Solving a System of Equations Using the Substitution Method

When faced with solving a system of equations, it's often helpful to simplify the process by introducing substitution. This article will walk you through the steps to solve a system of equations:

Problem Statement

Given the following system of equations:

[ left{begin{array}{l} frac{2}{a} - frac{5}{b} - 20 0 frac{5}{a} frac{2}{b} - 21 0 end{array} right. ]

We will solve this using the substitution method. The substitution method involves defining new variables to represent the reciprocals of the original variables, which simplifies the system into a linear system that is easier to solve. Let us proceed step by step.

Introducing Substitution

Let us define the following new variables:

x ( frac{1}{a} ) y ( frac{1}{b} )

Substituting these new variables into our system:

2x - 5y 20 5x 2y 21

Step-by-Step Solution

Step 1

First, solve Equation 1 for y:

2x - 5y 20 5y 2x - 20 y ( frac{2x - 20}{5} )

Step 2

Next, substitute the expression for y into Equation 2 and solve for x:

5x 2 left( frac{2x - 20}{5} right) 21 25x 2(2x - 20) 105 25x 4x - 40 105 29x - 40 105 29x 145 x 5

Step 3

Having found x 5, substitute this value back into the expression for y:

y frac{2(5) - 20}{5} frac{10 - 20}{5} frac{-10}{5} -2

Step 4

Recall that:

x ( frac{1}{a} ) and y ( frac{1}{b} )

Thus:

a frac{1}{x} frac{1}{5} b frac{1}{y} -frac{1}{2}

Conclusion: The solutions to the system of equations are:

a frac{1}{5} and b -frac{1}{2}

Alternative Methods

Typically, there are multiple methods to solve systems of equations, such as substitution, elimination, or using matrices. For instance, one can also solve the system by manipulating the equations.

Alternate Method 1

To solve the system by subtracting the two equations, we can start with the original equations:

[ frac{2}{a} - frac{5}{b} - 20 0 ]

[ frac{5}{a} frac{2}{b} - 21 0 ]

Subtract the second equation from the first:

[ frac{2}{a} - frac{5}{b} - 20 - (frac{5}{a} frac{2}{b} - 21) 0 ]

[ frac{2}{a} - frac{5}{b} - 20 - frac{5}{a} - frac{2}{b} 21 0 ]

[ frac{2}{a} - frac{5}{a} - frac{5}{b} - frac{2}{b} 1 0 ]

[ -frac{3}{a} - frac{7}{b} -1 ]

[ frac{3}{a} frac{7}{b} 1 ]

This is another linear system that can be solved, but it ultimately leads to the same solution as the substitution method.

Alternate Method 2

Multiplying both equations by ab to clear the denominators, we obtain:

2b - 5a 20ab ......... (i)

5b - 2a 21ab ......... (ii)

Rewriting ab as a constant t:

2b - 5a 20t

5b - 2a 21t

From (i), we get:

b 5t

Substituting b 5t into (ii) and solving for a, we get:

5(5t) - 2a 21t

25t - 2a 21t

-2a -4t

a 2t

Since ab t, substituting a 2t and b 5t into the equation:

(2t)(5t) t

10t^2 t

t 0.1

Therefore:

a 2t 0.2

b 5t -0.5

This provides the same solution as the previous methods.

By utilizing various methods, you can solve systems of equations effectively, demonstrating the power and flexibility of algebraic techniques.