Problem 12 - Algebra 1

Question

Use the graph to decide whether the point lies on the graph of the line. Justify your answer algebraically. \(3 x-4 y=10\) a. (2,-1) b. (-1,2)

Step-by-Step Solution

Answer

Based on the calculations, the point (2,-1) lies on the graph of the line but the point (-1,2) does not.

Step 1: Substitute the Point into the Equation

To algebraically determine if a point lies on a line, one needs to substitute the x and y coordinates of the point into the equation of the line. For the point (2, -1), substitute \(x = 2\) and \(y = -1\) into the equation \(3 x - 4 y = 10\) to obtain \(3(2) - 4(-1)\).

Step 2: Evaluate the Resulting Expression

Perform the operations to see if the left side of the equation equals to the right side. Thus, obtain \(6 + 4\), which equals to 10. Since the right side of the equation is also 10, one can see that the point (2, -1) satisfies the equation.

Step 3: Repeat for the Second Point

Repeat the same process for the second point (-1, 2). Substitute \(x = -1\) and \(y = 2\) into the equation to obtain \(3(-1) - 4(2)\). Evaluating this results in \(-3 - 8\), which equals to -11. Since the right side of the equation is 10, one can see that the point (-1, 2) does not satisfy the equation.

Key Concepts

Points on a Line

Understanding if specific points on a line are a part of the given linear equation is a fundamental concept in algebra, particularly when graphing. The equation of a linear line represents all the points that line contains, and each of these points has coordinates that satisfy the equation.

For example, considering the linear equation provided, one can verify if a point like (2, -1) is on the line by substituting the x and y coordinates into the given equation. If after the substitution, the equation balances or holds true, then the point lies on the line depicted by the equation. If the equation does not hold true after the substitution, then the point is not part of the line.

Let's consider another example for better clarity. If you have an equation such as \( y = 2x + 3 \) and you want to check if the point (1, 5) lies on the line it represents, you substitute x with 1 and y with 5 to get \( 5 = 2(1) + 3 \) which simplifies to \( 5 = 5 \), thereby confirming that the point (1,5) indeed lies on the line.

Substitution Method

The substitution method is a critical algebraic tool for solving systems of linear equations but can also be used when verifying points on a graph. It involves replacing variables with their corresponding values to simplify the equation and reach a solution or verify a condition.

In the context of the given exercise, the substitution method was used to insert the values of the x and y coordinates from the points (2, -1) and (-1, 2) into the equation \(3x - 4y = 10\) to determine if these points satisfy the equation. This method is straightforward and provides a clear-cut decision on whether the given point lies on the line.

Here's how it works step by step: If you have the point (x, y) and the equation \( ax + by = c \), you substitute x with its value from the point and y with its y-value. Simplify the equation, and if both sides equal, the point lies on the line. Otherwise, it does not.

Solving Equations Algebraically

The process of solving equations algebraically often requires understanding how to manipulate and simplify expressions to find the unknown value of variables. It encompasses methods like substitution, elimination, and the use of the distributive property, among others.

When it comes to linear equations, such as the one presented in the exercise, the goal is to determine the values of x and y that make the equation true. Algebraically solving the equation can also involve techniques like graphing or using algebraic properties to isolate variables and find their values.

For instance, in step-by-step problem solving, the evaluative step performed after substitution is part of solving the equation algebraically. By replacing x and y with their known values and simplifying the resulting expression, we were effectively 'solving' it for the given values to see if the equation maintains its balance, in much the same way as one would solve for unknown values.