Problem 20 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{array}{c} x+2 y=1 \\ 5 x-4 y=-23 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution for the given system of equations is \(x=-3\) and \(y=2\).

Step 1: Isolate one variable in the first equation

Rearrange the first equation to isolate \(x\): \(x = 1 - 2y\).

Step 2: Substitute the expression of x into the second equation

Replace \(x\) in the second equation with the expression from Step 1: \(5(1 - 2y) - 4y = -23\) that simplifies into \( 5 - 10y - 4y = -23 \) and finally into \(-14y = -28\).

Step 3: Solve for y

Solve the equation from Step 2 for \(y\). This gives \(y = 2\).

Step 4: Substitute y=2 into x equation

From the expression of \(x\) derived in Step 1: \(x = 1 - 2*2\). This simplifies to \(x = -3\)

Step 5: Solution of the system

The solution of the system is the pair of values \((x, y) = (-3, 2)\). You may plot these equations graphically to verify that they intersect at the point (-3, 2)

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of linear equations. The fundamental principle behind this method is to solve one of the equations for one variable and then substitute this expression into the other equation. This process reduces the system to a single equation with one variable.

For example, let's consider the system given: \begin{align*} x + 2y &= 1 \ 5x - 4y &= -23 \right\brace\right.\br>First, we need to isolate one variable in one of the equations. In this case, isolating the variable 'x' from the first equation seems most straightforward: \(x = 1 - 2y\).Next, we substitute \(1 - 2y\) for 'x' in the second equation, which gives us a single equation in terms of 'y'. After solving for 'y', we substitute this value back into our expression for 'x'. This method is systematic and can be easier to understand compared to other methods, like elimination, especially for students who are new to algebraic solutions of linear equations.

Systems of Linear Equations

A system of linear equations consists of two or more linear equations that have common solutions. Each equation in the system represents a line when graphed on a coordinate plane, and the solution to the system is the point or points where the lines intersect.

Returning to our initial system: \[\begin{align*} x + 2y &= 1 \ 5x - 4y &= -23 \end{align*}\]It can be seen as two lines that need to be solved for their intersection point. When the substitution method is applied effectively, these points of intersection are revealed by the values of 'x' and 'y' that satisfy both equations simultaneously. Understanding how these equations interact on the graph can also provide a visual representation of the solution, adding an extra dimension to the comprehension of the system of equations.

Graphing Utility Verification

After solving equations algebraically, it's often recommended to verify solutions using a graphing utility. A graphing utility allows students to visualize how the equations form lines and how those lines relate to each other, providing a graphical inspection of the solution.

For our system \[\begin{align*} x + 2y &= 1 \ 5x - 4y &= -23\end{align*}\]a graphing utility would plot two lines, and the point of intersection if present, should be at (-3, 2), which is our algebraically derived solution. Verification using a graphing calculator or software not only confirms the accuracy of the solution but also strengthens understanding of the concept of system of equations as a geometrical intersection.

Algebraic Solutions

Algebraic solutions involve manipulating equations using algebra to find the value of unknown variables that satisfy all equations in the system. The process often involves simplifying equations, performing operations to isolate variables, and substituting these values into the other equations until the solution is found.

In our example, once we determined that \(y = 2\) from the second equation after substitution, we found \(x\) by substituting the value of \(y\) back into the isolated equation \(x = 1 - 2y\), calculating as \(x = 1 - 2(2)\) which simplifies to \(x = -3\). By working through these steps methodically, we're ensuring a clear and precise approach to solving for the unknowns. This algebraic approach provides students with essential skills for solving a variety of mathematical problems.