Problem 22 - Algebra 1: Concepts and Skills

Question

Estimate the solution of the linear system graphically. Then check the solution algebraically. $$ \begin{array}{r} {3 x+6 y=15} \\ {-2 x+3 y=-3} \end{array} $$

Step-by-Step Solution

Answer

The estimated solution is incorrect as it is does not hold true when substituting back into the second equation. Therefore, the estimation needs to be adjusted or checked for accuracy.

Step 1: Graph the Equations

Begin by re-writing the equations in slope-intercept form \(y = mx + b\). This gives \(y = -0.5x + 2.5\) and \(y = \frac{2}{3}x + 1\). Now plot these two equations on a graph.

Step 2: Estimate the Intersection Point

From the graph, estimate where the two lines intersect, this point is the solution to the system of equations. For these equations, it appears they intersect at (1,2).

Step 3: Check the Solution Algebraically

Take the estimated solution (1,2) and substitute it back into the original equations. If both equations hold true, then the estimated solution is correct. For the first equation, we get \(3(1) + 6(2) = 15\) which holds true. For the second equation we get \(-2(1) + 3(2) = -2 + 6 = 4 \neq -3\), which is not true.

Key Concepts

Slope-Intercept Form

Understanding the slope-intercept form of a line, which is expressed as \( y = mx + b \), is fundamental when graphing linear equations. In this form, \( m \) represents the slope or gradient of the line, indicating its steepness and direction. The slope is calculated by the rise over the run between any two points on the line. Meanwhile, \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

When graphing linear systems, converting equations to this form allows for quick and easy plotting. For the given problem, the two equations are initially in standard form. The first step is to solve each for \( y \).

For example, take the equation \( 3x + 6y = 15 \). To convert it to slope-intercept form, you first isolate \( y \) as follows:

• Subtract \( 3x \) from both sides to get \( 6y = -3x + 15 \).
• Divide everything by 6 to obtain \( y = -0.5x + 2.5 \).

This process reveals the slope \( m = -0.5 \) and the y-intercept \( b = 2.5 \). Repeating this for the second equation \( -2x + 3y = -3 \) gives us \( y = \frac{2}{3}x + 1 \), identifying a slope of \( \frac{2}{3} \) and a y-intercept of 1.

System of Equations

A system of equations is a set of two or more equations with the same variables which are solved together to find a common solution. Graphically, for two-dimensional cases with two variables, the solution is the point where the lines representing the equations intersect.

By graphing the slope-intercept forms of the given linear equations, as shown in the initial problem, we estimate where these lines cross. It's important to use accurate plotting and a well-scaled graph to minimize error in this step. In our case, the lines represented by \( y = -0.5x + 2.5 \) and \( y = \frac{2}{3}x + 1 \) seem to intersect at the point (1,2), implying that both x equals 1 and y equals 2 is a solution to the system. However, graphing provides only an estimated solution; an algebraic check is crucial to verify the exactness of this solution.

Algebraic Solution Check

To confidently confirm the solution of a linear system estimated by graphing, an algebraic check is performed. This involves substituting the x and y values of the estimated point into the original equations.

Following the estimated solution point (1,2), let's substitute into the equations:

• For the first equation \(3x + 6y = 15\), substituting x = 1 and y = 2 gives us \(3(1) + 6(2) = 3 + 12 = 15\), which is accurate.
• However, for the second equation \(-2x + 3y = -3\), the same substitution gives us \(-2(1) + 3(2) = -2 + 6 = 4\), which does not equal -3, indicating an error.

This check shows that the point (1,2) does not satisfy both original equations simultaneously, thus it is not the solution to the system. The discrepancy between the graphical estimate and algebraic check highlights the importance of precision in graphing and the necessity of follow-up verification. For precise solutions, one must solve the equations algebraically, perhaps by methods such as substitution or elimination, to find the exact point of intersection.