Question

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

Short answer

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

Step-by-step solution

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 \(b = 2a - 1\)

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 \(\frac{3}{5}(2a - 1) - a = 0\)

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 \(a = \frac{3}{5}\)

Solve for 'b'

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