Problem 88 - Algebra 2

Question

The line \((y-1)=\frac{2}{3}(x+1)\) contains point \(H(a,-3)\) . Find a. Show your work.

Step-by-Step Solution

Answer

The value of a is -7.

Step 1: Substitute the coordinates of point H into the equation

Since point H has coordinates \(a, -3\), substitute these into the line equation \(y-1)=\frac{2}{3}(x+1)\). Thus, \(y\) becomes \-3\ and \(x\) becomes \(a\).

Step 2: Solve for a

The resulting equation after substitution is \( (-3)-1)=\frac{2}{3}((a)+1)\). Simplify and solve this equation to find the value of \(a\).

Step 3: Simplify the equation

Combine like terms to get \( -4 = \frac{2}{3}(a) + \frac{2}{3}\). Next, subtract \frac{2}{3} from both sides of the equation to get \( -\frac{14}{3} = \frac{2}{3}a\).

Step 4: Find the value of a

Multiply both sides of the equation by \(\frac{3}{2}\) to solve for \(a\): \( a = -\frac{14}{3} * \frac{3}{2}\) which simplifies to \(a = -7\).

Key Concepts

Coordinate Substitution

One of the most useful techniques in solving equations in algebra is coordinate substitution. This technique is particularly helpful when dealing with equations that describe geometric figures like lines, parabolas, or circles. Coordinate substitution involves replacing the variables in an equation with the actual coordinates of a known point that lies on the figure.

In the exercise, we are given a line equation and a point H with the coordinates \(a, -3)\). To find the value of \(a\), we substitute \(-3\) for \(y\) and \(a\) for \(x\) in the line equation. It creates an equation where the only unknown is \(a\), which we can then solve. This practical method simplifies the problem and makes finding \(a\) straightforward once we perform the algebraic manipulation.

Algebraic Manipulation

Simplifying Expressions

Algebraic manipulation plays a crucial role in solving linear equations. It involves simplifying expressions and rearranging terms so we can isolate the variable we're trying to find. Key steps include combining like terms, adding or subtracting terms on both sides, and multiplying or dividing both sides by the same number.

In our problem, after substituting the coordinates, the equation \(-4 = \frac{2}{3}(a) + \frac{2}{3}\) is obtained. To further solve for \(a\), we subtract \(\frac{2}{3}\) from both sides to eliminate the fraction and isolate the term containing \(a\). This is a fundamental step to simplify the equation and prepare it for the final steps to solve for the variable.

Linear Equation Analysis

Mastering linear equation analysis is essential when dealing with problems related to straight lines. A linear equation typically has the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. The analysis involves understanding how changes in these values affect the graph of the line.

In the exercise, the equation of the line is given in a transformed format \((y-1)=\frac{2}{3}(x+1)\). Recognizing that the slope is \(\frac{2}{3}\) and the changed y-intercept is at \((0, 1)\) is pivotal. It helps us understand the relationship between the line and the points that lie on it. A meticulous analysis directs us to the next steps needed to solve for our variable with confidence.

Finding Variable Values

The Goal of Solving Equations

Finding the values of variables is the target of many algebraic problems. This process involves working through the equation step by step using algebraic manipulation until the variable is isolated and its value can be calculated.

In this case, we aim to find the value of \(a\) when a point \(H(a, -3)\) lies on a specific line. After simplifying and rearranging terms, we reach the last step where we multiply both sides by \(\frac{3}{2}\) to find that \(a = -7\). This result is crucial because it represents the exact point on the x-axis that, alongside the y-coordinate \(-3\), sits on the line described by the given equation. Successfully recovering the value of \(a\) demonstrates our ability to navigate the equations with precision.