Problem 41 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} a-b=1 \\ \frac{a}{3}+\frac{b}{5}=1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( a = \frac{9}{4} \) and \( b = \frac{5}{4} \).

Step 1: Substitute one variable

First, solve the equation 1 for one of the variables. From the first equation, we have: \[ a - b = 1 \] Solve for \( a \): \[ a = b + 1 \]

Step 2: Substitute into the second equation

Next, substitute the expression for \(a\) from Step 1 into the second equation: \[ \frac{a}{3} + \frac{b}{5} = 1 \] Replace \(a\) with \(b + 1\): \[ \frac{b + 1}{3} + \frac{b}{5} = 1 \]

Step 3: Eliminate the denominators

Find a common denominator to eliminate the fractions. The common denominator of 3 and 5 is 15: \[ \frac{15(b + 1)}{45} + \frac{15b}{75} = 15 \] Simplifying the denominators: \[ 5(b + 1) + 3b = 15 \]

Step 4: Combine like terms

Distribute and combine like terms: \[ 5b + 5 + 3b = 15 \] Combine the \(b\) terms: \[ 8b + 5 = 15 \]

Step 5: Solve for \(b\)

Subtract 5 from both sides of the equation to isolate the \(b\) term: \[ 8b = 10 \] Divide both sides by 8: \[ b = \frac{10}{8} \] Simplify the fraction: \[ b = \frac{5}{4} \]

Step 6: Back-substitute to find \(a\)

Use the original substitution equation \( a = b + 1 \) to find \(a\). Substitute \(b = \frac{5}{4}\): \[ a = \frac{5}{4} + 1 = \frac{5}{4} + \frac{4}{4} = \frac{9}{4} \]

Key Concepts

solving linear equations

Understanding how to solve linear equations is fundamental in algebra. A linear equation is any equation that can be written as \textbf{ax + by = c}\, where \(a,\ b,\text{ and }c\) are constants and \(x,\ y\) are variables. Solving linear equations involves finding the values of \(x\) and \(y\) that make the equation true.

Here's a simple example:
If we have the equation \(2x + 3 = 7\), we want to isolate \(x\).

• Subtract 3 from both sides: \(2x = 4\)
• Divide both sides by 2: \(x = 2\)

Now, try plugging \(x = 2\) back into the original equation to check. If both sides are equal, then you've solved it correctly!

Always ensure each step keeps the equation balanced. This is crucial for accuracy.

substitution method

The substitution method is a fantastic way to solve systems of equations. It involves isolating one variable in one of the equations and then substituting that expression into the other equation.

For example, consider the system:
\textbf{1) a - b = 1}\
\textbf{2) \frac{a}{3} + \frac{b}{5} = 1}\

First, solve Equation 1 for \(a\):

• \(a - b = 1\ Rightarrow a = b + 1\)

Next, substitute this expression for \(a\) into Equation 2:

• \(\frac{b+1}{3} + \frac{b}{5} = 1\)

This transforms your system into a single equation with one variable \(b\). Solve this resulting equation for \(b\). Finally, substitute \(b\) back into the first equation to find \(a\).

This method is particularly useful when one of the equations is easily solved for one of the variables.

algebraic fractions

Algebraic fractions can be tricky, but they are an essential part of solving equations. These fractions have variables in the numerator or the denominator. Simplifying and working with them involves finding a common denominator, which makes adding, subtracting, and solving them much more manageable.

In our earlier example, one of the steps involves the fraction \(\frac{a}{3} + \frac{b}{5} = 1\). Here's how to handle it:

• Identify the common denominator for the fractions (3 and 5, which is 15).
• Eliminate the fractions by multiplying each term by 15: \(15 \cdot \frac{a}{3} + 15 \cdot \frac{b}{5} = 15\ Rightarrow 5a + 3b = 15\).

Working with algebraic fractions also involves simplification of complex fractions and solving them step by step. The key is always to clear the denominators by multiplying both sides by the LCD (Lowest Common Denominator).

Practice makes perfect, so keep practicing these methods to become more confident!