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Chapter 6: Problem 17

Solve each system by substitution. $$\begin{aligned}&3 x-2 y=12\\\&5 x=4-2 y\end{aligned}$$

Short Answer

Expert verified
The solution is \(x = 2\) and \(y = -3\).

Step by step solution

01

Rearrange Second Equation

Start by solving the second equation for one of the variables. Given the equation \(5x = 4 - 2y\), let's solve for \(x\). \[x = \frac{4 - 2y}{5}\] This makes it prepared for substitution in the next step.
02

Substitute Expression in First Equation

Substitute the expression for \(x\) obtained in the previous step into the first equation, \(3x - 2y = 12\). This gives: \[3\left(\frac{4 - 2y}{5}\right) - 2y = 12\] Which simplifies to: \[\frac{12 - 6y}{5} - 2y = 12\] Multiply everything by 5 to eliminate the fraction: \[12 - 6y - 10y = 60\]
03

Solve for y

Combine like terms and solve the simplified equation for \(y\): \[12 - 16y = 60\] Rearranging gives: \[-16y = 60 - 12\] Thus: \[-16y = 48\] Dividing by \(-16\) gives: \[y = -3\] Now, we have the value of \(y\).
04

Substitute y Back to Find x

Use the value of \(y = -3\) and substitute it back into the expression for \(x\) from Step 1: \[x = \frac{4 - 2(-3)}{5}\] Simplifying gives: \[x = \frac{4 + 6}{5} = \frac{10}{5} = 2\] Thus, the value of \(x\) is 2.
05

Conclude the Solution

The solution to the system of equations is \(x = 2\) and \(y = -3\). This means the two lines intersect at the point (2, -3).

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Key Concepts

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

Substitution Method
The substitution method is a technique used to solve systems of linear equations by replacing one variable with an equivalent expression from another equation. This approach is particularly useful when one of the equations in the system can easily be rearranged to express a variable in terms of the other. Here is how you can approach this method:
  • First, take one of the linear equations and solve it for one of the variables.
  • Substitute this expression into the other equation, allowing you to solve for the remaining variable.
  • Once found, substitute the value back into the first equation to find the second variable.
This method simplifies a system of equations to a single equation with one variable, which is easier to manage. It works best when one of the equations is straightforward to isolate either variable.
Linear Equations
Linear equations are equations of the first degree, meaning they involve only the first power of the variables. They represent straight lines when graphed on a coordinate plane. Linear equations usually take the form:\[ ax + by = c \]where \( a \), \( b \), and \( c \) are constants. In a system of linear equations, two or more lines can either intersect, be parallel, or be the same line. Solutions to these systems correspond to points where the lines intersect. Understanding linear equations is key:
  • They form the basis of more complex mathematical concepts.
  • They can represent real-life situations such as speed-distance-time relationships.
Being comfortable with linear equations allows for solving systems efficiently, providing a solid foundation for understanding the relationships between variables.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to make them easier to solve. This skill is essential when using the substitution method and solving linear equations. Here are some principles:
  • Reorganize terms to isolate variables, often requiring addition or subtraction of terms.
  • Use multiplication or division to simplify fractions and eliminate denominators.
  • Combine like terms to make equations simpler and more manageable.
These techniques often involve balancing the equation by performing operations on both sides to maintain equality. Algebraic manipulation allows you to transform complex equations into simpler forms, ensuring an efficient way to find solutions in systems of equations.

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Most popular questions from this chapter

A wholesaler of party goods wishes to display her products at a convention of social secretaries in such a way that she gets the maximum number of inquiries about her whistles and hats. Her booth at the convention has 12 square meters of floor space to be used for display purposes. A display unit for hats requires 2 square meters, and one for whistles requires 4 square meters. Experience tells the wholesaler that she should never have more than a total of 5 units of whistles and hats on display at one time. If she receives three inquiries for each unit of hats and two inquiries for each unit of whistles on display, how many of each should she display in order to get the maximum number of inquiries? What is that maximum number?

Use a graphing calculator to solve each system. Express solutions with approximations to the nearest thousandth. $$\begin{aligned}&\pi x+e y=3\\\&e x+\pi y=4\end{aligned}$$

The total spending on Black Friday during 2011 and 2012 was 1858 million dollars. From 2011 to \(2012,\) spending increased by 226 million dollars. (Source: www.marketingcharts.com) (a) Write a system of equations whose solution represents the Black Friday spending in each of these years. Let \(x\) be the amount spent in 2012 and \(y\) be the amount spent in 2011. (b) Solve the system. (c) Interpret the solution.

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. (Refer to Exercise 92 .) Suppose that supply is related to price by \(p=\frac{1}{10} q\) and that demand is related to price by \(p=15-\frac{2}{3} q,\) where \(p\) is price in dollars and \(q\) is the quantity supplied in units. (a) Determine the price at which 15 units would be supplied. Determine the price at which 15 units would be demanded. (b) Determine the equilibrium price at which the quantity supplied and quantity demanded are equal. What is the demand at this price?

In certain parts of the Rocky Mountains, deer are the main food source for mountain lions. When the deer population \(d\) is large, the mountain lions ( \(m\) ) thrive. However, a large mountain lion population drives down the size of the deer population. Suppose the fluctuations of the two populations from year to year can be modeled with the matrix equation $$\left[\begin{array}{c} m_{n+1} \\ d_{n+1} \end{array}\right]=\left[\begin{array}{cc} 0.51 & 0.4 \\ -0.05 & 1.05 \end{array}\right]\left[\begin{array}{l} m_{n} \\ d_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of animals in the two populations after \(n\) years and \(n+1\) years, where the number of deer is measured in hundreds. (a) Give the equation for \(d_{n+1}\) obtained from the second row of the square matrix. Use this equation to determine the rate the deer population will grow from year to year if there are no mountain lions. (b) Suppose we start with a mountain lion population of 2000 and a deer population of \(500,000\) (that is, 5000 hundred deer). How large would each population be after 1 year? 2 years? (c) Consider part (b), but change the initial mountain lion population to \(4000 .\) Show that the populations would both grow at a steady annual rate of \(1 \$ 6\).
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