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Chapter 6: Problem 43

Solve each system by elimination. $$\begin{array}{l}\frac{2 x-1}{3}+\frac{y+2}{4}=4 \\\\\frac{x+3}{2}-\frac{x-y}{3}=3\end{array}$$

Short Answer

Expert verified
The solution is \(x = 5\) and \(y = 2\).

Step by step solution

01

Simplify the Equations

First, we need to simplify each equation by eliminating the fractions.For the first equation, multiply every term by 12 (the least common multiple of 3 and 4) to clear the fractions:\[12 \left( \frac{2x - 1}{3} \right) + 12 \left( \frac{y + 2}{4} \right) = 12 \times 4\]This simplifies to:\[4(2x - 1) + 3(y + 2) = 48\]Expanding the terms, we get:\[8x - 4 + 3y + 6 = 48\]Combine like terms:\[8x + 3y = 46\]For the second equation, multiply every term by 6 (the least common multiple of 2 and 3):\[6 \left( \frac{x + 3}{2} \right) - 6 \left( \frac{x - y}{3} \right) = 6 \times 3\]This simplifies to:\[3(x + 3) - 2(x - y) = 18\]Expanding the terms, we get:\[3x + 9 - 2x + 2y = 18\]Combine like terms:\[x + 2y = 9\]
02

Align the System for Elimination

We now have the simplified system of equations:1. \[8x + 3y = 46\]2. \[x + 2y = 9\]We need to align these equations for elimination. We will multiply the second equation by a factor that makes the coefficient of \(x\) the same as in the first equation, so we can eliminate \(x\) by subtraction. Multiply the second equation by 8:\[8(x + 2y) = 8(9)\]This gives us:\[8x + 16y = 72\]
03

Subtract to Eliminate x

Subtract the first equation from the modified second equation to eliminate \(x\):1. \[8x + 16y = 72\]2. \[8x + 3y = 46\]Subtract equation 2 from equation 1:\[(8x + 16y) - (8x + 3y) = 72 - 46\]This simplifies to:\[13y = 26\]
04

Solve for y

Divide both sides of the equation by 13 to find \(y\):\[y = \frac{26}{13}\]\[y = 2\]
05

Substitute to Solve for x

Now that we know \(y = 2\), substitute \(y\) into the second simplified equation to solve for \(x\):\[x + 2(2) = 9\]\[x + 4 = 9\]Subtract 4 from both sides:\[x = 5\]
06

Verify the Solution

Substitute \(x = 5\) and \(y = 2\) back into the original equations to verify the solution.First equation: \(\frac{2(5) - 1}{3} + \frac{2 + 2}{4} = 4\)Simplifies to \(\frac{9}{3} + \frac{4}{4} = 3 + 1 = 4\), which is true.Second equation: \(\frac{5 + 3}{2} - \frac{5 - 2}{3} = 3\)Simplifies to \(\frac{8}{2} - \frac{3}{3} = 4 - 1 = 3\), which is true.Since both equations are satisfied, the solution is verified.

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Key Concepts

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

Systems of Equations
A system of equations consists of multiple equations that share the same set of variables. In the given problem, we have two algebraic equations with two variables: \( x \) and \( y \). The goal is to find the values of these variables that satisfy all the equations simultaneously. Key points to remember about systems of equations are:
  • A solution to the system is any set of values for the variables that makes all the equations true.
  • Systems can have one solution (intersect at one point), no solution (parallel lines, no intersection), or infinitely many solutions (the same line).
  • The methods for solving systems include graphing, substitution, and elimination.
In this exercise, the elimination method is used to find the solution. The system first needs to be simplified by removing fractions, which makes it easier to apply the elimination technique.
Elimination Method
The elimination method is a technique for solving systems of equations. It involves removing one variable by adding or subtracting equations. This is achieved by aligning the coefficients of one of the variables so that they cancel each other out when the equations are combined. Here's how it works:
  • First, we manipulate the equations to eliminate fractions, making computations simpler.
  • We align the equations so that the coefficients of one variable are equal across the two equations by scaling one or both equations as needed.
  • The equations are then added or subtracted to eliminate one variable, reducing the system to a single equation in one variable.
  • Once one variable is found, its value is substituted back into one of the original equations to find the other variable.
In this example, we found a common coefficient for the \( x \) term by multiplying the second equation by 8, allowing us to subtract and eliminate the \( x \) variable. This leads directly to solving for \( y \), and subsequent substitution reveals \( x \).
Fractions in Equations
Fractions in algebraic equations can complicate the process of solving them. They require careful handling to simplify the problem. Here鈥檚 a simple approach to deal with fractions:
  • Identify a common denominator for all the fractions in the equations. This simplifies aligning and solving the equations.
  • Multiply every term of the equation (including the constant on the right side) by the common denominator to eliminate fractions.
  • This process transforms the equation into one with whole numbers, which is generally easier to manipulate.
In the example provided, the least common multiple (LCM) for the denominators in the first equation is 12, and for the second, it鈥檚 6. Multiplying through by these LCMs clears the fractions and simplifies further steps. This technique reduces errors and simplifies understanding, especially when using the elimination method.

