Question

Solve the system of linear equations. $$ \begin{array}{l} x+y=-1 \\ 2 x-3 y=8 \end{array} $$

Short answer

The solution is \( x = 1 \) and \( y = -2 \).

Step-by-step solution

- Label the Equations

Label the two equations as Equation 1 and Equation 2 for reference. Equation 1: \( x + y = -1 \) Equation 2: \( 2x - 3y = 8 \)

- Solve for x in Equation 1

Rearrange Equation 1 to express \( x \) in terms of \( y \). \[ x = -1 - y \]

- Substitute x in Equation 2

Substitute the expression for \( x \) from Step 2 into Equation 2 and solve for \( y \). Replace \( x \) with \( -1 - y \) in Equation 2:\[ 2(-1 - y) - 3y = 8 \] Simplify the equation:\[ -2 - 2y - 3y = 8 \] Combine like terms:\[ -2 - 5y = 8 \] Add 2 to both sides:\[ -5y = 10 \] Divide by -5:\[ y = -2 \]

- Substitute y back into Equation 1

Substitute \( y = -2 \) back into the rearranged Equation 1 to find \( x \). \[ x = -1 - (-2) \] Simplify:\[ x = -1 + 2 \] \[ x = 1 \]

- Conclusion

Combine the values obtained for \( x \) and \( y \) to write the solution for the system of equations. The solution is \( x = 1 \) and \( y = -2 \).

Key concepts

The Substitution Method

The substitution method is a powerful tool for solving systems of linear equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This method simplifies the problem by reducing the number of variables and making it easier to solve. In our original exercise, we started by solving Equation 1 for \( x \), which gave us \( x = -1 - y \). This step is crucial because it expresses one variable in terms of another, allowing for substitution.

• First, choose one of the equations to solve for one variable, usually the easier equation or the one that requires the least manipulation.
• Next, substitute the expression obtained into the second equation, effectively 'plugging' it in.
• This reduces the problem to a single-variable equation, which can be solved using basic algebraic techniques.

The key advantage of this method is that it avoids dealing with both variables simultaneously, simplifying the solution process and providing a clear path to the answer.

Understanding Linear Equations

Linear equations are mathematical expressions wherein each term is either a constant or the product of a constant and a single variable. They form straight lines when graphed on a coordinate plane, hence the name 'linear'. In the context of systems of equations, linear equations usually appear as a pair or more equations that share a set of variables.

The general form of a linear equation in two variables is \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are the variables representing unknown quantities. In the given problem:

• Equation 1 is \( x + y = -1 \), which is a simple form of a linear equation.
• Equation 2 is \( 2x - 3y = 8 \), illustrating how multiple coefficients affect the solution process.

Linear equations in systems can have one solution, no solution, or infinitely many solutions, depending on how the corresponding lines interact on a graph.

Techniques in Algebraic Manipulation

Algebraic manipulation involves the use of algebraic techniques to rearrange equations, isolate variables, and ultimately solve them. It is a foundational skill in algebra that simplifies complex problems and reduces them to more manageable parts. In this exercise, we used algebraic manipulation in several key ways:

• First, we rearranged Equation 1 to express \( x \) in terms of \( y \), making \( x = -1 - y \). This step involved isolating one variable by performing operations on both sides.
• Next, we substituted this expression into Equation 2, leading to a new equation with only one variable. This substitution simplified the problem by focusing the solution on a single variable.
• Finally, we combined like terms and used basic arithmetic to solve for \( y \), followed by back-substituting the found value into the modified Equation 1 to get \( x \).

The goal of algebraic manipulation is to isolate variables and solve equations efficiently without changing their equivalence. It is an essential skill for solving not only linear equations but a wide range of mathematical problems.