Consider the following system of equations:


1/3x + y = 3
2y = -2/3x + 10

A. Is (0, 3) a solution to the system of equations? Explain.

4 answers

To determine if the point (0, 3) is a solution to the system of equations, we can substitute \( x = 0 \) and \( y = 3 \) into each equation and check if both equations are satisfied.

Let's start with the first equation: \[ \frac{1}{3}x + y = 3 \] Substituting \( x = 0 \) and \( y = 3 \): \[ \frac{1}{3}(0) + 3 = 3 \implies 0 + 3 = 3 \implies 3 = 3 \] This equation is satisfied.

Now, let's check the second equation: \[ 2y = -\frac{2}{3}x + 10 \] Substituting \( x = 0 \) and \( y = 3 \): \[ 2(3) = -\frac{2}{3}(0) + 10 \implies 6 = 0 + 10 \implies 6 = 10 \] This equation is not satisfied.

Since (0, 3) satisfies the first equation but not the second, we conclude that:

A. No, (0, 3) is not a solution to the system of equations.

Consider the following system of equations:

1/3x + y = 3
2y = -2/3x + 10

Solve the system of equations using the elimination method.

To solve the given system of equations using the elimination method, we start with the two equations:

  1. \(\frac{1}{3}x + y = 3\)
  2. \(2y = -\frac{2}{3}x + 10\)

Step 1: Rearrange equations for elimination

Let's express both equations in standard form (Ax + By = C).

For the first equation:
Multiply through by 3 to eliminate the fraction: \[ x + 3y = 9 \] This can be labeled as Equation (1).

For the second equation:
First, rewrite the second equation: \[ 2y = -\frac{2}{3}x + 10 \] Multiply through by 3 to eliminate the fraction: \[ 6y = -2x + 30 \] Now rearranging gives: \[ 2x + 6y = 30 \] This can be labeled as Equation (2).

Step 2: Set the equations for elimination

We have the system now:

  1. \(x + 3y = 9\) (Equation 1)
  2. \(2x + 6y = 30\) (Equation 2)

Notice that Equation (2) is just \(2\) times Equation (1). Thus, we can see they represent the same line, which means every solution of one is also a solution of the other.

Step 3: Solve for one variable

Let's solve Equation (1) for \(y\): \[ y = 3 - \frac{1}{3}x \]

This shows that the system has infinitely many solutions along the line described by the original equations.

Step 4: Interpret the result

The system has infinitely many solutions, which can be expressed in parametric form. Setting \(x = t\) (where \(t\) is any real number) gives: \[ y = 3 - \frac{1}{3}t \] Thus, the solution set can be described as: \[ (x, y) = (t, 3 - \frac{1}{3}t) \quad \text{for any real number } t. \]

Summary

The system of equations is dependent and has an infinite set of solutions that can be expressed as the line: \[ y = 3 - \frac{1}{3}x. \]

Consider the following system of equations:

1/3x + y = 3
2y = -2/3x + 10

Could a system of two linear equations ever have an infinite number of solutions? If so,
how could you tell by looking at their equations? How could you tell from the graph? If not,
explain why.