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Chapter 4: Problem 32

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals. See Examples 2 through 6 $$ \left\\{\begin{array}{l} {2 x-\frac{3 y}{4}=-3} \\ {x+\frac{y}{9}=\frac{13}{3}} \end{array}\right. $$

Short Answer

Expert verified
The solution is \(x = 3\) and \(y = 12\).

Step by step solution

01

Clear Fractions

First, we clear the fractions in both equations. For the first equation, multiply every term by 4 to eliminate the fraction: \(4(2x - \frac{3y}{4}) = 4(-3)\). This simplifies to \(8x - 3y = -12.\) For the second equation, multiply every term by 9: \(9(x + \frac{y}{9}) = 9 \times \frac{13}{3}\). This simplifies to \(9x + y = 39.\) Now we have the system: \(\begin{cases} 8x - 3y = -12 \ 9x + y = 39 \end{cases}\).
02

Align the Equations

Write the equations to easily identify terms for elimination. We have: \(8x - 3y = -12\) and \(9x + y = 39\). The goal is to eliminate one variable by adding the equations.
03

Multiply to Prepare for Elimination

We want to eliminate \(y\). Multiply the second equation by 3 to prepare for elimination: \(3(9x + y) = 3(39)\). The new equation is \(27x + 3y = 117\). The updated system is: \(\begin{cases} 8x - 3y = -12 \ 27x + 3y = 117 \end{cases}\).
04

Add the Equations

Add the two equations to eliminate \(y\):\[ (8x - 3y) + (27x + 3y) = -12 + 117 \]This results in: \(35x = 105\).
05

Solve for \(x\)

Divide both sides of \(35x = 105\) by 35:\[ x = \frac{105}{35} \]This simplifies to \(x = 3\).
06

Substitute \(x\) Back

Substitute \(x = 3\) back into the second simplified equation to find \(y\):\[ 9(3) + y = 39 \]which simplifies to:\[ 27 + y = 39 \] Subtract 27 from both sides: \(y = 12\).
07

State the Solution

The solution to the system of equations is \(x = 3\) and \(y = 12\).

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Key Concepts

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

Clear Fractions
Clearing fractions from an equation may seem tricky at first, but it simplifies the problem, making it easier to solve. Fractions can complicate calculations, so starting with whole numbers offers a clearer path forward. Here's how you can clear fractions:
  • Identify the denominator in each term of the equation.
  • Find a common multiple, ideally the least common multiple (LCM), to multiply each term across the entire equation.
  • Multiply each term in the equation by this common multiple.
Take the system from our exercise:
  • First equation: \(2x - \frac{3y}{4} = -3\). The LCM of 4 is 4, so we multiply each term by 4, clearing the fraction: \(8x - 3y = -12\).
  • Second equation: \(x + \frac{y}{9} = \frac{13}{3}\). Here, the LCM is 9, so we multiply each term by 9, resulting in \(9x + y = 39\).
With these steps, we have new equations without fractions, simplifying further calculations.
Solve System of Equations
Solving a system of equations involves finding the values that satisfy all equations simultaneously. There are several methods for solving, with the addition method being one of the most popular.
  • First, align the equations vertically by like terms. This allows for easier manipulation and preparation for elimination of one variable.
  • Identify a variable to eliminate. This variable will be the same in both equations, but you'll change one equation so that they add to zero.
For example, in our equations \(8x - 3y = -12\) and \(9x + y = 39\), we want to eliminate \(y\). We multiply the second equation by 3 to align it with a multiple of the \(y\) term in the first equation. This leads to the system:
  • First equation: \(8x - 3y = -12\).
  • New second equation: \(27x + 3y = 117\).
By adding these equations, \(y\) is eliminated, allowing us to solve for \(x\). After finding \(x\), the solution for \(y\) follows by substituting the \(x\) value back into one of the original equations.
Substitution Method
The substitution method is another strategy to solve systems of equations. Unlike the addition method, substitution focuses on solving one equation for one variable and using that solution in the other equation. Here's how it works:
  • Solve one of the equations for one of its variables. Choose the one that seems easiest or has coefficients that simplify quickly.
  • Substitute this expression into the other equation. This will give you a single equation with one variable.
  • Solve for the variable in this new equation.
  • Once you have a value, substitute it back into the expression you first created to find the other variable.
Returning to the exercise, once \(x = 3\) was determined by the addition method, you can substitute \(x\) into the second equation \(9x + y = 39\). With \(x = 3\), substituting gives \(27 + y = 39\), which simplifies to \(y = 12\). Thus, the solution to the system is \(x = 3\), \(y = 12\), demonstrating how substitution complements our process.

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Most popular questions from this chapter

Solving systems involving more than three variables can be accomplished with methods similar to those encountered in this section. Apply what you already know to solve each system of equations in four variables. $$\left\\{\begin{aligned} x+y+z+w &=5 \\ 2 x+y+z+w &=6 \\ x+y+z &=2 \\ x+y &=0 \end{aligned}\right.$$

Given the cost function \(C(x)\) and the revenue function \(R(x)\), find the number of units \(x\) that must be sold to break even. See Example 6. $$ C(x)=75 x+160,000 R(x)=200 x $$

Without graphing, decide. See Examples 7 and \(8 .\) a. Are the graphs of the equations identical lines, parallel lines, or lines intersecting at a single point? b. How many solutions does the system have? $$ \left\\{\begin{array}{l} {2 y=x+2} \\ {y+2 x=3} \end{array}\right. $$

The highest scorer during the WNBA 2006 regular season was Diana Taurasi of the Phoenix Mercury. Over the season. Taurasi scored 116 more points than Seimone Augustus of the Minnesota Lynx. Together, Taurasi and Augustus scored 1604 points during the 2006 regular season. How many points did each player score over the course of the season? (Source: Women's National Basketball Association)

Macadamia nuts cost an astounding 16.50 dollars per pound, but research by an independent firm says that mixed nuts sell better if macadamias are included. The standard mix costs 9.25 dollars per pound. Find how many pounds of macadamias and how many pounds of the standard mix should be combined to produce 40 pounds that will cost 10 dollars per pound. Find the amounts to the nearest tenth of a pound.
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