The sum of the ages of two brothers is 6. The age of the older brother is two more than triple the square of the age of the younger brother. A. Select the two equations that represent this situation if x represents the age of the younger brother and y represents the age of the older brother. (2 points) Responses y=2+3x2 y is equal to 2 plus 3 x squared x+y=6 x plus y is equal to 6 y=2+3x y is equal to 2 plus 3 x y=6+x y is equal to 6 plus x Question 2 B. Solve the system from part A by graphing. Clearly label the solution(s) on the graph and then state the solution in words.

Answer 1

To tackle the problem, we first define the variables clearly:

• Let \( x \) be the age of the younger brother.
• Let \( y \) be the age of the older brother.

Part A:

From the problem statement, we derive the following two equations:

1. The sum of the ages of the two brothers is 6: \[ x + y = 6 \]

2. The age of the older brother is two more than triple the square of the age of the younger brother: \[ y = 2 + 3x^2 \]

Identifying the correct equations given the options:

• The correct equations would be:
  • \( x + y = 6 \)
  • \( y = 2 + 3x^2 \)

Part B:

To solve the system of equations by graphing:

1. Rearranging the first equation:

   • From \( x + y = 6 \), we can express \( y \): \[ y = 6 - x \]

2. Using the second equation directly: \[ y = 2 + 3x^2 \]

Graphing:

You can graph these two equations on a coordinate system.

• For \( y = 6 - x \):

  • When \( x = 0 \), \( y = 6 \) (point: \( (0, 6) \)).
  • When \( x = 6 \), \( y = 0 \) (point: \( (6, 0) \)).
  • This line will slope downward from \( (0, 6) \) to \( (6, 0) \).

• For \( y = 2 + 3x^2 \):

  • This is a parabola opening upwards.
  • When \( x = 0 \), \( y = 2 \) (point: \( (0, 2) \)).
  • When \( x = 1 \), \( y = 5 \) (point: \( (1, 5) \)).
  • When \( x = -1 \), \( y = 5 \) (point: \( (-1, 5) \)).
  • You can continue to calculate additional points, but these should be enough to see the trend.

Finding the intersection point:

To find the solution, set \( 6 - x = 2 + 3x^2 \): \[ 6 - x = 2 + 3x^2 \] \[ 3x^2 + x - 4 = 0 \]

Using the quadratic formula:

For \( ax^2 + bx + c = 0 \), \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3, b = 1, c = -4 \): \[ x = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6} \]

Calculating the two potential values for \( x \):

1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{-8}{6} \) (not a valid age since negative).

So, \( x = 1 \).

Substituting \( x = 1 \) back into \( y = 6 - x \): \[ y = 6 - 1 = 5 \]

Conclusion:

The solution is: The younger brother is 1 year old, and the older brother is 5 years old.

When graphing, you should label the intersection point, which represents this solution, at the coordinates \( (1, 5) \).