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Chapter 12: Problem 50

Solve each system by elimination (addition). $$ \left\\{\begin{array}{l} \frac{1}{3} x+\frac{1}{2} y=\frac{31}{6} \\ 0.3 x+0.2 y=3.9 \end{array}\right. $$

Short Answer

Expert verified
The solution is \(x = 11\), \(y = 3\).

Step by step solution

01

Eliminate Fractions and Decimals

First, eliminate fractions and decimals to simplify calculations. Multiply the first equation by 6 to remove fractions:\[2x + 3y = 31\]For the second equation, multiply through by 10 to remove decimals:\[3x + 2y = 39\]This gives us the equivalent system:\[\begin{align*}2x + 3y &= 31 \3x + 2y &= 39\end{align*}\]
02

Prepare for Elimination by Equalizing Coefficients

To eliminate one of the variables, we need their coefficients in both equations to be opposites. We'll eliminate \(y\). Multiply the first equation by 2:\[4x + 6y = 62\]And the second equation by 3:\[9x + 6y = 117\]Our system is now:\[\begin{align*}4x + 6y &= 62 \9x + 6y &= 117\end{align*}\]
03

Eliminate One Variable

Subtract the first equation from the second to eliminate \(y\):\[(9x + 6y) - (4x + 6y) = 117 - 62\]This simplifies to:\[5x = 55\]Now, solve for \(x\):\[x = 11\]
04

Solve for the Second Variable

Substitute \(x = 11\) back into the first original equation:\[2(11) + 3y = 31\]Calculate:\[22 + 3y = 31\]Subtract 22 from both sides:\[3y = 9\]Divide by 3:\[y = 3\]
05

Solution Verification

Substitute \(x = 11\) and \(y = 3\) into the second original equation to verify:\[3(11) + 2(3) = 39\]Calculate:\[33 + 6 = 39\]The solution is verified as correct.

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Key Concepts

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

Elimination Method
The elimination method is a technique used to solve systems of linear equations. The goal is to eliminate one of the variables so that you can easily solve for the other. Here's how it works:
  • Start with your system of equations, which might look like this:\[ \left\{\begin{array}{l} \frac{1}{3} x+\frac{1}{2} y=\frac{31}{6} \ 0.3 x+0.2 y=3.9 \end{array}\right.\]
  • First, adjust the equations to eliminate fractions and decimals. This makes the arithmetic simpler.
  • Next, make the coefficients of one variable equal in both equations. This may require multiplying one or both equations by certain numbers.
  • Subtract or add the equations to eliminate that variable. You will be left with a single equation that you can solve easily.
  • Substitute the found value back into one of the original equations to find the second variable.
The elimination method is particularly useful when one of the coefficients can easily be made equal to another by simple multiplication. It's often faster than solving by substitution, especially when dealing with linear equations with neat coefficients.
Fractions and Decimals
Fractions and decimals often appear in linear equations, especially in real-world problems. Before using an elimination method, it often helps to first eliminate these fractions and decimals from the equations. This simplification makes further calculations more straightforward.
  • Fractions: Multiply every term in the equation by the least common multiple (LCM) of the denominators. This removes the fractions by converting them into whole numbers. For example: \[\frac{1}{3}x + \frac{1}{2}y = \frac{31}{6} \rightarrow 2x + 3y = 31\]
  • Decimals: Treat decimals by multiplying the entire equation by powers of 10. This transforms the equation into one without decimals. For instance: \[0.3x + 0.2y = 3.9 \rightarrow 3x + 2y = 39\]
Eliminating fractions and decimals not only simplifies calculation but also reduces the chance of making errors during the elimination process. It streamlines the arithmetic, making it easier to spot calculation mistakes.
Solving Linear Equations
Once the system is set, solving linear equations with the elimination method becomes easier. Here's the process:
  • After eliminating fractions and decimals, establish a set of linear equations that are easier to manage:
    • The modified system might look like: \[ \begin{align*} 2x + 3y &= 31\ 3x + 2y &= 39 \end{align*} \]
  • With these equations, adjust them by multiplying each to align coefficients for elimination. For instance, to eliminate \(y\), you might adjust as follows: \[4x + 6y = 62, \quad 9x + 6y = 117\]
  • Subtract the two equations to eliminate one variable: \[\begin{align*} 9x + 6y - (4x + 6y) &= 117 - 62\ 5x &= 55 \end{align*}\]
  • You can now solve for the variable \(x\): \[x = 11\]
  • Substitute \(x\) back into one of the simplified equations to solve for \(y\): \[2(11) + 3y = 31 \rightarrow 3y = 9 \rightarrow y = 3\]
  • Verify the solution by checking whether the values satisfy the original equations. This ensures the calculation was correct.
These steps reveal the solution with clarity and precision. By verifying the solution back in the original equations, students can confidently know their work is accurate.

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Most popular questions from this chapter

Write a system of two equations in two variables to solve each problem. Snowmobiling. A man rode a snowmobile at the rate of 20 mph and then skied cross country at the rate of 4 mph. During the 6 -hour trip, he traveled 48 miles. How long did he snowmobile, and how long did he ski?

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 6 x+3 y=18 \\ y=-2 x+5 \end{array}\right. $$

Solve each system by elimination (addition). $$ \left\\{\begin{array}{l} \frac{5}{6} x+\frac{1}{2} y=12 \\ 0.3 x+0.5 y=5.6 \end{array}\right. $$

Solve each system by elimination (addition). $$ \left\\{\begin{array}{l} 3 x+4 y=-24 \\ 5 x+12 y=-72 \end{array}\right. $$

Solve each system by graphing, if possible. If a system is inconsistent or if the equations are dependent, state this. (Hint: Several coordinates of points of intersection are fractions.) $$ \left\\{\begin{array}{l} x=\frac{1-3 y}{4} \\ y=\frac{12+3 x}{2} \end{array}\right. $$
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