Problem 47 - College Algebra Essentials

Question

In Exercises \(47-48,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} \frac{x+2}{2}-\frac{y+4}{3}=3 \\ \frac{x+y}{5}=\frac{x-y}{2}-\frac{5}{2} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = \frac{4}{9}\) and \(y = - \frac{4}{3}\)

Step 1: Simplify the equations

Rewrite and simplify both equations to a more standard form. Also equalize both equations to zero: For first equation: \(2(\frac{x+2}{2}-\frac{y+4}{3}) = 2(3)\) which gives \(x+2 - \frac{2(y+4)}{3} = 6\) or \(x - \frac{2y}{3} = 6 - 2 - \frac{8}{3} = 2 - \frac{8}{3} = - \frac{4}{3}\) And for the second equation: \(5(\frac{x+y}{5} - \frac{x-y}{2} + \frac{5}{2}) = 0\) which gives \(x + y - \frac{5(x - y)}{2} + 5 = 0\) or \(\frac{5y - 3x}{2} = - 5\)

Step 2: Solve for one variable

Choose one equation and solve it for one variable. In this case, the first equation can be easily solved for x. It gives \(x = \frac{2y}{3} + \frac{4}{3}\)

Step 3: Substitute the value of x into the second equation

The expression for x from the first equation can be substituted into the second equation. This gives \(\frac{5y - 3(\frac{2y}{3} + \frac{4}{3})}{2} = -5 \) or \(\frac{3y - 6}{2} = -5 \) or \(3y - 6 = -10 \) or \(3y = -4 \) which solve to \(y = - \frac{4}{3}\).

Step 4: Substitute y into first equation

Now place the value of y into first equation to solve for x. This gives \(x = \frac{2(-\frac{4}{3})}{3} + \frac{4}{3} = -\frac{8}{9} + \frac{4}{3} = \frac{4}{9}\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions that form a straight line when graphed on a coordinate plane. They typically have variables raised to the power of one and are in the standard form of \(ax + by = c\). In the given system of equations:

• The equations involve linear terms like \(x\) and \(y\).
• The task is to find the values of \(x\) and \(y\) where both lines intersect, essentially solving the equations simultaneously.

Understanding linear equations is crucial because they are foundational to algebra. Recognizing their form helps in deciding the best method to apply when solving systems involving them.

Substitution Method

The substitution method is a technique used to solve systems of equations by solving one equation for one variable. Then, this expression is substituted into the other equation. This reduces the system of equations into a single equation with one unknown, simplifying the solving process.

• In our example, once the first equation was simplified to \(x = \frac{2y}{3} + \frac{4}{3}\), we could substitute this into the second equation to find \(y\).
• The goal is to express one variable completely in terms of the other and make the substitution.

This method is particularly beneficial when one equation can be easily arranged to isolate one variable, streamlining the solving process.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to make them more workable for solving. This includes simplifying expressions, factoring, or otherwise reconfiguring an equation.

• It allows you to isolate variables and simplifies equations so they can be solved more easily.
• In our system, algebraic manipulation is evident when isolating \(x\) and \(y\) terms to simplify equations for substitution and further steps.
• It often involves combining like terms and performing operations such as multiplying or dividing both sides of an equation to maintain equality.

Through these operations, complex systems become more manageable, and the path to finding a solution becomes clearer.

Fractions in Equations

Handling fractions in equations can sometimes be challenging, yet it's crucial in algebra. When equations contain fractions, the aim is often to clear them to simplify the solving process.

• This can be done by finding a common denominator or multiplying through by a suitable value to eliminate the fractions.
• In our exercise, the manipulation of terms \(\frac{x+2}{2}\) and \(\frac{y+4}{3}\) involved using multiplication to clear fractions from the linear equations.
• Accurate arithmetic with fractions ensures correct equation simplification, leading to the correct solution.

Mastering fractions in equations is a vital skill, as it allows for the solution of complex equations with precision and clarity.