Problem 42 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} \frac{a}{4}-4 b=2 \\ \frac{a}{8}-5 b=2 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( a = \frac{8}{3} \) and \( b = -\frac{1}{3} \).

Step 1 - Simplify the Equations

First, simplify the given equations for easier manipulation:\[ \frac{a}{4} - 4b = 2 \]\[ \frac{a}{8} - 5b = 2 \]

Step 2 - Eliminate the Fractions

To eliminate the fractions, multiply the first equation by 8 and the second equation by 8 to get rid of the denominators:\[ 2a - 32b = 16 \]\[ a - 40b = 16 \]

Step 3 - Substitute or Eliminate

Using the second simplified equation, solve for \(a\): \[ a = 40b + 16 \]Substitute \( a = 40b + 16 \) into the first simplified equation:\[ 2(40b + 16) - 32b = 16 \]

Step 4 - Solve for \(b\)

Expand and solve for \(b\):\[ 80b + 32 - 32b = 16 \]Combine like terms:\[ 48b + 32 = 16 \]Subtract 32 from both sides:\[ 48b = -16 \]Divide by 48:\[ b = -\frac{1}{3} \]

Step 5 - Solve for \(a\)

Substitute \(b = -\frac{1}{3}\) into \( a = 40b + 16 \):\[ a = 40(-\frac{1}{3}) + 16 \]Calculate the value:\[ a = -\frac{40}{3} + 16 = -\frac{40}{3} + \frac{48}{3} = \frac{8}{3} \]

Key Concepts

linear equations

Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable.
They form straight lines when plotted on a graph. Linear equations have the general form of: $$ ax + by = c $$
These equations are fundamental in algebra and are used to model and solve many real-world problems.
You often encounter them in systems, where two or more such equations are solved together to find common solutions. In this exercise, we are dealing with a system of linear equations involving fractions, which is slightly more complex than simple linear equations.

elimination method

The elimination method is a standard technique to solve systems of linear equations. The goal is to eliminate one of the variables by adding or subtracting the equations. Here’s how it works:

• Align the equations for easy manipulation.
• Multiply one or both equations to get the coefficients of one variable to be the same (or opposites).
• Add or subtract the equations to eliminate that variable.
• Solve the resulting single-variable equation.

In our exercise, we first eliminated the fractions by multiplying the entire equations by a common number. Then we used elimination to solve the simplified equations for one variable.

substitution method

In the substitution method, you isolate one variable in one of the equations and then substitute that expression into the other equation. Here are the steps:

• Solve one equation for one variable in terms of the other variable.
• Substitute this expression into the other equation.
• Solve the resulting equation for the single variable.
• Substitute the found value back into the original equation to get the value of the other variable.

In the provided exercise, we used the substitution method after simplifying the equations. We isolated the variable 'a' and substituted it back into the other equation to solve for 'b'. This made it easier to solve the system step-by-step.

fractions in equations

Fractions in equations can complicate the solving process, but they can be effectively managed by clearing the denominators. Here’s how you deal with them:

• Find a common multiple for the denominators of the fractions.
• Multiply every term in the equation by this common multiple to eliminate the fractions.

For example, in our exercise, we had the equations \( \frac{a}{4} - 4b = 2 \) and \( \frac{a}{8} - 5b = 2 \). By multiplying the first equation by 8 and the second equation by 8, we eliminated the fractions, simplifying the equations to: \[ 2a - 32b = 16 \]\[ a - 40b = 16 \]
Once the fractions are removed, solving the equations becomes straightforward using either the elimination or substitution method as shown.