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Chapter 6: Problem 41

Solve each system by elimination. $$\begin{array}{l}\frac{x}{2}+\frac{y}{3}=8 \\\\\frac{2 x}{3}+\frac{3 y}{2}=17\end{array}$$

Short Answer

Expert verified
The solution is \((x, y) = (12, 6)\).

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 LCM of 2 and 3 is 6, and for the second equation, it's also 6.\[6\left(\frac{x}{2}\right) + 6\left(\frac{y}{3}\right) = 6 \times 8\] which simplifies to \[3x + 2y = 48\]. For the second equation: \[6\left(\frac{2x}{3}\right) + 6\left(\frac{3y}{2}\right) = 6 \times 17\] simplifies to \[4x + 9y = 102\].
02

Set Up for Elimination

Choose a variable to eliminate; here, we'll eliminate \(x\). Multiply the first equation by 4 and the second equation by 3 to set up for elimination.\[4(3x + 2y) = 4(48)\] gives \[12x + 8y = 192\]; \[3(4x + 9y) = 3(102)\] gives \[12x + 27y = 306\].
03

Eliminate a Variable

Subtract the first modified equation from the second to eliminate \(x\).\[(12x + 27y) - (12x + 8y) = 306 - 192\] results in \[19y = 114\].
04

Solve for \(y\)

Divide both sides by 19 to solve for \(y\).\[y = \frac{114}{19}\] gives \(y = 6\).
05

Substitute to Find \(x\)

Substitute \(y = 6\) back into one of the original equations, \(3x + 2y = 48\). Substitute \(y\) to find \(x\): \[3x + 2(6) = 48\] results in \[3x + 12 = 48\].
06

Solve for \(x\)

Subtract 12 from 48 and then divide by 3 to solve for \(x\).\[3x = 36\] results in \(x = 12\).

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Key Concepts

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

Systems of Equations
Systems of equations involve finding the values of variables that satisfy multiple equations simultaneously. They are a key concept in algebra and can be represented with several different types of equations including linear, quadratic, and more.

For example, the system given in the exercise consists of two linear equations. Solving such systems helps us find the intersection point of two lines on a graph, which corresponds to a set of variables' values applicable to both equations.

There are several methods to solve systems of equations, including substitution, graphing, and elimination. The elimination method, which we explore here, involves adding or subtracting equations to cancel out one of the variables. This simplifies the problem to a single equation with one variable, which is easier to solve. Once you find the value of one variable, you can substitute it back into one of the original equations to find the other variable's value.
Linear Equations
Linear equations are equations of the first degree, meaning they involve variables raised to the first power. They can often be written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

These equations graph as straight lines in the Cartesian coordinate system. The solution to a system of linear equations corresponds to the point where the lines intersect. In the given exercise, both equations are linear, lending themselves well to the elimination method.

A crucial first step is ensuring each equation is free of fractions. This standardizes the approach and reduces mistakes, making it easier to work with coefficients during elimination or other solving techniques. Linear equations have consistent properties, such as a constant slope, further simplifying the process of prediction and calculation when solving.
Fraction Elimination
Fractions can complicate solving systems of equations, so it's often beneficial to eliminate them early in the process. In this exercise, the approach was to multiply each equation by the least common multiple (LCM) of the denominators.

This transformation results in an equivalent system of equations but without the fractions, making further manipulation straightforward.

For example, to clear fractions from the first equation, multiply the entire equation by 6, the LCM of 2 and 3. The equation becomes \(3x + 2y = 48\). Performing similar steps for the second equation simplifies it as well, removing fractional coefficients and resulting in equations that are simpler to handle in the elimination process.

By handling fractions this way, you ensure clarity and reduce potential errors, paving the way for a smoother solution process.

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Most popular questions from this chapter

Solve each application. A company produces two models of bicycles: model \(A\) and model \(B\). Model \(A\) requires 2 hours of assembly time, and model \(B\) requires 3 hours of assembly time. The parts for model \(A\) cost \(\$ 25\) per bike; those for model \(B\) cost \(\$ 30\) per bike. If the company has a total of 34 hours of assembly time and \(\$ 365\) available per day for these two models, what is the maximum number of each model that can be made in a day and use all of the available resources?

To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$\left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of females in the three age groups after \(n\) years and \(n+1\) years. Multiplying the matrices yields the following. $$\begin{aligned} &j_{n+1}=0.33 a_{n}\\\ &s_{n+1}=0.18 j_{n}\\\ &a_{n+1}=0.71 s_{n}+0.94 a_{n} \end{aligned}$$ (Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, "A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape," Conservation Biology, Vol. \(6, \text { No. } 4 .)\) (a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the \(3 \times 3\) matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to \(0.3 .\) Rework part (a) under this new assumption.

Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \(A+B=B+A\) (commutative property)

Draw a sketch of the two graphs described with the indicated number of points of intersection. (There may be more than one way to do this.) A circle and a parabola; four points.

Farmer Jones raises only pigs and geese. She wants to raise no more than 16 animals, with no more than 12 geese. She spends \(\$ 50\) to raise a pig and \(\$ 20\) to raise a goose. She has \(\$ 500\) available for this purpose. Find the maximum profit she can make if she makes a profit of \(\$ 80\) per goose and \(\$ 40\) per pig. Indicate how many pigs and geese she should raise to achieve this maximum.
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