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Most popular questions from this chapter

The table shows weight \(W,\) neck size \(N,\) overall length \(L,\) and chest size \(C\) for four bears. $$\begin{array}{|c|c|c|c|} \hline W \text { (pounds) } & N \text { (inches) } & L \text { (inches) } & C \text { (inches) } \\ \hline 125 & 19 & 57.5 & 32 \\\ 316 & 26 & 65 & 42 \\ 436 & 30 & 72 & 48 \\ 514 & 30.5 & 75 & 54 \end{array}$$ A. We can model these data with the equation $$ W=a+b N+c L+d C $$ where \(a, b, c,\) and \(d\) are constants. To do so, represent a system of linear equations by a \(4 \times 5\) augmented matrix whose solution gives values for \(a, b, c,\) and \(d\) B. Solve the system. Round each value to the nearest thousandth. C. Predict the weight of a bear with \(N=24, L=63\) and \(C=39 .\) Interpret the result.

Explain how one can determine whether a system is inconsistent or has dependent equations when using the substitution or elimination method.

As the price of a product increases, businesses usually increase the quantity manufactured. However, as the price increases, consumer demand-or the quantity of the product purchased by consumers-usually decreases. The price we see in the market place occurs when the quantity supplied and the quantity demanded are equal. This price is called the equilibrium price and this demand is called the equilibrium demand. The supply of a certain product is related to its price by the equation \(p=\frac{1}{3} q,\) where \(p\) is in dollars and \(q\) is the quantity supplied in hundreds of units. (a) If this product sells for 9 dollars, what quantity will be supplied by the manufacturer? (b) Suppose that consumer demand for the same product decreases as price increases according to the equation \(p=20-\frac{1}{5} q .\) If this product sells for 9 dollars, what quantity will consumers purchase? How does this compare with the quantity being supplied by the manufacturer at this price? (c) On the basis of parts (a) and (b), what should happen to the price? Explain. (d) Determine the equilibrium price at which the quantity supplied and quantity demanded are equal. What is the demand at this price?

Fawn Population To model spring fawn count \(F\) from adult antelope population \(A\), precipitation \(P\), and severity of winter \(W,\) environmentalists have used the equation. $$F=a+b A+c P+d W$$ where the coefficients \(a, b, c,\) and \(d\) are constants that must be determined before using the equation. The table lists the results of four different (representative) years. $$\begin{array}{|c|c|c|c|} \text { Fawns } & \text { Adults } & \begin{array}{c} \text { Precip. } \\ \text { (in inches) } \end{array} & \begin{array}{c} \text { Winter } \\ \text { Severity } \end{array} \\ \hline 239 & 871 & 11.5 & 3 \\ 234 & 847 & 12.2 & 2 \\ 192 & 685 & 10.6 & 5 \\ 343 & 969 & 14.2 & 1 \end{array}$$ A. Substitute the values for \(F, A, P,\) and \(W\) from the table for each of the four years into the given equation \(F=a+b A+c P+d W\) to obtain four linear equations involving \(a, b, c,\) and \(d\) B. Write a \(4 \times 5\) augmented matrix representing the system, and solve for \(a, b, c,\) and \(d\) C. Write the equation for \(F\), using the values from part (b) for the coefficients. D. If a winter has severity \(3,\) adult antelope population 960 and precipitation 12.6 inches, predict the spring fawn count. (Compare this with the actual count of \(320 .\) )

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 line and a circle; two points.
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