he 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.

Question

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

x+y=6
𝑥
+
𝑦
=
6
x plus y is equal to 6

y=2+3x2
𝑦
=
2
+
3
𝑥
2
y is equal to 2 plus 3 x squared

y=6+x
𝑦
=
6
+
𝑥
y is equal to 6 plus x

y=2+3x
𝑦
=
2
+
3
𝑥
y is equal to 2 plus 3 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. graph it

Answer 1

To solve the given problem, first, we need to identify the two correct equations from the provided options and then graph the system of equations.

Part A: Selecting the Equations

1. We know the sum of the ages of the two brothers: \[ x + y = 6 \] (where \(x\) is the age of the younger brother and \(y\) is the age of the older brother.)

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

So the correct equations that represent this situation are:

• Equation 1: \(x + y = 6\)
• Equation 2: \(y = 2 + 3x^2\)

Part B: Solving the System by Graphing

Now, let's graph these two equations and find the intersection, which would give us the solution.

1. Graph of \(x + y = 6\):

   • To find the y-intercept, set \(x = 0\): \[ y = 6 \]
   • To find the x-intercept, set \(y = 0\): \[ x = 6 \]
   • Points: (0, 6) and (6, 0). Draw a straight line between these two points.

2. Graph of \(y = 2 + 3x^2\):

   • This is a parabola opening upwards.
   • Find y-values for different x-values:
     • When \(x = 0\): \[ y = 2 + 3(0)^2 = 2 \quad \text{(Point: (0, 2))} \]
     • When \(x = 1\): \[ y = 2 + 3(1)^2 = 5 \quad \text{(Point: (1, 5))} \]
     • When \(x = -1\): \[ y = 2 + 3(-1)^2 = 5 \quad \text{(Point: (-1, 5))} \]
     • When \(x = 2\): \[ y = 2 + 3(2)^2 = 14 \quad \text{(Point: (2, 14))} \]
   • Sketch the parabola through the points calculated.

Finding the Intersection:

To find where these two graphs intersect, we can solve for \(x\) and \(y\) using either algebraically or graphically by locating the point of intersection.

Algebraic Method:

Substituting \(y\) from the first equation into the second: \[ 6 - x = 2 + 3x^2 \]

Rearranging gives: \[ 3x^2 + x - 4 = 0 \]

Now solving the quadratic: Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Where \(a = 3\), \(b = 1\), and \(c = -4\):

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \] \[ x = \frac{-1 \pm \sqrt{1 + 48}}{6} \] \[ x = \frac{-1 \pm \sqrt{49}}{6} \] \[ x = \frac{-1 \pm 7}{6} \]

Calculating the two possible solutions for \(x\):

1. \( x = 1 \)
2. \( x = -\frac{4}{3} \) (not valid since age cannot be negative)

Therefore:

\(x = 1 \) gives us the younger brother's age. Calculating \(y\): \[ y = 6 - x = 6 - 1 = 5 \]

Solution:

In words: The younger brother is 1 year old and the older brother is 5 years old.

Graph:

To visualize the solution graphically, you would plot both equations on a graph with \(x\) on the horizontal axis and \(y\) on the vertical axis, indicating the intersection point \((1, 5)\).

If you need a graphical representation, you can use graphing software or tools like Desmos to create the graph based on the equations provided.