Problem 20 - College Algebra

Question

Solve each system by substitution. $$ \begin{array}{l} -\frac{1}{4} x+\frac{3}{2} y=11 \\ -\frac{1}{8} x+\frac{1}{3} y=3 \end{array} $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (-8, 6)\).

Step 1: Solve for x in the first equation

The first equation is \(-\frac{1}{4}x + \frac{3}{2}y = 11\). Solve for \(x\) to use it in the substitution method. First, isolate \(x\):\[\frac{3}{2}y = 11 + \frac{1}{4}x\]Then, solve for \(x\):\[x = \frac{3}{2}y \times 4 - 44 = 6y - 44\]

Step 2: Substitute x in the second equation

Substitute \(x = 6y - 44\) into the second equation: \(-\frac{1}{8}x + \frac{1}{3}y = 3\). Replace \(x\) with the expression we found in Step 1:\[ -\frac{1}{8}(6y - 44) + \frac{1}{3}y = 3\]. Simplify this equation.

Step 3: Simplify and solve for y

Expand and simplify the equation from Step 2:\[-\frac{3}{4}y + \frac{11}{2} + \frac{1}{3}y = 3\]. Find a common denominator and simplify further. Combine the terms to solve for \(y\):\[-\frac{3}{4}y + \frac{1}{3}y = 3 - \frac{11}{2}\].\[ -\frac{9}{12}y + \frac{4}{12}y = -\frac{5}{2}\]. Simplifying gives \(-\frac{5}{12}y = -\frac{5}{2}\). Thus, \(y = 6\).

Step 4: Substitute y back to find x

Substitute \(y = 6\) back into the expression \(x = 6y - 44\) from Step 1 to find \(x\). Compute \(x = 6(6) - 44\). Simplify to get \(x = 36 - 44 = -8\).

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations, especially useful when dealing with linear equations. The idea is to solve one of the equations for one variable, and then substitute that expression into the other equation. This allows you to reduce the system to a single equation with one variable.

This method works well when one of the equations can be easily manipulated to express one variable in terms of the other. In our solution, we start by isolating \(x\) in the first equation. This step-by-step approach simplifies the process:

• Solve one equation for one variable.
• Substitute the expression into the other equation.
• Simplify and solve the resulting equation.

Through these clear steps, the substitution method not only helps in tackling complex systems but also reinforces an understanding of the relationships between the variables.

Linear Equations

Linear equations form the backbone of systems of equations. These equations have one or two variables with constant coefficients; typically, they appear in the form \(ax + by = c\). Each side of a linear equation forms a straight line when graphed, hence the name.

In the given system:

• The equation \(-\frac{1}{4}x + \frac{3}{2}y = 11\) has coefficients of \(-\frac{1}{4}\) for \(x\) and \(\frac{3}{2}\) for \(y\).
• The second equation \(-\frac{1}{8}x + \frac{1}{3}y = 3\) involves similar structure but different coefficients.

When solved graphically, these equations will intersect at a point representing the solution.
This intersection is a core principle in understanding systems of linear equations, showcasing that solutions represent points of overlap in their respective graphs.

Algebraic Solutions

Algebraic solutions involve manipulating algebraic expressions to find the values of variables in a system of equations. Solving systems with algebraic methods like substitution involves a series of steps that require careful handling of expressions.

In our example, after expressing \(x\) in terms of \(y\), we substitute to find \(y\). This involves:

• Simplifying by finding common denominators so both \(\frac{3}{4}y\) and \(\frac{1}{3}y\) could be combined.
• Handling fractions with care to avoid mistakes in arithmetic.
• Backing up the final answer by placing the solved value back into the equation - ensuring consistency of the result.

This approach not only gives the right answer but also helps develop algebraic thinking.
Understanding how variables change and interact algebraically is powerful, laying a foundation for more advanced mathematical concepts.