/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Write each system of equations i... [FREE SOLUTION] | 91Ó°ÊÓ

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Write each system of equations in the form \(\left\\{\begin{array}{l}A x+B y=E \\\ C x+D y=F\end{array}\right.\) and then solve the system. $$\left\\{\begin{aligned} -4 y &=x+5 \\ \frac{x}{3}+\frac{y}{2} &=1 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -\frac{1}{2}\), \(y = \frac{2}{13}\).

Step by step solution

01

Rewrite the equations in standard form

The equations provided are: \(-4y = x + 5\) and \(\frac{x}{3} + \frac{y}{2} = 1\) and these need to be re-arranged into the standard form \(Ax + By = E\) and \(Cx + Dy = F\).Starting with the first equation, swap the x and y terms on the left side to give: \(x = -4y - 5\). Multiply through by -1 for the standard form to get:\(-x = 4y + 5\).For the second equation, we need to eliminate the fractions. Multiply through by 3 and 2 to get: \(2x + 3y = 6\).This provides the desired standard form equations:\[-x + 4y = -5\]\[2x + 3y = 6\]
02

Solve the system of equations

There are several ways to solve a system of linear equations. One of the most common is substitution. As we already have \(-x\) in the first equation, we can substitute \(-x\) from the first equation into the \(2x\) in the second equation.This gives: \(2(-x + 4y) + 3y = 6\).Simplify to get: \(-2x + 8y + 3y = 6\).Combine like terms to get: \(-2x + 11y = 6\).Isolate x in the equation: \(x = -\frac{11y - 6}{2}\).Substituting into the first equation we get: \(-(-\frac{11y - 6}{2}) + 4y = -5\).Solving for y yields: \(y = \frac{2}{13}\).Substituting \(y = \frac{2}{13}\) into the equation \(x = -\frac{11y - 6}{2}\) yields: \(x = -\frac{1}{2}\).
03

Check the solution

Substitute \(x = -\frac{1}{2}\) and \(y = \frac{2}{13}\) into the original equations to verify that they satisfy both equations.Substituting into \(-4y = x + 5\), we get: \(-4(\frac{2}{13}) = -\frac{1}{2} + 5\), which simplifies to \(-\frac{8}{13} = -\frac{1}{2} + 5\). This simplifies to \(-\frac{8}{13} = -\frac{13}{26}\), which satisfies the first equation.Substituting into \(\frac{x}{3} + \frac{y}{2} = 1\), we get \(\frac{-\frac{1}{2}}{3} + \frac{\frac{2}{13}}{2} = 1\), which simplifies to \(-\frac{1}{6} + \frac{1}{13} = 1\). This simplifies to \(-\frac{13}{78} + \frac{39}{78} = 1\), which satisfies the second equation.The solution \(x = -\frac{1}{2}\), \(y = \frac{2}{13}\) is therefore correct.

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Key Concepts

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

Linear Equations
A linear equation is an algebraic expression where each term is either a constant or the product of a constant with one variable. In simple terms, it's an equation that forms a straight line when graphed on a coordinate plane. Understanding this concept is essential because it forms the building blocks for solving systems of equations. You typically encounter linear equations in the form of \(y = mx + b\), but they can also be rewritten to fit other forms, such as the standard form. Linear equations contain no exponents on variables, making them quite predictable and straightforward to solve.
  • The variables are raised only to the first power.
  • Graphically illustrated as a straight line on a Cartesian plane.
  • They can be written in multi-variable forms, such as \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants.
These equations are used extensively in numerous fields, including economics, engineering, and ecology. In solving multi-variable equations, techniques like substitution or elimination help determine the values that simultaneously satisfy each equation in the system.
Substitution Method
The substitution method is a straightforward technique to solve a system of equations, particularly helpful when dealing with two equations and two variables. The key objective of substitution is to express one variable in terms of the other. This allows you to substitute back into the other equation to find precise values for each variable.
  • First, solve one of the equations for one variable in terms of the other.
  • Substitute this expression into the other equation.
  • Solve for the remaining variable.
For example, in our exercise, we took the equation \(-x = 4y + 5\) and expressed \(x\) in terms of \(y\). Once we substituted \(-x\) into the second equation, \(2x + 3y = 6\), it simplified down to just one variable. This simplification is the crux of the substitution method.
While effective, this strategy requires careful algebraic manipulation, ensuring equations are rearranged and substituted correctly. This method can be very efficient but can sometimes result in complex expressions, especially with fractions or negative signs, requiring attention to detail.
Standard Form
The standard form of a linear equation is expressed as \(Ax + By = C\), where \(A\), \(B\), and \(C\) are real numbers, and \(x\) and \(y\) are variables. This form is particularly useful in systems of equations as it allows for easier application of solving techniques like substitution or elimination.
  • Rewriting equations into standard form helps to simplify calculations by eliminating fractions or decimals.
  • For graphical interpretations, it helps identify intercepts easily.
  • In our example, reshaping complex equations into standard form clarified the relationships between variables.
Transforming an equation like \(-4y = x + 5\) into \(-x + 4y = -5\) sets the stage for efficient problem-solving. The strategic reformatting of equations reveals a clearer path to finding solutions, making them less prone to algebraic errors during manual calculations. Understanding this format is crucial in not just graphing lines but also in comprehending and solving comprehensive algebraic problems.

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Most popular questions from this chapter

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. Privately owned, single-family homes in a small town were heated with gas, electricity, or oil. The percentage of homes heated with electricity was 9 times the percentage heated with oil. The percentage of homes heated with gas was 40 percentage points higher than the percentage heated with oil and the percentage heated with electricity combined. Find the percentage of homes heated with each type of fuel.

Tara is planning a party for at least 100 people. She is going to serve two types of appetizers: mini pizzas and mini quiche. Each mini pizza costs \(\$ .50\) and each mini quiche costs \(\$ .60 .\) Tara thinks that each person will eat only one item, either a mini pizza or a mini quiche. She also estimates that she will need at least 60 mini pizzas and at least 20 mini quiche. How many mini pizzas and how many mini quiche should Tara order to minimize her cost?

Consider the following system of equations.$$\left\\{\begin{array}{r}x+y=3 \\\\-x+y=1 \\\2 x+y=4\end{array}\right.$$ Use Gauss-Jordan elimination to find the solution, if it exists. Interpret your answer in terms of the graphs of the given equations.

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{ll}1 & -3 \\\1 & -2\end{array}\right]$$

Keith and two of his friends, Sam and Cody, take advantage of a sidewalk sale at a shopping mall. Their purchases are summarized in the following table. $$\begin{array}{lc|c|c|} \hline& {3}{|}\text { Quantity } \\\\\hline\text { Name } & \text { Shirt } & \text { Sweater } & \text { Jacket } \\\\\hline \text { Keith } & 3 & 2 & 1 \\\\\text { Sam } & 1 & 2 & 2 \\\\\text { Cody } & 2 & 1 & 2\\\\\hline\end{array}$$ The sale prices are \(\$ 14.95\) per shirt, \(\$ 18.95\) per sweater, and \(\$ 24.95\) per jacket. In their state, there is no sales tax on purchases of clothing. Use matrix multiplication to determine the total expenditure of each of the three shoppers.

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