Problem 106 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} y=-\frac{2}{3} x+5 \\ 2 x+3 y=11 \end{array}\right. $$

Step-by-Step Solution

Answer

The system of equations has no solution.

Step 1 - Express One Variable in Terms of Another

The first equation is already solving for y: \( y = -\frac{2}{3}x + 5 \).This expression can now be used to substitute y in the second equation.

Step 2 - Substitute for y in the Second Equation

Substitute \( y = -\frac{2}{3}x + 5 \) into the second equation: \( 2x + 3(-\frac{2}{3}x + 5) = 11 \)

Step 3 - Simplify the Equation

Distribute and simplify: \[ 2x + 3(-\frac{2}{3}x + 5) = 11 \] \[ 2x + (-2x) + 15 = 11 \] \[ 2x - 2x + 15 = 11 \]}, {

Step 4 - Solve for x

Since the coefficients of x cancel out: \( 15 = 11 \). This is a contradiction, so there are no solutions to this system of equations.

Key Concepts

Substitution Method

Solving systems of equations using the substitution method involves replacing one variable in one equation with an expression derived from another equation. This method is useful when one equation is already solved for a particular variable.
For our problem, the first equation is given as: \( y = -\frac{2}{3}x + 5 \). This equation is already solved for \( y \), making it easy to substitute into the second equation. By plugging this expression into the second equation, we simplify our problem to a single variable.

• This helps get rid of one variable.
• We then solve for the remaining variable.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operators (like + and -). In our exercise, the expression \( y = -\frac{2}{3}x + 5 \) is what we substitute into the second equation.

• An expression represents a value and can be simplified or manipulated.
• Substituting \( y \) in the second equation changes it into an equation involving only \( x \).

This substitution changes the second equation to \( 2x + 3(-\frac{2}{3}x + 5) = 11 \). Simplifying this will allow us to solve for \( x \).

Contradictions in Algebra

Sometimes while solving systems of equations, you encounter a contradiction. A contradiction occurs when simplification leads to a statement that is always false. In our exercise, after simplifying, we get \( 2x + (-2x) + 15 = 11 \) or \( 15 = 11 \). Since 15 is not equal to 11, we have reached a contradiction.
This means our system of equations has no solution. Contradictions indicate that the equations represent two parallel lines that never intersect, meaning there's no common solution.

Linear Equations

Linear equations are equations of the first degree, meaning the variables are not raised to any power other than one. They represent straight lines in a graph. For example, the system given \( y = -\frac{2}{3}x + 5 \) and \( 2x + 3y = 11 \) are both linear equations.
When solving such systems:

• The goal is to find the point where the lines intersect.
• If the lines intersect at one point, there's a unique solution.
• If they are parallel and never intersect, there's no solution, as seen in our example.
• If the lines coincide, there are infinitely many solutions.