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

Is 0,3 a solution? Explain why or why not

7 answers

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

The equations are:

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

Let's substitute (0, 3) into the first equation:

  1. \( \frac{1}{3}(0) + 3 = 3 \) \[ 0 + 3 = 3 \] This is true.

Now, let's substitute (0, 3) into the second equation:

  1. \( 2(3) = -\frac{2}{3}(0) + 10 \) \[ 6 = 0 + 10 \] \[ 6 = 10 \] This is false.

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

Solve the system of equations using the elimination method.

To solve the system of equations using the elimination method, let's start with the original equations:

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

First, it's best to eliminate the fractions in these equations for easier manipulation. We can do this by multiplying each equation by a common factor.

Step 1: Remove the Fractions

For the first equation, multiply the whole equation by 3: \[ 3\left(\frac{1}{3}x + y\right) = 3 \cdot 3 \ x + 3y = 9 \quad \text{(Equation 1')} \]

For the second equation, multiply the whole equation by 3 to eliminate the fraction: \[ 3\left(2y\right) = 3\left(-\frac{2}{3}x + 10\right) \ 6y = -2x + 30 \ 2x + 6y = 30 \quad \text{(Equation 2')} \]

Step 2: The New System

Now we have the following system of equations:

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

Step 3: Elimination

Notice that Equation 2' can be simplified by dividing by 2: \[ x + 3y = 15 \quad \text{(Equation 2'')} \]

Now the new system is:

  1. \( x + 3y = 9 \)
  2. \( x + 3y = 15 \)

Step 4: Subtract the Equations

Next, we'll subtract Equation 1' from Equation 2'': \[ (x + 3y) - (x + 3y) = 15 - 9 \ 0 = 6 \]

This leads us to a contradiction (0 cannot equal 6). This means that the two equations represent parallel lines, which implies that they do not intersect.

Conclusion

Since there is no solution to the system of equations, we conclude that the system is inconsistent. The two lines represented by the equations are parallel and thus do not meet at any point in the coordinate plane. Therefore, there are no values of \( x \) and \( y \) that satisfy both equations simultaneously.

Simplify what you said

Certainly! Here’s a simplified explanation:

To solve the system of equations:

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

Step 1: Eliminate Fractions

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

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

Step 2: Simplify

Next, simplify Equation 2' by dividing everything by 2: \[ x + 3y = 15 \quad \text{(Equation 2'')} \]

Step 3: Solve the System

Now we have:

  1. \( x + 3y = 9 \)
  2. \( x + 3y = 15 \)

Now, we can see that the left sides are identical, but the right sides are not, meaning one equation states that \( x + 3y \) equals 9 and the other says it equals 15.

Conclusion

This contradiction means there is no solution; the equations represent parallel lines that never intersect. Thus, the system is inconsistent.

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 in a two-dimensional space.

How to Tell from the Equations

  1. Identical Equations: If both equations are exactly the same, they will have an infinite number of solutions (all points on the line).

    • Example: \( y = 2x + 3 \) and \( 2y = 4x + 6 \) (the second equation simplifies to the first).

  2. Proportional Coefficients: If the equations are different but represent the same line, they will have proportional coefficients throughout.

    • For example, if you can manipulate one equation to become the other by multiplying or dividing:
      • Equation 1: \( 2x + 4y = 10 \)
      • Equation 2: \( x + 2y = 5 \)
    • Here, if you multiply the second equation by 2, you get the first.

In general, for two linear equations of the form \( Ax + By = C \) and \( Dx + Ey = F \), the system has infinitely many solutions if: \[ \frac{A}{D} = \frac{B}{E} = \frac{C}{F} \] (with \( D \) and \( E \) not equal to zero).

How to Tell from the Graph

  1. Graphing the Lines: If you graph both equations and they lie on top of each other, then they represent the same line, indicating an infinite number of solutions.

  2. Overlap: When one line completely overlaps the other, every point on that line is a solution to both equations.

Conclusion

In summary, for two linear equations to have an infinite number of solutions, they must represent the same line. You can determine this by looking for identical equations or proportional coefficients in their standard forms, and graphically by observing that the lines overlap perfectly.