Problem 69 - Algebra 1

Question

Solve the system of linear equations if possible. Does the system have exactly one solution, no solution, or infinitely many solutions? $$ \begin{aligned}&10 x-6 y=-5\\\&3 y=5 x+2\end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -9/20, y=1/4\). The system has exactly one solution

Step 1: Convert the second equation

First, convert the second equation from \(3y = 5x + 2\) into a similar form as the first one by subtracting \(5x\) from both sides. This gives us \(-5x + 3y = 2\)

Step 2: Equate the coefficients

To solve this system of equations, we can match the coefficients in both equations. Multiply the second equation by 2 to match the coefficients with the first equation. This gives us: \(-10x + 6y = 4\)

Step 3: Subtract the equations

Subtract the second equation from the first one. That is \((10x - 6y) - (-10x + 6y) = -5 - 4\). Simplification gives \(20x = -9\)

Step 4: Solve for x

Divide both sides of the equation by 20 gives us \(x = -9/20\)

Step 5: Solve for y

Substitute x into the second equation gives us \(3y = 5(-9/20) + 2\), which solves to give \( y = 1/4\)

Step 6: Number of solutions

Since we obtained unique solutions for x and y, this system has exactly one solution

Key Concepts

Solutions of Linear Equations

Systems of linear equations consist of two or more equations with multiple variables. The solutions of these equations satisfy all the equations in the system at the same time. A solution can be a single unique point where all lines (formed by the equations) intersect.
Another possibility is no solution, where the lines never intersect, or infinitely many solutions, where all lines coincide.
The main goal in solving these systems is to find the values of the variables that satisfy all the included linear equations. By doing so, you can determine whether the system has a single solution, no solution, or infinite solutions.

Substitution Method

The substitution method is a popular algebraic technique used to solve systems of linear equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation(s). Here’s the step-by-step process:

• Solve one of the equations for one variable in terms of the others.
• Substitute this expression into the other equation(s) to eliminate one variable.
• Solve the resulting equation for the other variable.
• Substitute back into the original equation to find the value of the first variable.

In the exercise, by transforming the equations and using these steps, a solution for both variables was found, which demonstrates how effective the substitution method can be.

Unique Solution

A unique solution occurs when there is exactly one set of values for the variables that satisfies all the equations in a system. For linear equations, this typically means the lines intersect at just one point. In our exercise, the system of equations was able to be reduced to specific values for \(x\) and \(y\), meaning that the lines intersect at exactly one point in the plane.
When dealing with two linear equations in two variables, if the determinant of the coefficients matrix is non-zero, it usually indicates a unique solution. This situation is preferable as it provides a clear and definite result for the given system.

Linear Equation Transformation

Transforming linear equations can simplify the process of finding solutions. This generally involves operations like adding, subtracting, or multiplying entire equations to make them easier to solve.
In the exercise, transforming the equation by making the coefficients similar helped isolate the variables.
Remember that any operation performed must preserve the equality, so changes are legal as long as both sides of the equation are manipulated similarly. Transformations can also include substituting equivalent expressions, adding multiples of an equation, or scaling an equation up or down. This ensures the transformation retains the original relationships between the variables, paving the way for easier solution finding.