/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 In Exercises \(1-34\), find all ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 27

In Exercises \(1-34\), find all real solutions of the system of equations. If no real solution exists, so state. $$\left\\{\begin{aligned} x^{2}-(y-3)^{2} &=7 \\ x^{2}+& y^{2}=16 \end{aligned}\right.$$

Short Answer

Expert verified
The real solutions of the system are \((\sqrt{3.25}, 4.5), (-\sqrt{3.25}, 4.5), (\sqrt{11.25}, 1.5)\) and \(-\sqrt{11.25}, 1.5\).

Step by step solution

01

Express \(x^2\) from one equation in terms of other variables

Rearrange the first equation to express \(x^2\) in terms of \(y\): \(x^2 = (y-3)^2 + 7\).
02

Substitute \(x^2\) in the other equation

Substitute \(x^2 = (y-3)^2 + 7\) in second equation \(x^2 + y^2 = 16\), this gives: \((y-3)^2 + 7 + y^2 = 16\).
03

Simplify the last equation

Simplify the equation to isolate \(y\), which will give: \(2y^2 - 6y + 7 = 16\), further simplifies to \(2y^2 - 6y - 9 = 0\).
04

Solve for \(y\)

Now divide the equation by 2 and solve the quadratic equation \(y^2 - 3y - 4.5 = 0\) for \(y\). The solutions are \(y = 3 + \sqrt{9+4.5} = 4.5\) and \(y = 3 - \sqrt{9+4.5} = 1.5\).
05

Substituting \(y\) values for calculating \(x\)

Substitute the value of \(y\) in \(x^2 = (y-3)^2 + 7\) to find \(x\). Substituting \(y = 4.5\), we find \(x = \pm \sqrt{3.25}\). Similarly, substituting \(y = 1.5\), we find \(x = \pm \sqrt{11.25}\).
06

Finalize the solutions

So the solutions to the given system of equations are \((\sqrt{3.25}, 4.5), (-\sqrt{3.25}, 4.5), (\sqrt{11.25}, 1.5)\) and \(-\sqrt{11.25}, 1.5\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Quadratic Equations
Quadratic equations form the backbone of many algebraic problems and are characterized by the highest degree of the variable being two鈥攖ypically in the form of ax^2 + bx + c = 0. These equations can have zero, one, or at most two real solutions depending on the discriminant value b^2 - 4ac.

In the context of solving systems, when one equation in the system is quadratic and the other is linear, or both are quadratic, the approach involves solving for one variable in terms of the other and then substituting to find the solution. Students must be comfortable with manipulating these equations through various techniques including factoring, completing the square, and using the quadratic formula.
Real Solutions
The term 'real solutions' refers to answers that do not involve the imaginary unit i, which satisfies i^2 = -1. In real-life terms, these solutions can represent countable quantities, lengths, and other measurable attributes. Therefore, the existence of real solutions is essential when applying mathematical models to the physical world.

It's crucial to identify whether a quadratic equation has real solutions by examining its discriminant. If the discriminant is greater than zero, there are two different real solutions; if it鈥檚 equal to zero, there鈥檚 exactly one real solution (a repeated root); and if it's less than zero, there are no real solutions, meaning the parabola does not intersect the x-axis.
Algebraic Manipulation
Algebraic manipulation encompasses various techniques used to rearrange, simplify, and solve equations. Fundamental operations such as addition, subtraction, multiplication, and division, along with factoring, expanding, and combining like terms, are all part of this skill set.

When faced with a system of equations, algebraic manipulation can drastically simplify the problem. By rearranging one equation to express one variable in terms of another鈥攁s shown in the example problem鈥攁nd substituting into the second equation, the system newly configured into a single variable equation can be resolved more straightforwardly.
Substitution Method
The substitution method is an effective technique employed to solve systems of equations. This strategy involves isolating one variable in one equation and substituting that expression into the other equation of the system. The benefit of the substitution method is that it reduces a system of equations to a single equation with one variable, allowing for a more straightforward approach to finding a solution.

The key to successful substitution is clear algebraic manipulation to ensure that the substituted variable's expression is correctly implemented into the second equation. After solving for one variable, it is then used to find the other variable's values, ultimately providing the pairs that make up the system's solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For what value(s) of \(b\) does the following system of equations have two distinct, real solutions? $$\left\\{\begin{array}{l} y=-x^{2}+2 \\ y=x+b \end{array}\right.$$

For the given matrices \(A, B,\) and \(C,\) evaluate the indicated expression. $$\begin{aligned}&A=\left[\begin{array}{ll}4 & 1 \\\0 & 2 \\\5 & 1\end{array}\right] ; \quad B=\left[\begin{array}{rr}4 & 3 \\\\-6 & 2 \\\3 & -1\end{array}\right]\\\&C=\left[\begin{array}{rrr}1 & 2 & 3 \\\\-2 & -3 & -1 \\\3 & 1 & 2\end{array}\right] ; \quad C(B-A)\end{aligned}$$

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

Let \(I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] .\) Show that \(I A=A I,\) where \(A\) is any \(2 \times 2\) matrix.

Find \(A^{2}\) (the product \(A A\) ) and \(A^{3}\) (the prod\(\left.u c t\left(A^{2}\right) A\right)\). $$A=\left[\begin{array}{rrr}3 & 0 & 0 \\\0 & 1 & 1 \\\\-4 & 1 & 0\end{array}\right]$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.