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Chapter 1: Problem 8

What is the product of \(x y\) given the system of equations below? $$ \begin{array}{l} 4+y=32 x \\ y=2 x+2 \end{array} $$ (A) \(\frac{6}{25}\) (B) \(\frac{6}{25}\) (C) 12 (D) 15

Short Answer

Expert verified
The product \( xy \) is \( \frac{12}{25} \).

Step by step solution

01

Express y in terms of x

The second equation is given as \( y = 2x + 2 \). This expression already gives \( y \) in terms of \( x \), so nothing needs to be changed here.
02

Substitute y in the first equation

Substitute \( y = 2x + 2 \) into the first equation \( 4 + y = 32x \). This gives us: \[ 4 + (2x + 2) = 32x \]which simplifies to:\[ 6 + 2x = 32x \]
03

Solve for x

To get \( x \), move all terms involving \( x \) to one side of the equation:\[ 6 = 32x - 2x \]which simplifies to:\[ 6 = 30x \]Dividing both sides by 30 gives:\[ x = \frac{1}{5} \]
04

Solve for y

Now that we have \( x = \frac{1}{5} \), substitute this back into the equation for \( y \):\[ y = 2\left(\frac{1}{5}\right) + 2 \]This simplifies to:\[ y = \frac{2}{5} + 2 \]Converting 2 to fifths, we have:\[ y = \frac{2}{5} + \frac{10}{5} = \frac{12}{5} \]
05

Calculate the product xy

Using our values \( x = \frac{1}{5} \) and \( y = \frac{12}{5} \), the product \( xy \) can be found by multiplying:\[ xy = \frac{1}{5} \times \frac{12}{5} = \frac{12}{25} \]

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Key Concepts

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

System of Equations
A system of equations is a set of equations with multiple variables. Solving a system of equations means finding values for the variables that make all equations true. In this exercise, you have two equations:
  • Equation 1: \(4 + y = 32x\)
  • Equation 2: \(y = 2x + 2\)
The goal is to find values for both \(x\) and \(y\) that satisfy both equations at the same time. Once you achieve this, you can determine the product \(xy\). A system can have one solution, no solution, or infinitely many solutions. Here, we are working in a linear system with exactly one solution.
You often use methods like substitution and elimination to solve these systems, ensuring you manipulate the equations properly to find accurate values.
Solving for Variables
The process of solving for variables involves isolating each variable on one side of the equations. In this exercise, we first focus on making one variable easy to substitute. Since Equation 2 already expresses \(y\) in terms of \(x\), you can easily use this expression in Equation 1.
When solving for \(x\), you need to:
  • Substitute \(y = 2x + 2\) into the first equation.
  • Simplify the equation to combine like terms and isolate \(x\).
  • Rearrange the terms to isolate \(x\) on one side.
This approach allows you to effectively find the value of \(x\) from Equation 1, which is necessary before determining the value of \(y\). Knowing the value of \(x\), usually leads to a straightforward calculation of \(y\) using the already structured expression in Equation 2.
Substitution Method
The substitution method is a reliable technique in algebra for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. In the given exercise:
  • The variable \(y\) is expressed from Equation 2: \(y = 2x + 2\).
  • This \(y\) value is substituted into Equation 1, creating a new equation with one variable, \(x\).
  • You then solve the resulting equation to find \(x\).
Substitution simplifies the system by reducing the number of variables in one equation, making it easier to solve for one variable at a time. It is particularly effective when one of the equations in the system is already solved for a single variable, as seen here.
This method ensures precise calculations, especially when dealing with linear equations, helping you find the exact values of the variables involved.

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