/*! 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 7 Solve each system. $$\left\\{\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 7

Solve each system. $$\left\\{\begin{array}{l} x+y=5 \\ x-y=-1 \end{array}\right.$$

Short Answer

Expert verified
The solution of the system of equations is \(x = 2\) and \(y = 3\).

Step by step solution

01

Add the Two Equations

Given the equations \(x + y = 5\) and \(x - y = -1\). Add the two equations. Ideal for this system as the variable y in the two equations will cancel each other when performing addition.
02

Solve for x

After the addition of the equations in Step 1, the result is \(2x = 4\). To isolate the variable x, divide both sides of the equation by 2. This yields the value of x.
03

Substitute x-value to find y

After finding the value of x, substitute this value into either of the original equations to find the value of y. Choosing the first equation from the system \(x + y = 5\), replacing x with 2 gives \(2 + y = 5\). Solving for y gives us the value of y.

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.

Addition Method
The addition method, also known as the elimination method, is a technique used for solving systems of linear equations. This method can efficiently eliminate one variable and solve for the other. Here's how it works:
  • Identify two equations in a system that, when added together, will eliminate one of the variables. This is possible when the coefficients of one variable are opposites. For example, in the system \(x + y = 5\) and \(x - y = -1\), the y-terms \(+y\) and \(-y\) will cancel each other out when added.
  • Add the two equations together. For the given system, adding the equations yields \(2x = 4\).
  • Solve the resulting equation for the remaining variable. In this case, dividing both sides by 2 will isolate \(x\), resulting in \(x = 2\).
The main advantage of the addition method is its ability to quickly eliminate a variable, thus simplifying the equation system. It's a powerful tool when dealing with larger systems, especially when the system is designed for easy elimination.
Substitution Method
The substitution method is another way to tackle systems of linear equations, providing a clear path to finding solutions by isolating and substituting variables. To use it effectively, follow these steps:
  • Start by solving one of the equations for one variable. This is done by isolating the variable on one side of the equation. For example, from the equation \(x + y = 5\), we can solve for \(y\), resulting in \(y = 5 - x\).
  • Substitute this expression into the other equation. Replace \(y\) in the second equation \(x - y = -1\) with \(5 - x\), resulting in \(x - (5 - x) = -1\).
  • Simplify and solve for the remaining variable. This operation leads you to \(2x - 5 = -1\), and solving for \(x\) gives \(x = 2\).
  • Finally, substitute the value of the solved variable back into the expression from the first step to find the other variable. Here, substituting \(x = 2\) back gives \(y = 3\).
The substitution method is especially useful when one of the equations is easy to solve for a single variable. It offers a straightforward path to uncover the values of unknowns in the system.
Linear Equations
Linear equations are mathematical statements expressing equality between two expressions involving constants and variables. They represent straight lines when graphed on a coordinate plane. Here's what characterizes them:
  • They have no exponents higher than one on any variable. This means equations like \(x + y = 5\) and \(x - y = -1\) are linear because variables are not squared, cubed, or involved in higher powers.
  • They can be written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Constants can be any real number, and \(a\) or \(b\) can be zero, but not both at the same time.
  • The solutions to linear equations form a line on the graph, indicating infinitely many solutions for a single linear equation. In systems with two equations, finding the point where the two lines intersect solves the system.
Linear equations are foundational in algebra, laying the groundwork for solving complex systems and understanding geometric interpretations of algebraic statements. They're simple yet powerful tools for translating real-world problems into solvable mathematical forms.

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

Simplify each expression. Assume that all variables represent positive values. Assume no division by 0. SEE EXAMPLE 6. (OBJECTIVE 3) $$a^{3 / 5} a^{-1 / 2}$$

Simplify. Assume that all variables in the radicand of an even root represent positive values. Assume no division by 0. Express each answer with positive exponents only. $$\left(2^{1 / 2} 3^{1 / 2}\right)^{2}$$

If \(x>y,\) which is the larger number in each pair? $$\left(\frac{1}{2}\right)^{x} \cdot\left(\frac{1}{2}\right)^{y}$$

Simplify. Assume that all variables in the radicand of an even root represent positive values. Assume no division by 0. Express each answer with positive exponents only. $$81^{3 / 4}$$

Simplify each expression. All variables of square root expressions represent positive numbers. Assume no division by 0. $$-3 x y z \sqrt{18 x^{3} y^{5}}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.