Problem 26
Question
\(21-48=\) Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example \(6 .\) $$ \left\\{\begin{array}{r}{x+y=7} \\ {2 x-3 y=-1}\end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution is \((4, 3)\).
1Step 1: Write Down the System of Equations
The given system of equations is: 1. \( x + y = 7 \) 2. \( 2x - 3y = -1 \)
2Step 2: Solve the First Equation for y
Rearrange the first equation to solve for \( y \): \[ y = 7 - x \]
3Step 3: Substitute y in the Second Equation
Substitute \( y = 7 - x \) into the second equation: \[ 2x - 3(7 - x) = -1 \] Simplify to: \[ 2x - 21 + 3x = -1 \] Combine like terms to get: \[ 5x - 21 = -1 \]
4Step 4: Solve for x
Add 21 to both sides to isolate \( x \): \[ 5x = 20 \] Now divide by 5: \[ x = 4 \]
5Step 5: Substitute x into the Expression for y
Now that you have \( x = 4 \), substitute it back into \( y = 7 - x \): \[ y = 7 - 4 = 3 \]
6Step 6: Write the Solution as an Ordered Pair
The solution to the system of equations is the ordered pair \( (4, 3) \).
Key Concepts
Linear EquationsSubstitution MethodOrdered Pair SolutionAlgebraic Manipulation
Linear Equations
Linear equations are mathematical expressions showing the relationship between two variables in a straight line when plotted on a graph. They are represented in the standard form as \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. In our exercise, we have two linear equations:
- \(x + y = 7\)
- \(2x - 3y = -1\)
Substitution Method
The substitution method is a straightforward approach to solving systems of equations. It involves solving one equation for a single variable and then substituting the expression into the other equation. In this exercise, the substitution method is implemented in several stages:
- First, solve one of the equations for a variable. Here, we solve \(x + y = 7\) for \(y\): \(y = 7 - x\).
- Then, substitute this expression for \(y\) in the second equation \(2x - 3y = -1\). This allows us to work with only one variable, simplifying the problem.
Ordered Pair Solution
An ordered pair solution is a pair of values for the variables \(x\) and \(y\) that satisfy both equations in the system simultaneously. In this exercise, after we find specific values for \(x\) and \(y\), we write them as \((x, y)\). For our solution, the ordered pair is \((4, 3)\), meaning \(x = 4\) and \(y = 3\). This is crucial because an ordered pair explicitly shows the point of intersection for the two lines represented by the equations. Always ensure that this pair satisfies both original equations by substituting back into the initial linear equations to verify the solution.
Algebraic Manipulation
Algebraic manipulation refers to the use of various techniques to simplify or rearrange algebraic expressions to solve equations. In our example, several manipulations are crucial:
- Rearranging the first equation, \(x + y = 7\), to solve for \(y\). This rearrangement is a foundational step to isolate variables.
- Substituting expressions into other equations to eliminate variables like our substitution step for \(y = 7 - x\) in \(2x - 3y = -1\).
- Simplifying expressions, combining like terms, and using addition or division to isolate the variable \(x\).
Other exercises in this chapter
Problem 26
\(19-32\) . Find all solutions of the system of equations. $$ \left\\{\begin{array}{l}{x+\sqrt{y}=0} \\ {y^{2}-4 x^{2}=12}\end{array}\right. $$
View solution Problem 26
\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{9 x^{2}-9 x+6}{2 x^{3}-x^{2}-8 x+4} $$
View solution Problem 26
\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} 2 y+z &=3 \\ 5 x+4 y+3 z &=-1 \\ x-3 y
View solution Problem 26
The matrices \(A, B, C, D, E, F, G\) and \(H\) are defined as follows. $$ A=\left[\begin{array}{rr}{2} & {-5} \\ {0} & {7}\end{array}\right] \quad B=\left[\begi
View solution