/*! 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 74 Solve each system by the substit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 74

Solve each system by the substitution method. $$\begin{aligned}&y=2 x\\\&15 x-10 y=-2\end{aligned}$$

Short Answer

Expert verified
The solution is \( x = \frac{2}{5} \) and \( y = \frac{4}{5} \).

Step by step solution

01

Substitute y in the second equation

Given the system of equations: 1) \( y = 2x \) 2) \( 15x - 10y = -2 \), Substitute the expression for y from the first equation into the second equation. This means replacing y with 2x in the second equation: \( 15x - 10(2x) = -2 \).
02

Simplify the second equation

Simplify the equation obtained from the substitution: \( 15x - 20x = -2 \) which simplifies to \( -5x = -2 \).
03

Solve for x

Solve for x by dividing both sides of the equation by -5: \( x = \frac{-2}{-5} \) which simplifies to \( x = \frac{2}{5} \).
04

Substitute x back into the first equation

Substitute the value of x back into the first equation to find y: \( y = 2x \) \( y = 2 \left( \frac{2}{5} \right) \) \( y = \frac{4}{5} \).
05

Verify the solution

Verify the solution by substituting x and y into the original equations to ensure they are satisfied: 1) For \( y = 2x \), with \( x = \frac{2}{5} \), \( y = \frac{4}{5} \) satisfies the first equation. 2) For \( 15x - 10y = -2 \), with \( x = \frac{2}{5} \) and \( y = \frac{4}{5} \), \[ 15 \left( \frac{2}{5} \right) - 10 \left( \frac{4}{5} \right) = 6 - 8 = -2 \] which satisfies the second equation.

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.

Solving Systems of Equations
When tackling systems of equations, we aim to find a set of values that satisfy all equations. In this problem, we have two equations with two variables:
  • \( y = 2x \)
  • \( 15x - 10y = -2 \)
To solve for both variables, we use the substitution method. This involves replacing one variable with an expression involving the other variable, making it easier to solve step-by-step.
Linear Equations
A linear equation is one of the simplest types of equations in algebra. Each term in a linear equation is either a constant or a product of a constant and a single variable. In this exercise, our equations are linear:
  • \( y = 2x \)
  • \( 15x - 10y = -2 \)
These equations represent straight lines when graphed on a coordinate plane. Our goal is to find the intersection point of these lines, which represents the solution that satisfies both equations.
Algebraic Substitution
The substitution method in algebra involves replacing one variable with its equivalent expression from another equation. Let's see this in action with our given equations.
Starting with \( y = 2x \), we substitute \( y \) in the second equation \( 15x - 10y = -2 \). This becomes:
\( 15x - 10(2x) = -2 \)
Simplifying, we get: \( 15x - 20x = -2 \)
This reduces to: \( -5x = -2 \)
Solving for \( x \), we divide both sides by -5:
\( x = \frac {2}{5} \).
Next, we substitute this \( x \) value back into the first equation to find \( y \):
\( y = 2 \times \frac {2}{5} = \frac {4}{5} \).
Verification of Solutions
Verifying solutions ensures that our values of \( x \) and \( y \) satisfy both original equations. For our solutions \( x = \frac{2}{5} \) and \( y = \frac{4}{5} \), we substitute them back into the original equations:
  • First Equation: \( y = 2x \)
    \( \frac {4}{5} = 2 \times \frac{2}{5} \) checks out.
  • Second Equation: \( 15x - 10y = -2 \)
    \( 15 \times \frac {2}{5} - 10 \times \frac {4}{5} = 6 - 8 = -2 \) is also correct.
Thus, our solutions \( x = \frac{2}{5} \) and \( y = \frac{4}{5} \) satisfy both equations, confirming their accuracy.

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 each system using the Gauss-Jordan elimination method. $$ \begin{array}{r} x+y=3 \\ y=6 \end{array} $$

Determine the row operation that was used to convert each given augmented matrix into the equivalent augmented matrix that follows it. $$ \left[\begin{array}{rr|r} 3 & 2 & 12 \\ 1 & -1 & -1 \end{array}\right],\left[\begin{array}{rr|r} 1 & -1 & -1 \\ 3 & 2 & 12 \end{array}\right] $$

Solve each problem using a system of linear equations and the Gauss-Jordan elimination method. Two numbers. The sum of two numbers is 12 and their difference is \(2 .\) Find the numbers.

Solve each problem by using a system of three equations in three unknowns. In three days, Katy traveled 146 miles down the Mississippi River in her kayak with 30 hours of paddling. The first day she averaged 6{mph}, the second day 5{mph}, and the third day 4 {mph}. If her distance on the third day was equal to her distance on the first day, then for how many hours did she paddle each day?

Write the augmented matrix for each system of equations. $$ \begin{array}{r} x-y=4 \\ 2 x+y=3 \end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

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.