Question

If x is partly constant and partly varies as the square of y when y = 1 and x = four whole number one over four and when y = 7 x = one hundred and ninety six whole number one over four , find the equation connecting x and y

Answer 1

To find the equation that connects \( x \) and \( y \), given that \( x \) is partly constant and partly varies as the square of \( y \), we can express this relationship in the form:

\[
x = k + m y^2
\]

where \( k \) is a constant and \( m \) is a coefficient that dictates how \( x \) varies with \( y^2 \).

We are given the following data points:

1. When \( y = 1 \), \( x = 4.25 \) (which is \( 4 \frac{1}{4} = \frac{17}{4} \)).
2. When \( y = 7 \), \( x = 196.25 \) (which is \( 196 \frac{1}{4} = \frac{785}{4} \)).

Now we can set up two equations using these points.

1. For \( y = 1 \):
\[
\frac{17}{4} = k + m(1)^2
\]
which simplifies to:
\[
\frac{17}{4} = k + m \quad \text{(Equation 1)}
\]

2. For \( y = 7 \):
\[
\frac{785}{4} = k + m(7)^2
\]
which simplifies to:
\[
\frac{785}{4} = k + 49m \quad \text{(Equation 2)}
\]

Now we have a system of equations:
1. \( k + m = \frac{17}{4} \)
2. \( k + 49m = \frac{785}{4} \)

We can solve this system step by step.

**Step 1: Solve for \( k \) in terms of \( m \) using Equation 1:**
\[
k = \frac{17}{4} - m
\]

**Step 2: Substitute \( k \) into Equation 2:**
\[
\frac{17}{4} - m + 49m = \frac{785}{4}
\]
\[
\frac{17}{4} + 48m = \frac{785}{4}
\]

**Step 3: Isolate \( m \):**
\[
48m = \frac{785}{4} - \frac{17}{4}
\]
\[
48m = \frac{785 - 17}{4} = \frac{768}{4} = 192
\]
\[
m = \frac{192}{48} = 4
\]

**Step 4: Substitute \( m \) back into Equation 1 to find \( k \):**
\[
k + 4 = \frac{17}{4}
\]
\[
k = \frac{17}{4} - 4 = \frac{17}{4} - \frac{16}{4} = \frac{1}{4}
\]

Putting the values of \( k \) and \( m \) back into the equation:

\[
x = k + m y^2
\]
\[
x = \frac{1}{4} + 4y^2
\]

In conclusion, the equation connecting \( x \) and \( y \) is:
\[
x = 4y^2 + \frac{1}{4}
\]