Problem 22
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 \(6 .\) $$ \left\\{\begin{array}{l}{x-y=3} \\ {x+3 y=7}\end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution is \( (4, 1) \).
1Step 1: Identify the System Equations
The given system of equations is \( x - y = 3 \) and \( x + 3y = 7 \). Our task is to find values of \( x \) and \( y \) that satisfy both equations.
2Step 2: Solve One Equation for One Variable
Let's solve the first equation for \( x \). From \( x - y = 3 \), we get \( x = y + 3 \).
3Step 3: Substitute Into the Second Equation
Substitute \( x = y + 3 \) into the second equation \( x + 3y = 7 \). This gives \((y + 3) + 3y = 7\).
4Step 4: Simplify and Solve for y
Simplify the equation: \( y + 3 + 3y = 7 \) becomes \( 4y + 3 = 7 \). Solve for \( y \) by subtracting 3 from both sides: \( 4y = 4 \). Divide by 4 to find \( y = 1 \).
5Step 5: Substitute back to Find x
Use \( y = 1 \) in the equation \( x = y + 3 \) to find \( x = 1 + 3 = 4 \).
6Step 6: Verify the Solution
Substitute \( x = 4 \) and \( y = 1 \) back into both original equations to verify. For the first equation: \( 4 - 1 = 3 \) holds true. For the second equation: \( 4 + 3(1) = 7 \) holds true. Both verifications confirm the solution is correct.
Key Concepts
Linear EquationsSubstitution MethodSolution VerificationMathematics Education
Linear Equations
Linear equations form the backbone of many mathematics problems. They describe relationships between variables using a straight line when graphed. A general linear equation has the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables. In this exercise, we deal with a system of two linear equations: \( x-y=3 \) and \( x+3y=7 \).
Such equations are fundamental in algebra and involve finding the point, or points, at which they intersect. If two lines intersect at a single point, as often seen in these problems, there is one unique solution, which is the coordinates of the intersection. Systems of linear equations can also have no solution if the lines are parallel, or infinitely many solutions if the lines overlap completely.
Understanding how linear equations work prepares students for more complex areas of mathematics, such as calculus and matrix algebra.
Such equations are fundamental in algebra and involve finding the point, or points, at which they intersect. If two lines intersect at a single point, as often seen in these problems, there is one unique solution, which is the coordinates of the intersection. Systems of linear equations can also have no solution if the lines are parallel, or infinitely many solutions if the lines overlap completely.
Understanding how linear equations work prepares students for more complex areas of mathematics, such as calculus and matrix algebra.
Substitution Method
The substitution method is a popular technique for solving systems of equations. In this method, you solve one of the equations for one variable and then substitute that expression into the other equation. This reduces the system to a single equation with one variable, making it much easier to handle.
In our solved exercise, we used this method by first solving the equation \( x-y=3 \) for \( x \). We found \( x = y + 3 \). We then substituted this expression for \( x \) into the second equation \( x+3y=7 \), leading to \((y + 3) + 3y = 7\).
This approach is especially effective for systems that offer easy manipulation of one equation into another. The method simplifies many algebra problems and builds a strong foundation for solving even more complicated equations in future studies.
In our solved exercise, we used this method by first solving the equation \( x-y=3 \) for \( x \). We found \( x = y + 3 \). We then substituted this expression for \( x \) into the second equation \( x+3y=7 \), leading to \((y + 3) + 3y = 7\).
This approach is especially effective for systems that offer easy manipulation of one equation into another. The method simplifies many algebra problems and builds a strong foundation for solving even more complicated equations in future studies.
Solution Verification
Solution verification is a crucial step in mathematical problem-solving. Even after deriving a solution, it's essential to verify it to ensure accuracy. This step confirms that the solution satisfies all the original equations given in the problem statement.
In our specific case, after finding \( x = 4 \) and \( y = 1 \), we plugged these values back into both original equations. For the first equation \( x-y=3 \), substituting gives \( 4-1 = 3 \), confirming it holds true. Likewise, substituting into \( x+3y=7 \) gives \( 4+3(1) = 7 \), verifying the second equation.
Verification helps prevent simple errors and emphasizes understanding. It is a practice that ensures the robustness of solutions and strengthens problem-solving skills for students.
In our specific case, after finding \( x = 4 \) and \( y = 1 \), we plugged these values back into both original equations. For the first equation \( x-y=3 \), substituting gives \( 4-1 = 3 \), confirming it holds true. Likewise, substituting into \( x+3y=7 \) gives \( 4+3(1) = 7 \), verifying the second equation.
Verification helps prevent simple errors and emphasizes understanding. It is a practice that ensures the robustness of solutions and strengthens problem-solving skills for students.
Mathematics Education
Mathematics education prioritizes teaching problem-solving skills, critical thinking, and analytical reasoning. Learning methods such as the substitution method during algebra studies prepares students for tackling various real-world problems. This exercise offers a hands-on way to engage with linear equations and understand their use and importance.
Incorporating such exercises into the curriculum promotes active learning. It encourages students to derive solutions themselves while teachers facilitate understanding through further explanation and repetition of the concepts.
Ultimately, mathematics training aims to cultivate lifelong learners who can apply these methods beyond the classroom into practical situations, fostering innovation and technical proficiency in their future careers.
Incorporating such exercises into the curriculum promotes active learning. It encourages students to derive solutions themselves while teachers facilitate understanding through further explanation and repetition of the concepts.
Ultimately, mathematics training aims to cultivate lifelong learners who can apply these methods beyond the classroom into practical situations, fostering innovation and technical proficiency in their future careers.
Other exercises in this chapter
Problem 22
\(19-32\) . Find all solutions of the system of equations. $$ \left\\{\begin{array}{l}{y=4-x^{2}} \\ {y=x^{2}-4}\end{array}\right. $$
View solution Problem 22
The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x+y+z
View solution Problem 22
\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{8 x-3}{2 x^{2}-x} $$
View solution Problem 22
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rrr}{3} & {-2} & {0} \\ {5} & {1} & {1} \\ {2} & {-2} & {0}\end{array}\right] $$
View solution