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Chapter 7: Problem 37

For the following exercises, use any method to solve the nonlinear system. $$x^{2}+y^{2}-6 y=7$$ $$\quad\quad x^{2}+y=1$$

Short Answer

Expert verified
The solution for \( y \) is \( y = \frac{7 \pm \sqrt{73}}{2} \). Corresponding \( x \) values should be found numerically.

Step by step solution

01

Rewrite Equations

First, we'll rewrite the given system of equations:\[ x^2 + y^2 - 6y = 7 \] and \[ x^2 + y = 1 \] so that we can work more easily with them. Notice the second equation can be rearranged to isolate \( x^2 \): \[ x^2 = 1 - y \]
02

Substitute for \(x^2\)

Now that we have \( x^2 = 1 - y \), substitute this expression into the first equation: \[ 1 - y + y^2 - 6y = 7 \]
03

Simplify the Substituted Equation

Simplify the equation from Step 2: \[ y^2 - 6y + 1 - y = 7 \]Combine the \(y\) terms:\[ y^2 - 7y + 1 = 7 \]Subtract 7 from both sides:\[ y^2 - 7y - 6 = 0 \]
04

Solve the Quadratic Equation

Use the quadratic formula to solve \( y^2 - 7y - 6 = 0 \) where \(a = 1\), \(b = -7\), and \(c = -6\):\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]\[ y = \frac{7 \pm \sqrt{49 + 24}}{2} \]\[ y = \frac{7 \pm \sqrt{73}}{2} \]
05

Find Corresponding \(x\) Values

We found solutions for \( y \). Use these in \( x^2 = 1 - y \) to find \( x \):1. For \( y = \frac{7 + \sqrt{73}}{2} \): \[ x^2 = 1 - \left(\frac{7 + \sqrt{73}}{2}\right) \]2. For \( y = \frac{7 - \sqrt{73}}{2} \): \[ x^2 = 1 - \left(\frac{7 - \sqrt{73}}{2}\right) \] Solve each of these for \( x \).
06

Solve for \(x\) Numerically

To solve for \( x \) numerically use the values obtained for \( y \): 1. Substitute \( y = \frac{7 + \sqrt{73}}{2} \) into \( x^2 = 1 - y \) and simplify.2. Substitute \( y = \frac{7 - \sqrt{73}}{2} \) into \( x^2 = 1 - y \) and simplify.Finally, obtain the square roots to find the potential \( x \) values corresponding to each value of \( y \).

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which has the standard form: \[ ax^2 + bx + c = 0 \]When dealing with quadratic equations, the quadratic formula offers a way to find solutions by solving for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This equation helps resolve problems involving any quadratic equation by identifying when solutions exist and what they are.Key Points:
  • The term \( b^2 - 4ac \) inside the square root is called the discriminant and dictates the nature of the roots.
  • If the discriminant is positive, the quadratic equation has two distinct real roots.
  • If it is zero, there is one real double root.
  • If the discriminant is negative, the roots are complex and non-real.
In our given exercise, applying the quadratic formula was crucial in determining the possible values for \( y \) by plugging \( a = 1 \), \( b = -7 \), and \( c = -6 \) into the equation.
Substitution Method
The substitution method is an effective strategy for solving systems of equations, especially when one of the equations can easily be solved for one variable. It involves replacing a variable in one equation with an expression derived from another equation.For example, in our exercise:
  • We start with two equations: \[ x^2 + y^2 - 6y = 7 \] and \[ x^2 + y = 1 \].
  • Rearrange the second equation to isolate \( x^2 \): \[ x^2 = 1 - y \].
  • Substitute \( 1 - y \) for \( x^2 \) in the first equation, leading to a quadratic in \( y \).
This approach converts a system of nonlinear equations to a problem that can be tackled using techniques for solving quadratic equations, such as the quadratic formula. It reduces complexity by focusing on one variable at a time, simplifying the path to the solution.
Solving Quadratic Equations
When solving quadratic equations, like the transformed \( y^2 - 7y - 6 = 0 \) from our exercise, several methods can be applied, including factoring, using the quadratic formula, and completing the square. Our exercise uses the quadratic formula due to its wide applicability.Here鈥檚 a basic outline of solving quadratic equations:
  • Factoring: Break down the equation to factors that multiply to the original quadratic. But it may not always be possible.
  • Quadratic Formula: As explained earlier, it provides an algebraic solution when factoring is not straightforward.
  • Completing the Square: Rewriting the quadratic in a perfect square form to make it easier to solve.
In our exercise, solving \( y^2 - 7y - 6 = 0 \) using the quadratic formula gave the potential \( y \) values. These values were then used to find corresponding \( x \) values to solve the nonlinear system, completing the solution process.

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Most popular questions from this chapter

For the following exercises, use Gaussian elimination to solve the system. $$ \begin{aligned} \frac{x-3}{4}-\frac{y-1}{3}+2 z &=-1 \\ \frac{x+5}{2}+\frac{y+5}{2}+\frac{z+5}{2} &=8 \\ x+y+z &=1 \end{aligned} $$

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. At a competing cupcake store, \(\$ 4,520\) worth of cupcakes are sold daily. The chocolate cupcakes cost \(\$ 2.25\) and the red velvet cupcakes cost \(\$ 1.75 .\) If the total number of cupcakes sold per day is \(2,200,\) how many of each flavor are sold each day?

For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer鈥檚 Rule. You invest \(\$ 10,000\) into two accounts, which receive 8\(\%\) interest and 5\(\%\) interest. At the end of a year, you had \(\$ 10,710\) in your combined accounts. How much was invested in each account?

For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate- covered cashews. The nutritional information for these items is shown in Table 1 . $$ \begin{array}{|c|c|c|c|} \hline & \text { Fat (g) } & \text { Protein (g) } & \text { Carbohydrates (g) } \\ \hline \text { Almonds (10) } & 6 & 2 & 3 \\ \hline \text { Cranberries (10) } & 0.02 & 0 & 8 \\ \hline \text { Cashews (10) } & 7 & 3.5 & 5.5 \\ \hline \end{array} $$ For the special "low-carb" trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is \(425 \mathrm{~g},\) and the total amount of fat is \(570.2 \mathrm{~g}\). If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix?

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. You invested \(\$ 2,300\) into account \(1,\) and \(\$ 2,700\) into account 2 . If the total amount of interest after one year is \(\$ 254\), and account 2 has 1.5 times the interest rate of account 1 , what are the interest rates? Assume simple interest rates.
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