Question

x is partly constant and partly varies with y. When x = 7, y = 3, and when x = 9, y = 6.
a) Find the relationship between x and y
b) Find x when y = 4

Answer 1

To find the relationship between \( x \) and \( y \), we can assume that \( x \) can be expressed as a linear function of \( y \) and a constant.

Let's denote the linear relationship as \( x = a + by \), where \( a \) is the constant part and \( b \) is the coefficient that expresses how \( x \) varies with \( y \).

We are given two points:
1. When \( y = 3 \), \( x = 7 \).
2. When \( y = 6 \), \( x = 9 \).

From these points, we can set up a system of equations:

1. \( 7 = a + b(3) \)
2. \( 9 = a + b(6) \)

Now we can solve for \( a \) and \( b \):

From the first equation:
\[
7 = a + 3b \quad \text{(1)}
\]

From the second equation:
\[
9 = a + 6b \quad \text{(2)}
\]

Now we'll subtract equation (1) from equation (2):
\[
9 - 7 = (a + 6b) - (a + 3b)
\]
\[
2 = 3b
\]
\[
b = \frac{2}{3}
\]

Now, we can substitute \( b \) back into equation (1) to find \( a \):
\[
7 = a + 3\left(\frac{2}{3}\right)
\]
\[
7 = a + 2
\]
\[
a = 7 - 2 = 5
\]

Now we found \( a \) and \( b \):
- \( a = 5 \)
- \( b = \frac{2}{3} \)

Thus, the relationship between \( x \) and \( y \) can be expressed as:
\[
x = 5 + \frac{2}{3}y
\]

### b) Find \( x \) when \( y = 4 \).

Now we can substitute \( y = 4 \) into the equation we found:
\[
x = 5 + \frac{2}{3}(4)
\]
\[
x = 5 + \frac{8}{3}
\]
Converting 5 into a fraction with a denominator of 3:
\[
5 = \frac{15}{3}
\]
Thus,
\[
x = \frac{15}{3} + \frac{8}{3} = \frac{23}{3}
\]

Therefore, when \( y = 4 \), \( x = \frac{23}{3} \) or approximately \( 7.67 \).

### Summary:
a) The relationship between \( x \) and \( y \) is \( x = 5 + \frac{2}{3}y \)
b) When \( y = 4 \), \( x = \frac{23}{3} \) (approximately \( 7.67 \)).