Understanding Infinite Solutions for Linear Equations

A system of linear equations with two variables, say \(A_1x + B_1y = C_1\) and \(A_2x + B_2y = C_2\), can have one unique solution, no solution, or infinitely many solutions. For the system to have infinite solutions, the lines represented by the equations must be coincident (the same line). This condition is met when the ratios of the corresponding coefficients are equal:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Identifying Coefficients in the Given System

The given system of equations is:

1. \(2x - (a - 4)y = 2b + 1\)
2. \(4x - (a - 1)y = 5b - 1\)

Comparing these equations to the standard form \(Ax + By = C\), we can identify the coefficients:

• From equation 1: \(A_1 = 2\), \(B_1 = -(a - 4)\), \(C_1 = 2b + 1\)
• From equation 2: \(A_2 = 4\), \(B_2 = -(a - 1)\), \(C_2 = 5b - 1\)

Setting Up Ratios for Infinite Solutions

For the system to have infinite solutions, we apply the condition \( \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \):

\[ \frac{2}{4} = \frac{-(a - 4)}{-(a - 1)} = \frac{2b + 1}{5b - 1} \]

The first ratio simplifies to \(\frac{2}{4} = \frac{1}{2}\). The second ratio simplifies to \(\frac{a - 4}{a - 1}\) because the negative signs cancel out.

So, we have the condition:

\[ \frac{1}{2} = \frac{a - 4}{a - 1} = \frac{2b + 1}{5b - 1} \]

This gives us two separate equations to solve to find the values of \(a\) and \(b\).

Solving for the Value of 'a'

We use the first two parts of the ratio equality:

\[ \frac{1}{2} = \frac{a - 4}{a - 1} \]

To solve for \(a\), we can cross-multiply:

\[ 1 \times (a - 1) = 2 \times (a - 4) \]

\[ a - 1 = 2a - 8 \]

Now, gather the 'a' terms on one side and the constant terms on the other:

\[ a - 2a = -8 + 1 \]

\[ -a = -7 \]

\[ a = 7 \]

So, the value of \(a\) for which the system can have infinite solutions is 7.

Solving for the Value of 'b'

Next, we use the first and third parts of the ratio equality:

\[ \frac{1}{2} = \frac{2b + 1}{5b - 1} \]

Again, we cross-multiply to solve for \(b\):

\[ 1 \times (5b - 1) = 2 \times (2b + 1) \]

\[ 5b - 1 = 4b + 2 \]

Now, gather the 'b' terms on one side and the constant terms on the other:

\[ 5b - 4b = 2 + 1 \]

\[ b = 3 \]

So, the value of \(b\) for which the system can have infinite solutions is 3.

Conclusion for Infinite Solutions

For the given system of equations to have infinite solutions, both conditions derived from the ratio equality must be satisfied simultaneously. We found that \(a = 7\) and \(b = 3\).

Let's check these values:

If \(a = 7\) and \(b = 3\), the equations become:

1. \(2x - (7 - 4)y = 2(3) + 1 \implies 2x - 3y = 7\)
2. \(4x - (7 - 1)y = 5(3) - 1 \implies 4x - 6y = 14\)

Now, check the ratios:

\[ \frac{A_1}{A_2} = \frac{2}{4} = \frac{1}{2} \]

\[ \frac{B_1}{B_2} = \frac{-3}{-6} = \frac{1}{2} \]

\[ \frac{C_1}{C_2} = \frac{7}{14} = \frac{1}{2} \]

Since \(\frac{1}{2} = \frac{1}{2} = \frac{1}{2}\), the condition for infinite solutions is satisfied when \(a = 7\) and \(b = 3\).

Revision Table: System of Linear Equations

Additional Information on Solving Systems

Apart from the condition for infinite solutions using coefficient ratios, systems of linear equations can be solved using various methods:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites, then add the equations to eliminate that variable.
• Graphical Method: Graph both equations on the same coordinate plane. The intersection point, if any, is the solution. For infinite solutions, the lines would be the same.
• Matrix Methods: Using augmented matrices and row operations (like Gaussian elimination) or using Cramer's rule (for unique solutions) are advanced methods.

Understanding when a system has infinite solutions, no solution, or a unique solution is crucial before attempting to solve it, as the method and outcome depend on these conditions. The coefficient ratio test is a quick way to determine the nature of the solution without solving the system directly.