Problem 37 - Algebra A Combined Function

Question

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals. $$ \left\\{\begin{array}{l} \frac{3}{5} x-y=-\frac{4}{5} \\ 3 x+\frac{y}{2}=-\frac{9}{5} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{2}{3}\) and \(y = \frac{2}{5}\).

Step 1: Clear the Fractions

To make it easier to solve, first eliminate the fractions in each equation. Multiply the first equation by 5 and the second equation by 10 to get rid of the fractions: First Equation:\[5 \left( \frac{3}{5} x \right) - 5y = 5 \times -\frac{4}{5}\]This simplifies to:\[3x - 5y = -4\]Second Equation:\[10 \left( 3x + \frac{y}{2} \right) = 10 \times -\frac{9}{5}\]This simplifies to:\[30x + 5y = -18\]

Step 2: Add the Equations

Now add the simplified equations together to eliminate one of the variables.Take:\[3x - 5y = -4\]\[30x + 5y = -18\]Add them together:\[(3x + 30x) + (-5y + 5y) = -4 - 18\]This results in:\[33x = -22\]

Step 3: Solve for x

Now solve the equation from Step 2 for \(x\).Divide both sides by 33:\[x = \frac{-22}{33}\]Simplify the fraction:\[x = -\frac{2}{3}\]

Step 4: Substitute x into One of the Original Equations

Use the value of \(x\) found in Step 3 and substitute it back into one of the original equations to solve for \(y\). Using the first equation:\[\frac{3}{5}(-\frac{2}{3}) - y = -\frac{4}{5}\]Calculate:\[-\frac{2}{5} - y = -\frac{4}{5}\]Add \(\frac{2}{5}\) to both sides:\[-y = -\frac{4}{5} + \frac{2}{5}\]This simplifies to:\[-y = -\frac{2}{5}\]Therefore:\[y = \frac{2}{5}\]

Step 5: Final Solution

The solution to the system of equations is \(x = -\frac{2}{3}\) and \(y = \frac{2}{5}\). Therefore, the point where both lines intersect is \[\left(-\frac{2}{3}, \frac{2}{5}\right)\].

Key Concepts

Understanding the Addition Method

The addition method, also known as the elimination method, is a popular technique used to solve systems of equations. This method is especially useful when the coefficients of one of the variables can be eliminated directly or made to cancel each other out through addition.
In essence, the goal is to add the two equations together to form a new equation with only one variable. Here's how it works:

• First, adjust the equations if necessary so that adding them will result in the elimination of one variable. This often involves multiplying one or both equations by a constant.
• Next, add the equations together, aiming to cancel out one of the variables. This will leave you with a single equation in one variable.
• Solve the resulting equation for the remaining variable.

This method is effective because it simplifies a system of equations into a single-variable problem, making the solution more straightforward.

Clearing Fractions in Equations

When working with systems of equations, especially those that contain fractions, it can be beneficial to first eliminate these fractions. Doing so simplifies the equations, making the addition method much easier to apply. Here's a simple way to clear fractions:

• Identify the least common multiple (LCM) of the denominators in the equation.
• Multiply each term in the equation by this LCM. This will convert the equation to one without fractions, leaving you with whole numbers instead.

For example, if your equation has terms with denominators 5 and 10, multiplying the entire equation by 10 will eliminate the fractions. Once the fractions are cleared, the resulting equation is generally simpler to manage and to solve using methods like addition or substitution.

Applying the Substitution Method

The substitution method is another technique used to solve systems of equations. It involves solving one of the equations for one variable and then substituting this solution into the other equation. Here's how you can use the substitution method:

• Choose one of the equations and solve it for one of the variables. This is usually the variable with the simplest coefficient to isolate.
• Take the expression for this variable and substitute it into the other equation. This will result in an equation with one variable.
• Solve this equation for the unknown variable.
• Use the value found to substitute back into the equation from the first step to find the second variable.

The substitution method is particularly useful when one variable is already isolated or can be easily isolated. It provides a direct approach to reaching a solution by systematically reducing the number of variables you are working with.