Problem 154 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} x+\frac{1}{2} y=\frac{3}{2} \\ \frac{1}{5} x-\frac{1}{5} y=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is x = -3 and y = 9.

Step 1: Solve for one variable in one equation

Choose the first equation to solve for one of the variables. From the first equation, solve for x: \[ x + \frac{1}{2}y = \frac{3}{2} \] Subtract \( \frac{1}{2} y \) from both sides: \[ x = \frac{3}{2} - \frac{1}{2} y \]

Step 2: Substitute into the second equation

Substitute \( x = \frac{3}{2} - \frac{1}{2} y \) into the second equation: \[ \frac{1}{5} (\frac{3}{2} - \frac{1}{2} y) - \frac{1}{5} y = 3 \]

Step 3: Simplify and solve for y

Distribute \( \frac{1}{5} \) in the second equation: \[ \frac{1}{5} \cdot \frac{3}{2} - \frac{1}{5} \cdot \frac{1}{2} y - \frac{1}{5} y = 3 \] \[ \frac{3}{10} - \frac{1}{10} y - \frac{1}{5} y = 3 \] Combine like terms (convert \( \frac{1}{5} \) to \( \frac{2}{10} \)): \[ \frac{3}{10} - \frac{3}{10} y = 3 \] Isolate \( y \) by subtracting \( \frac{3}{10} \) from both sides and dividing by \( -\frac{3}{10} \): \[ -\frac{3}{10} y = 3 - \frac{3}{10} \] Convert 3 to a fraction with a denominator of 10: \[ 3 = \frac{30}{10} \] \[ 3 - \frac{3}{10} = \frac{30}{10} - \frac{3}{10} = \frac{27}{10} \] \[ -\frac{3}{10} y = \frac{27}{10} \] \[ y = -\frac{27}{10} \cdot \frac{10}{-3} = -\frac{27}{10} \cdot -\frac{10}{3} = 9 \] So, \( y = 9 \)

Step 4: Substitute back to find x

Use the value of \( y \) to find \( x \) from the equation \( x = \frac{3}{2} - \frac{1}{2} y \): \[ x = \frac{3}{2} - \frac{1}{2} \cdot 9 \] \[ x = \frac{3}{2} - \frac{9}{2} \] \[ x = \frac{3 - 9}{2} \] \[ x = \frac{-6}{2} \] \[ x = -3 \].

Key Concepts

Substitution Method

The substitution method is a way to solve systems of linear equations. It involves isolating one variable in one of the equations and substituting this expression into the other equation.
This helps to transform a system of equations into a single equation with one variable.
Here's a quick rundown of how the substitution method works:

• Solve one of the equations for one variable.
• Substitute the solved variable's expression into the other equation.
• Simplify and solve for the remaining variable.
• Substitute the found value back to solve for the other variable.

For example, consider the given system: ewline \[\begin{equation} \begin{cases} x + \frac{1}{2} y = \frac{3}{2} \ \frac{1}{5} x - \frac{1}{5} y = 3 \end{cases} \end{equation}\]. First, solve the first equation for x: \[\begin{equation} x = \frac{3}{2} - \frac{1}{2} y \end{equation}\]. Now, substitute this into the second equation and solve for y.

Linear Equations

A linear equation in two variables represents a straight line when graphed on a coordinate plane.
These equations have the general form \[\begin{equation} ax + by = c \end{equation}\], where a, b, and c are constants.
Here are some key characteristics:

• Each variable is raised to the power of 1.
• The graph is a straight line.
• They can have one solution, no solutions, or infinitely many solutions.

In the provided exercise, we have two linear equations: Solving these simultaneously helps us find the point where they intersect.

Solving for Variables

Solving for variables means isolating one variable on one side of the equation to express it in terms of the other variable(s).
This is a crucial step in the substitution method as it allows you to reduce the system of equations step by step.
In our example, we first isolate x in the first equation: \[\begin{equation} x = \frac{3}{2} - \frac{1}{2} y \end{equation}\]. Next, we substitute this expression in place of x in the second equation: \[\begin{equation} \frac{1}{5} (\frac{3}{2} - \frac{1}{2} y) - \frac{1}{5} y = 3 \end{equation}\].
This substitution allows us to solve for y. Once we have the value of y, we substitute it back into the expression for x to find its value.
This step-by-step isolation and substitution ensure that each variable is solved accurately.