - x \]
02

- Substitute into the first equation

Substitute \( y = 5 - x \) into the first equation: \[ x(5 - x) = 4 \]
03

- Expand and simplify

Expand the equation: \[ 5x - x^2 = 4 \] Rearrange into standard quadratic form: \[ x^2 - 5x + 4 = 0 \]
04

- Solve the quadratic equation

Solve the quadratic equation \( x^2 - 5x + 4 = 0 \). Factorize it as: \[ (x - 4)(x - 1) = 0 \]
05

- Find the solutions for x

Set each factor equal to zero and solve for x: \[ x - 4 = 0 \implies x = 4 \] \[ x - 1 = 0 \implies x = 1 \]
06

- Find corresponding y values

Substitute each value of x back into the equation \( y = 5 - x \) to find corresponding y values: For \( x = 4 \): \[ y = 5 - 4 = 1 \] For \( x = 1 \): \[ y = 5 - 1 = 4 \]
07

- Verify the solutions

Verify the solutions by substituting into the original equations: For the solution (4, 1): \[ 4 \times 1 = 4 \] \[ 4 + 1 = 5 \] For the solution (1, 4) \[ 1 \times 4 = 4 \] \[ 1 + 4 = 5 \] Both pairs satisfy the original equations.

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Key Concepts

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

Quadratic Equations
A quadratic equation is a second-order polynomial equation in a single variable with the general form: \[ ax^2 + bx + c = 0 \]where \(a 鈮� 0\). Solutions to quadratic equations can be found using various methods, such as factoring, completing the square, or the quadratic formula.

For example, consider the quadratic equation derived from our problem after substitution and simplification:\[ x^2 - 5x + 4 = 0 \]Here, \(a = 1\), \(b = -5\), and \(c = 4\).

This equation can be factored into:\[ (x - 4)(x - 1) = 0 \]Setting each factor to zero gives the solutions for \(x\):
  • \( x - 4 = 0 \rightarrow x = 4 \)
  • \( x - 1 = 0 \rightarrow x = 1 \)
These values of \(x\) are then used to find the corresponding \(y\) values.
Substitution Method
The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Let's see how this method works in our example:

We start with two equations:\[ xy = 4 \]\[ x + y = 5 \]First, we solve the second equation for \(y\):\[ y = 5 - x \]Next, we substitute this expression into the first equation:\[ x(5 - x) = 4 \]This substitution reduces our system to a single quadratic equation in terms of \(x\), which we can solve using the methods described earlier.

Finally, we use the values of \(x\) to find the corresponding values of \(y\).
  • For \( x = 4 \): \( y = 5 - 4 = 1 \)
  • For \( x = 1 \): \( y = 5 - 1 = 4 \)
Verification of Solutions
Verifying solutions is essential to check whether the found solutions satisfy the original equations. Let's verify our solutions \((4, 1)\) and \((1, 4)\):

For the solution \((4, 1)\):
  • Check the first original equation: \( 4 \times 1 = 4 \) 鉁�
  • Check the second original equation: \( 4 + 1 = 5 \) 鉁�
For the solution \((1, 4)\):
  • Check the first original equation: \( 1 \times 4 = 4 \) 鉁�
  • Check the second original equation: \( 1 + 4 = 5 \) 鉁�
Both solutions satisfy the original equations, confirming their correctness.

This step is crucial because it ensures that our algebraic manipulations were correct and that the solutions are indeed valid for the given system of equations.

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