Problem 32 - Algebra 1

Question

Use the substitution method to solve the linear system. $$\begin{aligned} &x+y=12\\\ &x+\frac{3}{2} y=\frac{3}{2} \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system is \(x=33, y=-21\).

Step 1: Rearrange the first equation

First, rearrange the first equation to express \(x\) in terms of \(y\). So, equation \(x+y=12\) will become \(x=12-y\).

Step 2: Substitution into the second equation

Plug \(x=12-y\) from the first equation into the second one. So now, instead of \(x+\frac{3}{2} y=\frac{3}{2}\), you will have \((12-y)+\frac{3}{2} y=\frac{3}{2}\).

Step 3: Solve for \(y\)

Solving the equation in Step 2 will give \(y\). So, \((12-y)+\frac{3}{2} y=\frac{3}{2}\) simplifies to \(\frac{1}{2}y=-\frac{21}{2}\), which gives \(y=-21\).

Step 4: Find \(x\)

Substitute \(y=-21\) back into the equation \(x=12-y\) from Step 1 to solve for \(x\). This gives \(x=12-(-21)\), which simplifies to \(x=33\).

Key Concepts

Understanding Linear Equations

Linear equations are equations of the first degree, which means they contain no exponents greater than one. They are typically in the form of \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. These equations produce straight lines when graphed on a coordinate plane.

• They represent relationships with constant rates of change.
• Solutions to linear equations are any ordered pair \((x, y)\) that makes the equation true when the values are substituted in.

In the given exercise, we deal with two linear equations: \(x + y = 12\) and \(x + \frac{3}{2}y = \frac{3}{2}\). These equations reflect linear relationships, each representing a line on a graph.
The intersection of these lines, or the point where they meet, is the solution that satisfies both equations simultaneously.

Exploring Systems of Equations

A system of equations consists of two or more equations that share the same set of variables. The solution to a system of equations is the set of variable values that satisfies all the equations in the system. There are several types of solutions a system can have:

• One solution: The lines intersect at a single point.
• No solution: The lines are parallel and never meet.
• Infinitely many solutions: The lines coincide completely, representing the same line.

In our specific problem, we have a linear system comprised of \(x+y=12\) and \(x+\frac{3}{2}y=\frac{3}{2}\). Our task was to find the point \((x, y)\) where both lines intersect, revealing the values that satisfy both equations at once.
By implementing the substitution method, we determined that this system has a unique solution, \(x = 33\) and \(y = -21\), where both lines meet exactly at that point.

Solving Equations Using the Substitution Method

The substitution method is an algebraic technique for solving systems of linear equations. It involves expressing one variable in terms of another and substituting this expression into another equation. This helps reduce the number of variables, making it easier to solve the system.
Here's a simple breakdown of the substitution method, using our original problem:

• Step 1: Solve one of the equations for one variable. We rearranged \(x + y = 12\) to obtain \(x = 12 - y\).
• Step 2: Substitute this expression into the other equation. We placed \(x = 12 - y\) into \(x + \frac{3}{2}y = \frac{3}{2}\), resulting in \((12 - y) + \frac{3}{2}y = \frac{3}{2}\).
• Step 3: Solve the resulting single-variable equation. After simplifying, we found \(y = -21\).
• Step 4: Substitute back. Finally, substituting \(y = -21\) back into \(x = 12 - y\), we found \(x = 33\).

Using substitution helps isolate variables efficiently and is especially useful when one of the equations in the system is easily rearranged. This method not only provides a systematic way to solve systems of linear equations but also reinforces the understanding of the interdependence of variables within a system.