Question

Solve the system by elimination Then state whether the system is consistent inconsistent. $$\left\\{\begin{array}{l}3 x+2 y=10 \\ 2 x+5 y=3\end{array}\right.$$

Short answer

The solution of the system of equations is \(x = 4\) and \(y = -1\), indicating the system is consistent.

Step-by-step solution

Multiply the equations by suitable numbers to make the coefficient of one variable same

In this case, it's easier to eliminate variable \(x\). Multiply the first equation by 2 and the second by 3 to achieve a common coefficient for \(x\). The equations will now be \(6x + 4y = 20\) and \(6x + 15y = 9\).

Subtract one equation from the other

Subtract the second equation from the first to eliminate \(x\). This results in \(-11y = 11\).

Solve for the remaining variable

Divide by -11 to solve for \(y\). So, \(y = -1\).

Substitute \(y\) back into one of the original equations and solve for \(x\)

Substitute \(y = -1\) back into the first original equation to get \(3x + 2*(-1) = 10\), which simplifies to \(3x - 2 = 10\). Solving for \(x\) by adding 2 to both sides and dividing by 3 gives \(x = 4\).

Determine the consistency of the system

The solution of this system of equations is \(x = 4\) and \(y = -1\), meaning that the system has at least one solution and is therefore consistent.

Key concepts

Elimination Method

The elimination method is a powerful technique used to solve systems of linear equations. It involves eliminating one of the variables from a pair of equations so that you can solve for the remaining variable. This approach simplifies the problem significantly and turns it into a series of manageable steps.

To effectively use the elimination method, begin by arranging the equations so that one variable can be easily eliminated. Often, this involves finding a common coefficient for one of the variables. In the given exercise, we aimed to eliminate the variable \(x\). To do this, multiply the entire first equation by 2 (giving us \(6x + 4y = 20\)) and the second equation by 3 (giving us \(6x + 15y = 9\)). This creates equations with matching \(x\) coefficients.

Once the coefficients match, subtract the second equation from the first. This operation will eliminate \(x\) and leave you with a single equation in terms of \(y\). Then, you simply solve for \(y\) as you would in a standard algebraic equation.

The elimination method is particularly useful because it allows one to skillfully navigate through even the most intricate systems, steering the process towards a solution by reducing complexity a step at a time.

Consistent System

In the world of linear equations, systems can either be consistent or inconsistent. A consistent system is one that has at least one solution—and in the case of linear equations, this often means an intersection point where the lines meet.

For the given system of equations, after solving for \(x\) and \(y\) using the elimination method as described, we found that the solution was \(x = 4\) and \(y = -1\). This means our system has a single solution where both equations coincide at a particular point.

But what makes a system inconsistent? If, after attempting to solve the system, you find yourself with a false statement such as \(0 = 5\), it means there are no solutions, indicating that the lines are parallel and do not intersect. This would render the system inconsistent.

Understanding whether a system is consistent helps refocus approaches to solving. It reassures that error-free calculation will lead to a definite answer, giving confidence in employing methods like elimination or substitution.

Solving for Variables

Solving for variables is a crucial step in resolving systems of equations. After one variable is eliminated using the elimination method, the next target is the isolated variable that remains.

In our exercise, after eliminating \(x\), we were left with the equation \(-11y = 11\). To find \(y\), divide both sides by \(-11\), yielding \(y = -1\). This part of the solution requires basic algebraic manipulation and attention to arithmetic accuracy.

Once \(y\) is determined, substitute this value back into one of the original equations to find the other variable, \(x\). For instance, substituting into the equation \(3x + 2y = 10\) where \(y = -1\) simplifies to \(3x - 2 = 10\). Solving it gives \(x = 4\).

These operations reinforce the importance of the basics like substitution and balancing equations—where each step is an integral component that works towards pinning down the exact values of the variables. In essence, solving for variables is the endpoint that ensures everything aligns precisely with the logic of the system equations.