Consider the following system of equations:

To determine if the point \((0,3)\) is a solution to the system of equations, we need to substitute \(x = 0\) and \(y = 3\) into both equations and see if they hold true.

1. For the first equation: \[ \frac{1}{3} x + y = 3 \] Substituting \(x = 0\) and \(y = 3\): \[ \frac{1}{3}(0) + 3 = 3 \] This simplifies to: \[ 0 + 3 = 3, \] which is true. Thus, \((0, 3)\) satisfies the first equation.

2. For the second equation: \[ 2y = -\frac{2}{3} x + 10 \] Substituting \(x = 0\) and \(y = 3\): \[ 2(3) = -\frac{2}{3}(0) + 10 \] This simplifies to: \[ 6 = 0 + 10, \] which is not true because \(6 \neq 10\). Thus, \((0, 3)\) does not satisfy the second equation.

Since \((0, 3)\) satisfies the first equation but not the second equation, we conclude that \((0, 3)\) is not a solution to the system of equations.

in a conclusion

In conclusion, the point \((0, 3)\) is not a solution to the system of equations because, while it satisfies the first equation \(\frac{1}{3} x + y = 3\), it does not satisfy the second equation \(2y = -\frac{2}{3} x + 10\). Therefore, \((0, 3)\) is not a solution to the given system of equations.

Solve the system of equations using the elimination method.

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

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

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

Let’s rewrite both equations in standard form (\(Ax + By = C\)) and then eliminate one variable.

Step 1: Rearranging the equations

From Equation 1:

\[ \frac{1}{3} x + y = 3 \]

Multiply through by 3 to eliminate the fraction:

\[ x + 3y = 9 \quad \text{(Equation 1')} \]

From Equation 2:

\[ 2y = -\frac{2}{3} x + 10 \]

Multiply through by 3 to eliminate the fraction:

\[ 6y = -2x + 30 \]

Rearranging gives us:

\[ 2x + 6y = 30 \quad \text{(Equation 2')} \]

Step 2: Multiply the first equation to align coefficients for elimination

We want to eliminate \(x\). We can multiply Equation 1' by 2:

\[ 2(x + 3y) = 2(9) \] \[ 2x + 6y = 18 \quad \text{(Equation 1'')} \]

Now our system looks like this:

1. \(2x + 6y = 18\) \quad (Equation 1'')

2. \(2x + 6y = 30\) \quad (Equation 2')

Step 3: Subtract equations to eliminate \(x\)

Now, we will subtract Equation 1'' from Equation 2':

\[ (2x + 6y) - (2x + 6y) = 30 - 18 \] \[ 0 = 12 \]

This results in a false statement, indicating that the lines represented by these equations are parallel and do not intersect. Therefore, the system has no solution.

Conclusion

The system of equations is inconsistent, meaning that there are no solutions since the two lines are parallel.

shorter

To solve the following system using the elimination method:

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

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

Step 1: Convert to standard form

1. Multiply the first equation by 3: \[ x + 3y = 9 \quad \text{(Equation 1')} \]

2. Multiply the second equation by 3 and rearrange: \[ 2x + 6y = 30 \quad \text{(Equation 2')} \]

Step 2: Eliminate \(x\) Multiply Equation 1' by 2: \[ 2x + 6y = 18 \quad \text{(Equation 1'')} \]

Now we have:

1. \(2x + 6y = 18\) \quad (Equation 1'')
2. \(2x + 6y = 30\) \quad (Equation 2')

Step 3: Subtract the equations Subtract Equation 1'' from Equation 2': \[ 0 = 12 \]

This results in a contradiction, indicating that the system has no solution. Thus, the lines represented by these equations are parallel.

simplified?

To solve the system of equations:

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

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

Step 1: Clear fractions and rearrange

Multiply the first equation by 3: \[ x + 3y = 9 \quad \text{(1)} \]

Multiply the second equation by 3: \[ 6y = -2x + 30 \implies 2x + 6y = 30 \quad \text{(2)} \]

Step 2: Eliminate one variable

Now, use (1) and (2):

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

Multiply (1) by 2: \[ 2x + 6y = 18 \quad \text{(3)} \]

Step 3: Subtract (3) from (2)

\[ (2x + 6y) - (2x + 6y) = 30 - 18 \] \[ 0 = 12 \]

Conclusion

This contradiction means there is no solution; the lines are parallel.

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.

Yes, a system of two linear equations can have an infinite number of solutions. This occurs when the two equations represent the same line, meaning they are dependent.

Identifying Infinite Solutions from the Equations

1. Identical Equations: If the two equations are exactly the same (e.g., \(y = 2x + 3\) and \(y = 2x + 3\)), they have infinitely many solutions.

2. Proportional Coefficients: Simplifying the equations can help identify equivalent lines. If you can manipulate one equation to obtain the other (by multiplying or dividing by a constant), then the lines represented by the equations are the same. For example:

   • Equation 1: \(2x + 4y = 8\)
   • Equation 2: \(x + 2y = 4\) (divide Equation 1 by 2) Here, both represent the same line.

Identifying Infinite Solutions from the Graph

Graphically, you can tell there are infinite solutions if the two lines overlap perfectly. This means they are the same line.

Summary

• Algebraically: Look for identical equations or proportional coefficients.
• Graphically: Observe if the two lines coincide (lie on top of each other), indicating infinite solutions.