Problem 71 - Algebra 1

Question

Use substitution to solve the linear system. $$\begin{aligned} &\frac{3}{5} b-a=0\\\ &1+b=2 a \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system is \(a = \frac{3}{5}\) and \(b = \frac{1}{5}\)

Step 1: Isolate a variable

The given system contains two equations. Let's start with the second equation, where it's already in a form that we can easily isolate 'b' without fractions. Isolate 'b' in the equation \(1 + b = 2a\). By doing this, we will get \(b = 2a - 1\)

Step 2: Substitute 'b' in the first equation

Next, substitute \(b = 2a - 1\) into the first equation given as \(\frac{3}{5}b - a = 0\). This will give \(\frac{3}{5}(2a - 1) - a = 0\)

Step 3: Solve for 'a'

Simplify and solve the equation \(\frac{3}{5}(2a - 1) - a = 0\) for 'a'. The expression simplifies to \(\frac{6}{5}a - \frac{3}{5} - a = 0\), which simplifies further to \(a = \frac{3}{5}\)

Step 4: Solve for 'b'

Substitute \(a = \frac{3}{5}\) into the equation \(b = 2a - 1\) to find 'b'. We get \(b = 2(\frac{3}{5}) - 1 = \frac{1}{5}\)

Key Concepts

Linear Systems

Linear systems involve multiple linear equations working together to find a common solution. Each equation in a linear system can be visualized as a line on a graph, and the solution corresponds to the point(s) where these lines intersect. In most common cases, especially with two variables, linear systems can be represented by two equations. Like the example:

• Equation 1: \( \frac{3}{5}b - a = 0 \)
• Equation 2: \( 1 + b = 2a \)

The goal is to find values for the variables that satisfy all equations simultaneously. In our exercise, we're tasked with finding the values of 'a' and 'b' that make both equations true at the same time. Solving a linear system often involves methods like substitution, elimination, or graphing. Our focus here is on using the substitution method.

Isolate a Variable

Isolating a variable means rewriting an equation to express one variable in terms of the other(s). This is a crucial step in the substitution method as it allows us to replace one variable in another equation. When isolating, we rearrange the equation so that one variable stands alone on one side of the equal sign. In the given task, we examined the second equation:

• Start with: \( 1 + b = 2a \)
• Rearrange to isolate \( b \): \( b = 2a - 1 \)

By isolating 'b', we can now substitute this expression into the other equation. This eliminates 'b' from our system temporarily, allowing us to solve for 'a'. This process not only makes equations simpler but also paves the way for uncovering the system's solution.

Solving Equations

Once variables are isolated and substituted, the final part of the method involves solving equations to find the value of the variables. After substituting \( b = 2a - 1 \) into the first equation \( \frac{3}{5}(2a - 1) - a = 0 \), it's time to perform algebraic manipulations. Follow these steps:

• Distribute the fraction: \( \frac{6}{5}a - \frac{3}{5} - a = 0 \)
• Combine like terms: \( \left(\frac{6}{5}a - a\right) = \frac{3}{5} \)
• Simplify further: \( a = \frac{3}{5} \)

Finally, with \( a = \frac{3}{5} \) known, substitute it back into \( b = 2a - 1 \) to find 'b': \( b = 2(\frac{3}{5}) - 1 \), which yields \( b = \frac{1}{5} \). By solving these equations, we determine that the solution to the linear system is \( a = \frac{3}{5} \) and \( b = \frac{1}{5} \).