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Chapter 2: Problem 19

Solve the given system by back substitution. $$\begin{array}{r} x-2 y=1 \\ y=3 \end{array}$$

Short Answer

Expert verified
x = 7, y = 3.

Step by step solution

01

Understand the System of Equations

We are given two equations: 1. The first equation is \(x - 2y = 1\).2. The second equation is \(y = 3\).Back substitution means we will substitute the value of \(y\) obtained from the second equation into the first equation.
02

Substitute the Value of y

From the second equation, we know that \(y = 3\). Substitute this value into the first equation, which is \(x - 2y = 1\). This gives us:\[ x - 2(3) = 1 \]
03

Simplify the Equation

Simplify the equation by multiplying and solving for \(x\):\[ x - 6 = 1 \]
04

Solve for x

To isolate \(x\), add 6 to both sides of the equation:\[ x = 1 + 6 \]\[ x = 7 \]
05

Present the Solution

The solution to the system of equations is \(x = 7\) and \(y = 3\).

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Key Concepts

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

System of Equations
A system of equations consists of two or more equations that have the same set of variables. Solving these equations involves finding the values that satisfy all the equations in the system simultaneously.
In our example, we have the system:
  • Equation 1: \(x - 2y = 1\)
  • Equation 2: \(y = 3\)
To solve this system, we aim to find the values of \(x\) and \(y\) that satisfy both equations.
The process we use in this case is back substitution, which leverages the known value of one variable to find the other.
Linear Equations
Linear equations are mathematical expressions that represent a straight line on a graph.
They are equations of the first order, which means they have no exponents higher than one.
In the context of our system:
  • First Equation: \(x - 2y = 1\) is a linear equation because it forms a straight line when plotted.
  • Second Equation: \(y = 3\) is also linear and represents a horizontal line where \(y\) is always 3, regardless of \(x\).
Since these equations are linear, they can intersect at a point, which represents the solution to the system.
Solution Methods
There are several methods to solve a system of linear equations, such as
  • Graphical Method
  • Substitution Method
  • Elimination Method
  • Back Substitution
For our specific set of equations, back substitution is used.
This method involves substituting the value of one variable into another equation to find the unknowns.
In our case, we took \(y = 3\) from the second equation and substituted it into the first equation \(x - 2y = 1\), solving it to find \(x\).
This approach is efficient when one of the equations already provides a direct solution for one variable.

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Most popular questions from this chapter

Balance the chemical equation for each reaction. \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{4}+\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{C}_{2} \mathrm{HCl}_{3}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Recall that the cross product of vectors u and v is a vector \(\mathbf{u} \times \mathbf{v}\) that is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). ( See Exploration: The Cross Product in Chapter 1.) If$$\mathbf{u}=\left[\begin{array}{l} u_{1} \\ u_{2} \\ u_{3} \end{array}\right] \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right]$$ show that there are infinitely many vectors $$\mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]$$ that simultaneously satisfy \(\mathbf{u} \cdot \mathbf{x}=0\) and \(\mathbf{v} \cdot \mathbf{x}=0\) and that all are multiples of $$\mathbf{u} \times \mathbf{v}=\left[\begin{array}{l} u_{2} v_{3}-u_{3} v_{2} \\ u_{3} v_{1}-u_{1} v_{3} \\ u_{1} v_{2}-u_{2} v_{1} \end{array}\right]$$

Show that \(\mathbb{R}^{2}=\operatorname{span}\left(\left[\begin{array}{l}1 \\\ 1\end{array}\right],\left[\begin{array}{r}1 \\ -1\end{array}\right]\right)\)

Find the partial fraction decomposition of the given form. (The capital letters denote constants.) $$\frac{3 x+1}{x^{2}+2 x-3}=\frac{A}{x-1}+\frac{B}{x+3}$$

Determine by inspection (i.e., without performing any calculations) whether a linear system with the given augmented matrix has a unique solution, infinitely many solutions, or no solution. Justify your answers. $$\left[\begin{array}{rrrr|r} 1 & 2 & 3 & 4 & 0 \\ 5 & 6 & 7 & 8 & 0 \\ 9 & 10 & 11 & 12 & 0 \end{array}\right]$$
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