# Solving for "a" and "b"

• May 7th 2012, 06:23 AM
Bashyboy
Solving for "a" and "b"
The problem is a line is represented by the equation $\displaystyle ax + by = 4$

I was able to answer the first few questions regardin this equation. But this problem left uncertain what to do. Th8e question is, "give the values for "a" and "b" such that the line has a slope of 5/8. How do I go about solving this?
• May 7th 2012, 06:30 AM
Prove It
Re: Solving for "a" and "b"
Quote:

Originally Posted by Bashyboy
The problem is a line is represented by the equation $\displaystyle ax + by = 4$

I was able to answer the first few questions regardin this equation. But this problem left uncertain what to do. Th8e question is, "give the values for "a" and "b" such that the line has a slope of 5/8. How do I go about solving this?

Rearrange the equation so it's in \displaystyle \displaystyle \begin{align*} y = mx + c \end{align*} form, then \displaystyle \displaystyle \begin{align*} m \end{align*} is the gradient.

\displaystyle \displaystyle \begin{align*} ax + by &= 4 \\ by &= -ax + 4 \\ y &= -\frac{a}{b}x + \frac{4}{b} \end{align*}

So the gradient is \displaystyle \displaystyle \begin{align*} -\frac{a}{b} \end{align*}. You can choose any values you like as long as you get \displaystyle \displaystyle \begin{align*} \frac{5}{8} \end{align*} when you divide and negate.
• May 7th 2012, 06:53 AM
Bashyboy
Re: Solving for "a" and "b"
So, you are saying -a/b = m = 5/8? And since it is a continuous function I can put any value I want into x and y, and solve for the remaining b?
• May 7th 2012, 07:10 AM
Prove It
Re: Solving for "a" and "b"
I'm saying you can choose any a and b you like, as long as -a/b = 5/8. You won't be substituting anything in for x or y.