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91Ó°ÊÓ

Solve each system. State whether it is inconsistent or has infinitely many solutions. If the system has infinitely many solutions, write the solution set with y arbitrary. $$\begin{array}{c} 2 x-6 y=0 \\ -7 x+21 y=10 \end{array}$$

Short Answer

Expert verified
The system is inconsistent and has no solutions.

Step by step solution

01

- Write the system of equations

The system of equations is:\[\begin{cases}2x - 6y = 0 \-7x + 21y = 10\end{cases}\]
02

- Simplify the equations

Notice the first equation can be simplified by dividing all terms by 2:\[x - 3y = 0\]
03

- Express x in terms of y

From the simplified first equation, express x in terms of y:\[x = 3y\]
04

- Substitute x into the second equation

Substitute \(x = 3y\) into the second equation:\[-7(3y) + 21y = 10\]\[-21y + 21y = 10\]which simplifies to 0 = 10, a contradiction.
05

- Determine the nature of the system

Since substituting x into the second equation gives a contradiction (0 = 10), the system is inconsistent and has no solutions.

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Key Concepts

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

Understanding Inconsistent Systems
Inconsistent systems occur when there are no solutions that satisfy all equations involved. This happens because the equations in the system represent parallel lines in a geometric sense.
Imagine two lines on a graph that never meet, they are parallel. When this happens in a system of equations, we have an inconsistent system.
For instance, in the provided exercise, after substituting the value of x from the first equation into the second, we received the impossible equation: 0 = 10.
This contradiction confirms that the two lines represented by the equations never intersect, making the system inconsistent.
When working with systems of equations, always look for such contradictions to determine if the system is inconsistent.
The Substitution Method
The substitution method is a powerful technique for solving systems of linear equations. It involves expressing one variable in terms of another and then substituting this expression into the remaining equations.
Here's a simplified approach:
  • Simplify one of the equations and solve for one variable in terms of the other.
  • Substitute this expression into the other equation(s).
  • Solve the resulting equation for the remaining variable.
  • Substitute back to find the other variable.
The step-by-step example in the exercise used this method:

1. First, we simplified the first equation to express x in terms of y as: \( x = 3y \).
2. Then, we substituted \( x = 3y \) into the second equation.
3. The result was \( -21y + 21y = 10 \), leading to the contradiction \( 0 = 10 \), indicating inconsistency.
Basics of Linear Equations
Linear equations are the foundation of algebra. They describe a straight line when graphed on a coordinate plane. A linear equation in two variables generally looks like \( ax + by = c \), where a, b, and c are constants.
Key features of linear equations include:
  • They have no exponents on the variables.
  • They graph as straight lines.
  • The solution of a linear equation can be a single point, a line (infinitely many solutions), or no point (inconsistent system).
In solving these equations, methods like substitution, elimination, and graphical analysis are commonly used.
The exercise we analyzed showed how linear equations could reveal inconsistent systems by simplifying and substituting to see if the equations made sense together.

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Most popular questions from this chapter

Solve each problem. Purchasing costs The Bread Box, a small neighborhood bakery, sells four main items: sweet rolls, bread, cakes, and pies. The amount of each ingredient (in cups, except for eggs) required for these items is given by matrix \(A\) Eggs lour Sugar Shortening Milk \(\left.\begin{array}{l|ccccc}\text { Rolls (doz) } & 1 & 4 & \frac{1}{4} & \frac{1}{4} & 1 \\ \text { Bread (loaf) } & 0 & 3 & 0 & \frac{1}{4} & 0 \\ \text { Cake } & 4 & 3 & 2 & 1 & 1 \\ \text { Pie (crust) } & 0 & 1 & 0 & \frac{1}{3} & 0\end{array}\right]=A\) The cost (in cents) for each ingredient when purchased in large lots or small lots is given by matrix \(B\) Large Lot Small Lot \(\left.\begin{array}{l|rr}\text { Eggs } & 5 & 5 \\ \text { Flour } & 8 & 10 \\\ \text { Sugar } & 10 & 12 \\ \text { Shortening } & 12 & 15 \\ \text { Milk } & 5 & 6\end{array}\right]=B\) (a) Use matrix multiplication to find a matrix giving the comparative cost per bakery item for the two purchase options. (b) Suppose a day's orders consist of 20 dozen sweet rolls, 200 loaves of bread, 50 cakes, and 60 pies. Write the orders as a \(1 \times 4\) matrix, and, using matrix multiplication, write as a matrix the amount of each ingredient needed to fill the day's orders. (c) Use matrix multiplication to find a matrix giving the costs under the two purchase options to fill the day's orders.

Given \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{rrr}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product when possible. $$C A$$

Use the determinant theorems to find the value of each determinant. $$\left|\begin{array}{rrr} 9 & 1 & 7 \\ 12 & 5 & 2 \\ 11 & 4 & 3 \end{array}\right|$$

Given \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{rrr}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product when possible. $$C B$$

Use the determinant theorems to find the value of each determinant. $$\left|\begin{array}{rrrr} 4 & 0 & 0 & 2 \\ -1 & 0 & 3 & 0 \\ 2 & 4 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right|$$

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