Problem 41 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}\frac{x}{2}-\frac{2 y}{5}=\frac{-23}{60} \\ \frac{2 x}{3}+\frac{y}{4}=\frac{-1}{4}\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{-1}{2}\), \(y = \frac{1}{3}\).

Step 1: Solve the First Equation for x

Start by rearranging the first equation to solve for \(x\). The equation is \(\frac{x}{2} - \frac{2y}{5} = \frac{-23}{60}\). Multiply every term by 60 to clear the fractions, yielding: \(30x - 24y = -23\). Solve for \(x\) to get \(x = \frac{24y - 23}{30}\).

Step 2: Substitute x into the Second Equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation \(\frac{2x}{3} + \frac{y}{4} = \frac{-1}{4}\). Substitute \(x = \frac{24y - 23}{30}\) into this equation and simplify. It becomes \(\frac{2}{3}(\frac{24y - 23}{30}) + \frac{y}{4} = \frac{-1}{4}\).

Step 3: Simplify the Substituted Equation

Simplify \(\frac{2}{3}(\frac{24y - 23}{30}) + \frac{y}{4} = \frac{-1}{4}\). This becomes \(\frac{48y - 46}{90} + \frac{y}{4} = \frac{-1}{4}\). Multiply through by 90 to eliminate the fractions: \(48y - 46 + 22.5y = -22.5\).

Step 4: Solve for y

Combine like terms in the equation from Step 3: \(70.5y - 46 = -22.5\). Add 46 to both sides: \(70.5y = 23.5\). Then divide by 70.5: \(y = \frac{23.5}{70.5}\). Simplifying gives \(y = \frac{1}{3}\).

Step 5: Substitute y back to find x

Substitute \(y = \frac{1}{3}\) into the expression for \(x\) found in Step 1: \(x = \frac{24(\frac{1}{3}) - 23}{30}\). Simplify this to find \(x = \frac{8 - 23}{30} = \frac{-15}{30} = \frac{-1}{2}\).

Step 6: Verify the Solution

Substitute \(x = \frac{-1}{2}\) and \(y = \frac{1}{3}\) back into the original equations to check the solution. The first equation becomes \(\frac{-1/2}{2} - \frac{2(1/3)}{5} = \frac{-23}{60}\) which simplifies to \(-\frac{1}{4} - \frac{2}{15} = \frac{-23}{60}\). Similarly, check the second equation. Both check out, confirming the solution is correct.

Key Concepts

Solving Systems of Equations

Solving systems of equations means finding a set of values for the variables that satisfy all the given equations in the system. In algebra, these systems often consist of two or more equations that need to be solved together.
A system can be simple, with just two variables, or complex, involving many more. Importantly, solving these helps in understanding how different relationships can coexist.
There are several methods to tackle these systems:

• Graphical Method: involves plotting each equation on a graph. The solution is where lines intersect.
• Substitution Method: solves one equation for a variable and substitutes it into another.
• Elimination Method: simplifies by removing a variable using addition or subtraction.

Each method has its own benefits, but the substitution method is especially useful when one equation is easily solvable for one variable.

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. These equations graph as straight lines and are foundational in algebra.
An important aspect is their format, often written as \[ ax + by = c \] where \( a \), \( b \), and \( c \) are constants.
These equations are straightforward to solve alone but can become more complex as part of a system. The simplicity in their structure makes them ideal for methods like substitution, where one can isolate a variable with ease. Understanding linear equations is crucial in solving larger systems effectively.

System of Linear Equations

A system of linear equations features multiple linear equations involving the same set of variables.
Typically, in a system like this: \[\begin{align*}\frac{x}{2} - \frac{2y}{5} &= \frac{-23}{60} \\frac{2x}{3} + \frac{y}{4} &= \frac{-1}{4}\end{align*}\] ytikzcd{stand alone in a latex environment.One must find a common solution for both equations.
The unique solution is where the graphs of these equations intersect.

• The solution is a single set of values for \( x \) and \( y \) that satisfy both equations.
• The solution can be checked by substituting back into the original equations to ensure consistency.

Understanding systems like these is foundational for solving real-world problems that involve various conditions and constraints.

Algebraic Solution Methods

Algebraic solution methods are strategies used to find solutions to equations and systems of equations. One of the most popular is the substitution method, used for its simplicity and effectiveness.
In the substitution method, you:

• First, solve one equation for one variable.
• Next, substitute this expression into the other equation.
• This reduces the system to a single equation with one variable, which is easier to solve.

The final step involves checking the solution by substituting back into the original system. This algebraic method is particularly useful when one equation is already solved for a variable or can easily be manipulated, simplifying the process significantly.