Problem 5 - Elementary Algebra

Question

Determine Whether an Ordered Pair is a Solution of a System of Equations. In the following exercises, determine if the following points are solutions to the given system of equations. $$ \left\\{\begin{array}{l} x+y=2 \\ y=\frac{3}{4} x \end{array}\right. $$ (a) \(\left(\frac{8}{7}, \frac{6}{7}\right)\) (b) \(\left(1, \frac{3}{4}\right)\)

Step-by-Step Solution

Answer

(\(\frac{8}{7}, \frac{6}{7}\)) is a solution. (1, \(\frac{3}{4}\)) is not a solution.

Step 1 - Substitute the Point into the First Equation

For each ordered pair, start by substituting the coordinates into the first equation: \(x + y = 2\). If the equation holds true, proceed to the next equation.

Step 1a - Check Point (\(\frac{8}{7}, \frac{6}{7}\)) against First Equation

Substitute \(x = \frac{8}{7}\) and \(y = \frac{6}{7}\) into the first equation: \(\frac{8}{7} + \frac{6}{7} = \frac{14}{7} = 2\). The point satisfies the first equation.

Step 1b - Check Point (\(1, \frac{3}{4}\)) against First Equation

Substitute \(x = 1\) and \(y = \frac{3}{4}\) into the first equation: \(1 + \frac{3}{4} = \frac{7}{4} ≠ 2\). The point does not satisfy the first equation.

Step 2 - Substitute the Point into the Second Equation

For points that pass the first equation, continue by substituting the coordinates into the second equation: \(y = \frac{3}{4}x\). If the equation holds true, then the point is a solution to the system of equations.

Step 2a - Check Point (\(\frac{8}{7}, \frac{6}{7}\)) against Second Equation

Substitute \(x = \frac{8}{7}\) and \(y = \frac{6}{7}\) into the second equation: \(\frac{6}{7} = \frac{3}{4} \cdot \frac{8}{7}\). Simplify the right hand side: \(\frac{6}{7} = \frac{24}{28} = \frac{6}{7}\). The point satisfies the second equation.

Step 6: Conclusion for (1, \(\frac{3}{4}\))

Since Point (1, \(\frac{3}{4}\)) does not satisfy the first equation, it is not a solution of the system of equations.

Step 7: Conclusion for (\(\frac{8}{7}, \frac{6}{7}\))

Since Point (\(\frac{8}{7}, \frac{6}{7}\)) satisfies both equations, it is a solution of the system of equations.

Key Concepts

ordered pairs

When working with systems of equations, an ordered pair represents a potential solution. An ordered pair is written as \( (x, y) \), meaning it has two components: the x-coordinate and the y-coordinate. These coordinates can either satisfy or not satisfy the system of equations. To determine if an ordered pair is a solution, you substitute the values for x and y into each equation in the system to see if they create true statements.

solution verification

One important skill is verifying if a given point is a solution to a system of equations. To verify, follow these steps:

• First, substitute the x and y values of the point into the first equation of the system.
• Check if the equation holds true. If it does, move on to the second equation.
• Substitute the x and y values into the second equation.
• If both equations are satisfied, you have verified that the point is a solution.

If at any step the equations do not hold true, the ordered pair is not a solution of the system. For example, in the system
\[ \left\{ \begin{array}{l} x+y=2 \ y=\frac{3}{4} x \end{array} \right. \]
we verified that the point \( \left ( \frac{8}{7}, \frac{6}{7} \right ) \) satisfies both equations, confirming it as a solution.

substitution method

The substitution method is a common technique for solving systems of equations. It involves isolating one variable in one equation and substituting its value into the other equation. This method simplifies the system into a single equation with one variable, making it easier to solve. For example, consider the system
\[ \left\{ \begin{array}{l} x+y=2 \ y=\frac{3}{4} x \end{array} \right. \]
Here, we can substitute the expression for y from the second equation into the first equation:
\[ x + \frac{3}{4}x = 2 \]
This combination will often resolve one variable, paving the way to find both the x and y values.

equation substitution

Equation substitution is a critical part of verifying solutions. It involves directly replacing a variable with its corresponding value or expression from another equation. For example, in our system:
\[ \left\{ \begin{array}{l} x+y=2 \ y=\frac{3}{4} x \end{array} \right. \]
We substitute the values of the ordered pair \( ( \frac{8}{7}, \frac{6}{7} ) \).
First, we substitute into the first equation: \[ \frac{8}{7} + \frac{6}{7} = 2 \]
Upon validation, if true, we then substitute into the second equation: \[ \frac{6}{7} = \frac{3}{4} \cdot \frac{8}{7} \]
If both substitutions produce true statements, the pair is a solution to the system.