Convert the following relationships into equations of format y=c + mx 
(where c= constant, and = y-axis intercept; m = gradient of straight line)
1.	 ax + by = d 


To convert the equation \( ax + by = d \) into the format \( y = c + mx \), follow these steps:

1. Start with the original equation:
   \[
   ax + by = d
   \]

2. Solve for \( y \) by isolating it on one side of the equation. First, subtract \( ax \) from both sides:
   \[
   by = d - ax
   \]

3. Divide through by \( b \) to solve for \( y \):
   \[
   y = \frac{d}{b} - \frac{a}{b}x
   \]

This is now in the form \( y = c + mx \), where:
- \( c = \frac{d}{b} \) (the constant or y-intercept),
- \( m = -\frac{a}{b} \) (the gradient or slope of the line).

### Final equation:
\[
y = \frac{d}{b} - \frac{a}{b}x
\]

Would you like further details or have any questions?

Here are some related questions to explore:

1. What happens when \( b = 0 \) in the original equation \( ax + by = d \)?
2. How can you interpret the slope \( m = -\frac{a}{b} \) in terms of the original equation?
3. What would the graph of this equation look like if \( a = 0 \)?
4. How does the value of \( d \) affect the position of the line on the graph?
5. How can you find the x-intercept from the equation \( ax + by = d \)?

**Tip**: Always isolate \( y \) first when trying to convert an equation into the slope-intercept form \( y = c + mx \).

Convert the following relationships into equations of format y=c + mx (where c = constant, and = y-axis intercept; m = gradient of straight line) 1. ax + by = d

Learn how to convert the linear equation ax + by = d into the slope-intercept form y = c + mx. This problem covers algebraic concepts such as linear equations and slope, making it ideal for students in Grades 9-10. Detailed step-by-step solution provided.

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