/*! 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 15 Solve each system by the substit... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each system by the substitution method. $$\begin{aligned} &x+y=1\\\ &x^{2}+x y-y^{2}=-5 \end{aligned}$$

Short Answer

Expert verified
The system of equations has no solution.

Step by step solution

01

Rearrange Equation

Rearrange the first equation \(x + y = 1\) to get \(x = 1 - y\). This equation is now in a form where x can be easily substituted into the second equation.
02

Substitute into Second Equation

Substitute \(x = 1 - y\) in the second equation \(x^{2} + xy - y^{2} = -5\). This yields \((1-y)^{2} + (1-y)y - y^{2} = -5\). Simplifying the equation gives \(y^{2} -3y + 6 = 0\).
03

Solve Quadratic Equation

Now, solve the quadratic equation by using the quadratic formula \(y = {3 \pm \sqrt {9 - 24}} \over 2\). However, it results in complex roots since the discriminant is negative.
04

Conclusion

Since we're dealing with real numbers only and the quadratic equation doesn't have real solutions, we can conclude that the original set of equations has no solution.

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Key Concepts

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

Systems of Equations
When dealing with systems of equations, the goal is to find the values of variables that satisfy all equations simultaneously. These systems can include two or more equations involving the same set of variables. In our exercise, we have two equations with two variables, which forms a simple system.

One common method to solve such systems is the substitution method, which involves expressing one variable in terms of the other and then substituting this expression into the other equation. This effectively reduces the system to a single equation with one variable. After finding the value of one variable, you can easily find the value of the other by substituting it back into any of the original equations.

The substitution method is particularly useful when one of the equations can be easily rearranged to isolate one variable, which is exactly what was done in the step-by-step solution.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The quadratic formula is expressed as \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). This formula allows us to find solutions by simply substituting the coefficients \(a\), \(b\), and \(c\) from the equation.

In the provided exercise, we apply the quadratic formula to solve for \(y\) after substituting \(x\) with \(1 - y\) and simplifying the equation. However, we encounter a negative value under the square root, which is known as the discriminant. In the case where the discriminant is negative, it tells us that the roots of the quadratic equation are complex numbers, indicating there are no real solutions to the equation.
Complex Numbers
When we encounter a negative discriminant using the quadratic formula, this leads us to the realm of complex numbers. A complex number is composed of a real part and an imaginary part and is usually represented in the form \(a + bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) is the imaginary unit with the property that \(i^2 = -1\).

In our exercise example, the discriminant is \(9 - 24\), which is negative. Using the quadratic formula, the roots of the quadratic equation become complex numbers. Given that the original problem is set within the realm of real numbers, complex solutions indicate that there is no solution to the system in the set of real numbers.

Understanding complex numbers is critical when working with quadratic equations, as they expand our ability to find solutions beyond the real number system. Complex numbers are used extensively in fields such as engineering, physics, and applied mathematics.

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Most popular questions from this chapter

The weekly demand and supply models for a particular brand of scientific calculator for a chain of stores are given by the demand model \(N=-53 p+1600,\) and the supply model \(N=75 p+320 .\) In these models, \(p\) is the price of the calculator and \(N\) is the number of calculators sold or supplied each week to the stores. a. How many calculators can be sold and supplied at \(\$ 12\) per calculator? b. Find the price at which supply and demand are equal. At this price, how many calculators of this type can be supplied and sold each week?

In Exercises \(5-18\), solve each system by the substitution method. $$ \begin{aligned} &2 x-3 y=-13\\\ &y=2 x+7 \end{aligned} $$

Graphing utilities can be used to shade regions in the rectangular coordinate system, thereby graphing an inequality in two variables. Read the section of the user's manual for your graphing utility that describes how to shade a region. Then use your graphing utility to graph the inequalities. $$ y \leq 4 x+4 $$

In Exercises \(43-46,\) let \(x\) represent one number and let \(y\) represent the other number. Use the given conditions to write a system of equations. Solve the system and find the numbers. The sum of two numbers is \(7 .\) If one number is subtracted from the other, their difference is \(-1 .\) Find the numbers.

Exercises \(47-50\) describe a number of business ventures. For each exercise, a. Write the cost function, \(C\). b. Write the revenue function, \(R\) c. Determine the break-even point. Describe what this means. A company that manufactures small canoes has a fixed cost of \(\$ 18,000 .\) It costs \(\$ 20\) to produce each canoe. The selling price is \(\$ 80\) per canoe. (In solving this exercise, let \(x\) represent the number of canoes produced and sold.)

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