Solving Linear Equations to Find α + β

We are given a system of two linear equations:

1. \(3x + 5y = 19\)
2. \(10x - 3y = 24\)

We need to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. We can use the elimination method to solve this system.

Applying the Elimination Method

To eliminate one of the variables, we can multiply the equations by suitable constants so that the coefficients of one variable become opposite in sign or equal. Let's aim to eliminate \(y\).

• Multiply Equation 1 by 3: \(3 \times (3x + 5y) = 3 \times 19\)
• This gives us \(9x + 15y = 57\) (Let's call this Equation 3)
• Multiply Equation 2 by 5: \(5 \times (10x - 3y) = 5 \times 24\)
• This gives us \(50x - 15y = 120\) (Let's call this Equation 4)

Now, we have the coefficient of \(y\) as \(+15\) in Equation 3 and \(-15\) in Equation 4. Adding these two equations will eliminate the \(y\) term.

Add Equation 3 and Equation 4:

\((9x + 15y) + (50x - 15y) = 57 + 120\)

\(9x + 50x + 15y - 15y = 177\)

\(59x = 177\)

Now, solve for \(x\):

\(x = \frac{177}{59}\)

\(x = 3\)

Finding the Value of y

Substitute the value of \(x = 3\) into either of the original equations. Let's use Equation 1:

\(3x + 5y = 19\)

\(3(3) + 5y = 19\)

\(9 + 5y = 19\)

Subtract 9 from both sides:

\(5y = 19 - 9\)

\(5y = 10\)

Now, solve for \(y\):

\(y = \frac{10}{5}\)

\(y = 2\)

So, the solution to the system of linear equations is \(x = 3\) and \(y = 2\).

Relating the Solution to α and β

The problem states that the solution is \(x = \frac{\alpha}{3}\) and \(y = \frac{\beta}{2}\). We found \(x = 3\) and \(y = 2\).

• Comparing the \(x\) values: \(3 = \frac{\alpha}{3}\)
• To find \(\alpha\), multiply both sides by 3: \(\alpha = 3 \times 3 = 9\)

• Comparing the \(y\) values: \(2 = \frac{\beta}{2}\)
• To find \(\beta\), multiply both sides by 2: \(\beta = 2 \times 2 = 4\)

So, we have \(\alpha = 9\) and \(\beta = 4\).

Calculating α + β

Finally, we need to find the value of \(\alpha + \beta\):

\(\alpha + \beta = 9 + 4 = 13\)

Therefore, the value of \(\alpha + \beta\) is 13.

Revision Table: Key Steps in Solving System of Equations

Additional Information on Linear Equation Systems

A system of linear equations is a set of two or more linear equations involving the same variables. The solution to a system of linear equations in two variables (\(x\) and \(y\)) is a pair of values \((x, y)\) that satisfies every equation in the system.

There are several methods to solve a system of linear equations:

• Substitution Method: Solve one equation for one variable in terms of the other, and then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method (or Addition Method): Multiply one or both equations by suitable constants so that the coefficients of one variable are opposites. Then, add the equations to eliminate that variable. This results in a single equation with one variable. This is the method used in the solution above.
• Graphical Method: Graph each linear equation on the same coordinate plane. The point of intersection of the lines represents the solution to the system. If the lines are parallel, there is no solution. If the lines are the same, there are infinitely many solutions.

The number of solutions a system of two linear equations in two variables can have is:

• Exactly one solution: The lines intersect at a single point. (Consistent and Independent system)
• No solution: The lines are parallel and distinct. (Inconsistent system)
• Infinitely many solutions: The lines are the same (coincident). (Consistent and Dependent system)

In this specific problem, we found exactly one solution (\(x=3, y=2\)), which is typical for many linear systems encountered in algebra.