Problem 88 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} x+2 y=-1 \\ 2 x+3 y=1 \end{array}\right. $$

Step-by-Step Solution

Answer

x = 5, y = -3

Step 1: Solve for one variable

Choose one of the equations and solve for one of the variables. Here, solve for x in the first equation: \(x + 2y = -1\) \(x = -1 - 2y\)

Step 2: Substitute into the other equation

Substitute the expression for x from Step 1 into the second equation: \(2x + 3y = 1\) Replace \(x\) with \(-1 - 2y\): \(2(-1 - 2y) + 3y = 1\)

Step 3: Simplify and solve for y

Solve the equation from Step 2 for y: \(2(-1 - 2y) + 3y = 1\) Distribute: \(-2 - 4y + 3y = 1\) Combine like terms: \(-2 - y = 1\) Add 2 to both sides: \(-y = 3\) Divide by -1: \(y = -3\)

Step 4: Substitute back to find x

Use the value of y found in Step 3 to find x. Substitute \(y = -3\) back into the expression for x: \(x = -1 - 2y\) Substitute \(y\): \(x = -1 - 2(-3)\) Simplify: \(x = -1 + 6\) \(x = 5\)

Key Concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In algebra, letters like \( x \) and \( y \) can represent numbers. This allows us to write equations and solve problems more generally rather than dealing with specific numbers.
For example, in the equation \( x + 2y = -1 \), \( x \) and \( y \) are variables that can take on different values. The goal in an algebra problem is usually to find the value of these variables that satisfy the equation.
This becomes crucial when solving more complex problems involving multiple equations, where we need to find a solution that works for all equations at the same time.

Substitution Method

The substitution method is one way to solve a system of equations. It involves solving one of the equations for one of the variables and then substituting that solution into the other equation.
Here’s a quick guide on how to use the substitution method:

• Isolate one of the variables in one of the equations.
• Substitute this expression into the other equation.
• Solve the new equation for the remaining variable.
• Substitute the found variable back into the original isolated variable equation to find the other variable.

For instance, solving \( x \) from \( x + 2y = -1 \) gives us \( x = -1 - 2y \). We then substitute \( x \) in the second equation \(2x + 3y = 1\). This turns it into an equation with one variable that we can solve.

Linear Equations

A linear equation is an equation that makes a straight line when it’s graphed. It generally has the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. Linear equations are fundamental in algebra.
They are easy to recognize because they only involve the first power of the variable (e.g., \( x \) and \( y \)). When you have two linear equations like \( x + 2y = -1 \) and \( 2x + 3y = 1 \), you can use methods like substitution to find their intersection point, i.e., the values of \( x \) and \( y \) that satisfy both equations simultaneously.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving a system means finding a set of values for the variables that satisfies all equations in the system.
There are different methods to solve a system of equations: substitution, elimination, and graphing.
In our given system:

1. \( x + 2y = -1 \)
2. \( 2x + 3y = 1 \)

We used the substitution method to find the solution. This involved first isolating \( x \) in the first equation and then substituting this expression into the second equation. Finally, we solved for \( y \) and then back-substituted to find \( x \). The solution \( (x, y) = (5, -3) \) is a point where both equations intersect when graphed.