/*! 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 19 Solve each system. $$\begin{al... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 19

Solve each system. $$\begin{aligned} &y=6 x^{2}-1\\\ &2 x^{2}+5 y=-5 \end{aligned}$$

Short Answer

Expert verified
The solution to the given system of equations is \(x = 0\) and \(y = -1\), which can be written as a coordinate point (0, -1).

Step by step solution

01

Identify the equations

We have a system of two equations as follows: 1) \(y = 6x^2 - 1\) 2) \(2x^2 + 5y = -5\) We will use the substitution method to solve the system.
02

Substitute equation 1 into equation 2

Substitute the expression of y from equation 1 into equation 2: \(2x^2 + 5(6x^2 - 1) = -5\)
03

Simplify and solve for x

Simplify the equation: \(2x^2 + 30x^2 - 5 = -5\) Combine like terms: \(32x^2 = 0\) Now, solve for x: \(x^2 = 0 \Rightarrow x = 0\) We find that \(x = 0\)
04

Substitute the value of x into equation 1

Now, substitute the value of x we found in step 3 into equation 1: \(y = 6(0)^2 - 1\) Solve for y: \(y = -1\) So, we find that \(y=-1\)
05

Solution

Thus, the solution to the given system of equations is \(x = 0\) and \(y = -1\), which can be written as a coordinate point (0, -1).

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 are polynomial equations of degree two. They are typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. In our system of equations:
  • The equation \(y = 6x^2 - 1\) is a quadratic equation in terms of \(x\).
Quadratic equations can take multiple forms, but they generally describe a parabola when graphed on a coordinate system. Here, "6x^2" indicates the "a" value of 6 signifies a parabolic shape opening upward. Understanding these basics helps in solving quadratic systems effectively.
Substitution Method
The substitution method provides a strategic approach to solving systems of equations by replacing variables with equivalent expressions. Let's break down the process:
  • You start with two equations, and select one to express one variable in terms of the other.
  • Next, substitute the expression into the other equation.
In our exercise:
  • From the equation \(y = 6x^2 - 1\), we express \(y\) in terms of \(x\).
  • Substituting \(y\) in the second equation gives us: \(2x^2 + 5(6x^2 - 1) = -5\).
This step transforms our system into a solvable quadratic equation. Using substitution is especially effective when one equation is already isolated for a variable.
Solving Equations
Solving equations involves finding the values for variables that satisfy all equations simultaneously. Here's how this applies to our system:
  • The substituted equation simplifies to \(2x^2 + 30x^2 - 5 = -5\).
  • Combine like terms to get \(32x^2 = 0\).
To solve this:
  • Divide both sides by 32 resulting in \(x^2 = 0\).
  • This leads to \(x = 0\).
To find \(y\), substitute \(x = 0\) back into \(y = 6x^2 - 1\), leading to \(y = -1\). So, the solution is \((0, -1)\), meeting the conditions of both equations.
Coordinate Systems
A coordinate system allows us to visually plot solutions and better understand the relationships between equations. In a 2D coordinate system:
  • Points are defined as \((x, y)\) coordinates.
  • The graphical representation shows how solutions interact.
For our system:
  • The quadratic equation \(y = 6x^2 - 1\) creates a parabola on the coordinate plane.
  • The solution \((0, -1)\) is the point where both equations intersect, showing the system's only solution graphically.
Understanding coordinate systems enriches our comprehension of how and where solutions manifest, making it a vital tool in solving and verifying systems of equations.

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

Solve the exponential equation algebraically. Then check using a graphing calculator. $$2 e^{x}=5-e^{-x}$$

Solve the system of equations. $$\begin{aligned} w+x+y+z &=2 \\ w+2 x+2 y+4 z &=1 \\ -w+x-y-z &=-6 \\ -w+3 x+y-z &=-2 \end{aligned}$$

Solve using any method. $$5^{2 x}-3 \cdot 5^{x}+2=0$$

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$2^{x^{2}-9 x}=\frac{1}{256}$$

Four solutions of the equation \(y=a x^{3}+b x^{2}+c x+d\) are given. Use a system of four equations in four variables to find the constants a, \(b, c,\) and \(d\) and write the equation. $$(-2,59),(-1,13),(1,-1), \text { and }(2,-17)$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

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.