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Chapter 10: Problem 44

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered pair form given in Example 6. $$\left\\{\begin{array}{l}\frac{3}{2} x-\frac{1}{3} y=\frac{1}{2} \\\2 x-\frac{1}{2} y=-\frac{1}{2}\end{array}\right.$$

Short Answer

Expert verified
The solution is \((5, 21)\).

Step by step solution

01

Clear Fractions

Multiply each equation by the least common multiple of the denominators to eliminate fractions. For the first equation, the denominators are 2 and 3, so multiply by 6:\[ 6 \left( \frac{3}{2}x - \frac{1}{3}y \right) = 6 \times \frac{1}{2} \] Resulting in:\[ 9x - 2y = 3 \]For the second equation, the denominator is 2, so multiply by 2:\[ 2 \left( 2x - \frac{1}{2}y \right) = 2 \times -\frac{1}{2} \] Resulting in:\[ 4x - y = -1 \]
02

Set Up for Elimination

Align the equations:\[ \begin{aligned} 9x - 2y &= 3 \ 4x - y &= -1 \end{aligned}\]
03

Eliminate Variable y

Multiply the second equation by 2 to align with the coefficients of \(y\):\[ \begin{aligned} 9x - 2y &= 3 \ 8x - 2y &= -2 \end{aligned}\]Subtract the second equation from the first:\[ (9x - 2y) - (8x - 2y) = 3 - (-2)\]\[ x = 5 \]
04

Substitute x Back

Substitute \(x = 5\) into one of the original equations, such as \(4x - y = -1\):\[ 4(5) - y = -1\]\[ 20 - y = -1\]Solve for \(y\):\[ -y = -21 \ y = 21\]
05

Solution Verification

Now that \(x = 5\) and \(y = 21\), verify by plugging back into both original equations:For \( \frac{3}{2} x - \frac{1}{3} y = \frac{1}{2} \) :\[ \frac{3}{2} imes 5 - \frac{1}{3} imes 21 = 7.5 - 7 = 0.5\]Correct! For \( 2x - \frac{1}{2} y = -\frac{1}{2} \):\[ 2 imes 5 - \frac{1}{2} imes 21 = 10 - 10.5 = -0.5\]Correct! Both checks confirm solutions.

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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 powerful tool for solving systems of linear equations. It involves manipulating the equations to eliminate one of the variables, making it easier to solve for the remaining variables.
In the given problem, we start by aligning the equations:
  • First equation: \(9x - 2y = 3\)
  • Second equation: \(4x - y = -1\)
We then multiply the second equation by 2 to match the coefficient of \(y\) in the first equation. This gives us:
  • First: \(9x - 2y = 3\)
  • Second modified: \(8x - 2y = -2\)
By subtracting the second from the first, the \(y\) terms cancel out, simplifying the solution to find \(x\) as 5. Thus, eliminating the \(y\) variable makes it straightforward to solve for \(x\) using subtraction.
Substitution Method
The substitution method is another effective way to solve linear systems. Once you have one variable isolated, you can substitute it back into one of the equations to find the other variable.
In this problem, after using the elimination method, we discovered \(x = 5\). We then substitute \(x = 5\) back into the second simplified equation, \(4x - y = -1\):
  • Substitute: \(4(5) - y = -1\)
  • Solve for \(y: 20 - y = -1\)
  • Therefore, \(y = 21\)
Substitution allows us to confirm and find the value of the second variable by knowing the first, simplifying the process even further after elimination.
Fractions in Equations
Fractions can make equations appear complex, but they are manageable when you clear them effectively. The key is to eliminate fractions at the start of the problem to work with whole numbers instead.
To clear the fractions in the given system:
  • For the first equation, multiply by the least common multiple of 2 and 3, which is 6:
    \(6 \left( \frac{3}{2}x - \frac{1}{3}y \right) = 6 \times \frac{1}{2}\)
  • This becomes \(9x - 2y = 3\).
  • For the second equation, multiply by 2:
    \(2 \left( 2x - \frac{1}{2}y \right) = 2 \times -\frac{1}{2}\)
  • This results in \(4x - y = -1\).
Managing and clearing fractions early simplifies the entire calculation process, setting up a clear path to solve the system using elimination or substitution.
Verifying Solutions
Verifying solutions is crucial to ensure that the obtained values satisfy the original equations. After calculating solutions with methods like elimination or substitution, always check your work.
For our problem, we found \(x = 5\) and \(y = 21\). To verify:
  • First equation: \(\frac{3}{2} \times 5 - \frac{1}{3} \times 21 = 7.5 - 7 = 0.5\)
  • This matches the right-hand side of the first equation: \(0.5\).
  • Second equation: \(2 \times 5 - \frac{1}{2} \times 21 = 10 - 10.5 = -0.5\)
  • This matches the right-hand side of the second equation: \(-0.5\).
Both values satisfy their respective original equations, confirming the solutions are correct. Always verify to prevent mistakes and ensure the accuracy of your solutions.

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Most popular questions from this chapter

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A nutritionist is studying the effects of the nutrients folic acid, choline, and inositol. He has three types of food available, and each type contains the following amounts of these nutrients per ounce. $$\begin{array}{|l|c|c|c|} \hline & \text { Type A } & \text { Type B } & \text { Type C } \\ \hline \text { Folic acid (mg) } & 3 & 1 & 3 \\ \text { Choline (mg) } & 4 & 2 & 4 \\ \text { Inositol (mg) } & 3 & 2 & 4 \\ \hline \end{array}$$ (a) Find the inverse of the matrix $$ \left[\begin{array}{lll} 3 & 1 & 3 \\ 4 & 2 & 4 \\ 3 & 2 & 4 \end{array}\right] $$ and use it to solve the remaining parts of this problem. (b) How many ounces of each food should the nutritionist feed his laboratory rats if he wants their daily diet to contain 10 mg of folic acid, 14 mg of choline, and 13 mg of inositol? (c) How much of each food is needed to supply 9 mg of folic acid, 12 mg of choline, and 10 mg of inositol? (d) Will any combination of these foods supply 2 mg of folic acid, 4 mg of choline, and 11 mg of inositol?
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