Problem 42 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{c}\frac{2 x}{3}-\frac{y}{2}=\frac{3}{5} \\\ \frac{x}{4}+\frac{y}{2}=\frac{7}{80}\end{array}\right)\)

Step-by-Step Solution

Answer

\( x = \frac{3}{4} \), \( y = -\frac{1}{5} \).

Step 1: Solve one equation for a variable

Let's solve the second equation for \( x \). The equation is:\[ \frac{x}{4} + \frac{y}{2} = \frac{7}{80} \]We can isolate \( x \) by first subtracting \( \frac{y}{2} \) from both sides:\[ \frac{x}{4} = \frac{7}{80} - \frac{y}{2} \]Next, multiply every term by 4 to solve for \( x \):\[ x = 4 \times \left( \frac{7}{80} - \frac{y}{2} \right) \]Simplify the expression:\[ x = \frac{28}{80} - 2y \] \[ x = \frac{7}{20} - 2y \]. Now, we have expressed \( x \) in terms of \( y \).

Step 2: Substitute into the first equation

Substitute \( x = \frac{7}{20} - 2y \) into the first equation:\[ \frac{2}{3} \left( \frac{7}{20} - 2y \right) - \frac{y}{2} = \frac{3}{5} \]Let’s distribute \( \frac{2}{3} \) in the expression:\[ \frac{2}{3} \times \frac{7}{20} - \frac{4y}{3} - \frac{y}{2} = \frac{3}{5} \]Calculate \( \frac{2}{3} \times \frac{7}{20} \):\[ \frac{14}{60} = \frac{7}{30} \]The equation now is:\[ \frac{7}{30} - \frac{4y}{3} - \frac{y}{2} = \frac{3}{5} \]

Step 3: Clear the fractions and solve for y

To simplify solving, find a common denominator for the fractions in the equation. The least common denominator for 30, 3, and 2 is 30. Multiply each term by 30:\[ 30 \times \frac{7}{30} - 30 \times \frac{4y}{3} - 30 \times \frac{y}{2} = 30 \times \frac{3}{5} \]This simplifies to:\[ 7 - 40y - 15y = 18 \]Combine like terms:\[ 7 - 55y = 18 \]Subtract 7 from both sides:\[ -55y = 11 \]Divide by -55:\[ y = -\frac{11}{55} = -\frac{1}{5} \]. Therefore, \( y = -\frac{1}{5} \).

Step 4: Substitute back to find x

Substitute \( y = -\frac{1}{5} \) into \( x = \frac{7}{20} - 2y \):\[ x = \frac{7}{20} - 2\left(-\frac{1}{5}\right) \]Calculate \( 2 \times -\frac{1}{5} \):\[ -2 \times -\frac{1}{5} = \frac{2}{5} \]Substitute this back:\[ x = \frac{7}{20} + \frac{2}{5} \]Convert \( \frac{2}{5} \) to have a common denominator of 20:\[ \frac{2}{5} = \frac{8}{20} \]So,\[ x = \frac{7}{20} + \frac{8}{20} = \frac{15}{20} = \frac{3}{4} \]. Thus, \( x = \frac{3}{4} \).

Step 5: Conclusion: Provide final solution

After following all the substitution steps, the solution to the system is:\( x = \frac{3}{4} \) and \( y = -\frac{1}{5} \).

Key Concepts

Systems of Equations

Systems of equations are sets of two or more equations that have common variables and are solved together to find a solution that satisfies all equations in the system. Identifying a system of equations is crucial to seeing the connections between variables, which can then be used to find solutions using various methods.

In the exercise we are discussing, we have a system of two linear equations involving the variables \(x\) and \(y\):

• \(\frac{2x}{3} - \frac{y}{2} = \frac{3}{5}\)
• \(\frac{x}{4} + \frac{y}{2} = \frac{7}{80}\)

Such systems are linear because each equation graphs as a straight line. The solution is found at the point where these lines intersect in a coordinate plane. Understanding how these equations represent lines helps in visualizing how different solving methods, like substitution, find this point of intersection.

Algebraic Manipulation

Algebraic manipulation is a crucial skill in solving systems of equations, especially when using methods like substitution or elimination. It involves rearranging and simplifying equations to isolate or eliminate variables. This often requires understanding various algebraic operations such as addition, subtraction, multiplication, division, and finding common denominators.

Let’s look at the substitution method applied here. We first solve one equation for one variable (like solving for \(x\) in \(\frac{x}{4} + \frac{y}{2} = \frac{7}{80}\)) and then use it to substitute in the other equation:

• Subtract \(\frac{y}{2}\) to isolate \(\frac{x}{4}\)
• Multiply every term by 4 to solve for \(x\)
• Carefully handle fractions as you substitute back and forth, finding common denominators to simplify the operation

Breaking down equations into simpler forms allows you to see the relationships between variables clearly, aiding in pinpointing the solution.

Solving Linear Systems

To solve linear systems using substitution, you need to follow a structured approach, which involves substituting one equation into another to reduce the number of variables. This process usually involves:

• Solving one equation for a variable
• Substituting this expression for the variable into the other equation
• Solving the resulting single-variable equation
• Substituting back to find the other variable

In our example, after solving for \(x\) from the second equation, we substituted it into the first equation. By doing this, we transform the system into a single linear equation in terms of \(y\). Solving this gives us \(y = -\frac{1}{5}\). Subsequently, substituting \(y\) back into the expression for \(x\) yields \(x = \frac{3}{4}\). This step-by-step reduction clarifies the solution, making systems of equations manageable and solvable.