/*! 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 11 Solve each system by the substit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each system by the substitution method. \(\left\\{\begin{array}{c}5 x+2 y=0 \\ x=3 y-0\end{array}\right.\)

Short Answer

Expert verified
The system has two solutions: (-3,0) and (5,4)

Step by step solution

01

Isolate one variable from one of the equations

Rewrite the second equation \(2 y-x-3\) isolating \(x\). Doing so gives: \(x = 2y - 3\)
02

Substitute the isolated variable into the other equation

Now substitute \(x\) from the first step into the first equation: \(y^{2} - (2y - 3)^{2} - 9 = 0\)
03

Simplify the equation

Carry out the exponentiation and simplify to get a quadratic equation in terms of y: \(y^{2} - (4y^{2} - 12y + 9) - 9 = 0 \). This further simplifies to \(-3y^{2} + 12y = 0\)
04

Solve for y

Factor out \(y\) from each term to get \(y (-3y +12) = 0\). Solving this gives two solutions for \(y\); \(y=0\) and \(y=4\)
05

Substitute y-values to get x-values

Substitute each y-value from the fourth step into the isolated equation from step one to find the corresponding x-values: when \(y=0\), \(x = 2(0) - 3 = -3\) and when \(y=4\), \(x = 2(4) - 3 = 5\)

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 91Ó°ÊÓ!

Key Concepts

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

Systems of Equations
A system of equations is a collection of two or more equations that involve the same set of variables. The goal when solving such systems is to find the values of the variables that satisfy all equations simultaneously. There are multiple methods to tackle systems of equations, including graphing, substitution, elimination, and matrix methods. The substitution method, which is central to our current exercise, involves rearranging one of the equations to isolate one variable and then substituting that expression into the other equation(s). This method is particularly useful when one of the equations can be easily manipulated to isolate a variable. Once all equations are solved, the variables will have specific values that provide a solution to the entire system. It's important for each step to be handled with care to avoid mistakes that can lead to incorrect solutions.

When in a classroom or studying alone, visual aids such as graphing the equations can offer a better understanding of the system's solution. Graphically, the solution(s) represent the point(s) at which the graphs of the equations intersect. Understanding this concept is beneficial because it helps to visualize and confirm the solutions obtained algebraically.
Solving Quadratic Equations
Quadratic equations are second-degree polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not equal to zero. These equations are fundamental in algebra and can be solved by various methods such as factoring, completing the square, using the quadratic formula, or graphing.

The aforementioned exercise leads us to a quadratic equation after substituting and simplifying. To solve such equations when factoring is possible, look for two numbers that multiply to the constant term (when set to zero) and add up to the coefficient of the linear term. In the given example, the equation simplifies to \( -3y^2 + 12y = 0 \), which can be factored to yield solutions for \( y \). For students, understanding the methods for solving quadratic equations is vital, not just in algebra, but in various applications that include physics, engineering, and economics. A firm grasp of this allows one to solve a wide range of practical problems. Additionally, real-world problems often yield quadratic equations that do not factor nicely, hence knowledge of the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), becomes valuable.
Algebraic Substitution
Algebraic substitution is a technique where we replace one variable with another expression to simplify a problem or solve an equation. This technique is often used in conjunction with systems of equations as seen in the given exercise. The substitution method is a form of algebraic substitution, where you express one variable in terms of another and then replace this variable in other equation(s).

In our example, the variable \( x \) is isolated in the second equation and then substituted in the first equation for further simplification. This substitution simplifies the process of finding the solution since it reduces the system of equations to a single variable equation. For better understanding, it is useful to note that the goal of substitution is to reduce the complexity of the problem step by step. Students should practice this technique to become fluent in recognizing opportunities to apply algebraic substitution, which can transform a complex equation into a more solvable form. It's a powerful tool not only in algebra but also in calculus and other higher-level mathematics.

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

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. You invest in a new play. The cost includes an overhead of \(\$ 30,000,\) plus production costs of \(\$ 2500\) per performance. A sold-out performance brings in \(\$ 3125 .\) ( In solving this exercise, let \(x\) represent the number of sold-out performances.)

Consider the objective function \(z-A x+B y \quad(A>0\) and \(B>0\) ) subject to the following constraints: \(2 x+3 y \leq 9, x-y \leq 2, x \geq 0,\) and \(y \geq 0 .\) Prove that the objective function will have the same maximum value at the vertices \((3,1)\) and \((0,3)\) if \(A-\frac{2}{3} B\).

Bottled water and medical supplies are to be shipped to survivors of an earthquake by plane. The bottled water weighs 20 pounds per container and medical kits weigh 10 pounds per kit. Each plane can carry no more than \(80,000\) pounds. If \(x\) represents the number of bottles of water to be shipped per plane and \(y\) represents the number of medical kits per plane, write an inequality that models each plane's \(80,000\)-pound weight restriction.

This will help you prepare for the material covered in the next section. In each exercise, graph the linear function. $$2 x-3 y-6$$

The group should write four different word problems that can be solved using a system of linear equations in two variables. All of the problems should be on different topics. The group should turn in the four problems and their algebraic solutions.

See all solutions

Recommended explanations on Math Textbooks

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.