Problem 36 - Algebra 1: Concepts and Skills

Question

Find three ordered pairs that are solutions of the equation. $$ 3 x-5 y=15 $$

Step-by-Step Solution

Answer

The three ordered pairs that are solutions of the equation are (0,-3), (1,-2.4), and (2,-1.8).

Step 1: Solve for y

Rearrange the equation to solve for \(y\) in terms of \(x\). Start by isolating \(y\). Do this by subtracting \(3x\) from both sides of the equation: \(3x - 5y = 15 \rightarrow -5y = 15 - 3x \). Then, divide everything by -5: \( y = -3 + \frac{3}{5}x \).

Step 2: Substitute values for x

Now that we have \(y\) in terms of \(x\), we can choose various values for \(x\) and substitute them into the equation to get the corresponding \(y\) values. Let's choose \(x = 0\), \(x = 1\), and \(x = 2\).

Step 3: Calculate corresponding y values

Substitute \(x = 0\) into the equation: \(y = -3 + \frac{3}{5}(0) = -3\). So, when \(x = 0\), \(y = -3\) and one ordered pair is (0,-3). Substitute \(x = 1\) into the equation: \(y = -3 + \frac{3}{5}(1) = -2.4\). So, when \(x = 1\), \(y = -2.4\) and another ordered pair is (1,-2.4). Finally, substitute \(x = 2\) into the equation: \(y = -3 + \frac{3}{5}(2) = -1.8\). So, when \(x = 2\), \(y = -1.8\) and the final ordered pair is (2,-1.8).

Key Concepts

Ordered Pairs

Imagine you're on a treasure hunt with a map that uses a grid to locate the treasure. On this map, ordered pairs are like the 'X marks the spot,' giving you the specific location of the treasure. In algebra, ordered pairs are coordinates that pinpoint where a line crosses a point on a graph. They always come in the form of \( (x, y) \) where \( x \) represents the horizontal position and \( y \) represents the vertical position.

When you are given a linear equation, finding ordered pairs involves choosing different values for \( x \) and then solving for \( y \) to see where the line would cross the graph at that particular \( x \) value. To confirm that a pair is a solution, you can plug both \( x \) and \( y \) values into the original equation and see if the equation holds true. If it does, congratulations, you've found the treasure—or at least one of its coordinates!

Algebraic Solutions

When you solve a puzzle, finding the right pieces to fill the gaps is crucial. Similarly, algebraic solutions are the 'pieces' that complete the mathematical puzzle called an equation. These are the specific numbers that make the equation true, like finding the one key that fits a lock.

For linear equations, algebraic solutions are the values of the variables that satisfy the equation. To find these values, you might rearrange the equation, perform operations to isolate the variable, or use methods like substitution or elimination if there are multiple variables involved. In short, an algebraic solution is the final answer that makes the equation balance perfectly, like that last piece snugly fitting into your puzzle.

Linear Equations

The foundations of algebra are built on linear equations, much like how a sturdy house needs a strong base. A linear equation is a straight line when graphed on a coordinate plane and typically looks like \( ax + by = c \) where \( a \) and \( b \) are coefficients, and \( c \) is a constant. The 'linear' part means that the variables \( x \) and \( y \) are to the first power only—no squares, cubes, or higher dimensions involved.

Understanding these equations involves recognizing their patterns, predicting where the line will go on a graph, and knowing how changes to the numbers \( a \) and \( b \) will tilt or shift the line. Finding solutions to these equations is like decoding a message where each solution reveals a bit more about the mysterious line on your graph.

Substitution Method

In cooking, you might swap out an ingredient for another—say, using honey instead of sugar. This is like the substitution method in algebra, where you exchange one variable with an expression that includes another variable, leading you to a tasty final result—the solution. This method is especially useful when you have more than one equation in a system and want to find where the two lines intersect.

You start by solving one equation for one of the variables. Then, you 'substitute' that expression into the other equation, replacing the original variable. This lets you work with just one variable at a time, simplifying your calculations. Once you've found the value for one variable, you plug it back into one of the original equations to find the other variable. And voila, you end up with the perfect 'dish'—an ordered pair that solves the system.