Okay, Mat wanted me to go over this so here I am. Before I continue, you must know what a linear equation is. The definition is as follows: A linear equation is a mathematical expression that has an equal sign and linear expressions.
One of the most common equations is y=mx+b. m represents the slope or gradient of the line, while the b represents the y-intercept or where the line crosses the verticle or y axis.
Examples of linear equations with two variables:
x+2y=10
3a+472c=10c+37
2x+y-5=-7x+4y+3
Now, there are several different forms in which all the equations above can be written into.
There is General form where the equations looks like Ax+By+C=0 In this form, A is most likely not going to be negative.
The next form is Standard form. Ax+By=C A and B are both not zero (if they were, it would be pointless because then the y or x would become zero, as well, and it's just not possible). A, B, and C are all integers whose greatest common factor is one and A is non-negative.
Slope-Intercept form.
y-axis formula is y=mx+b I went over this above.
x-axis formula of this form is x=y/m+c. m cannot equal zero or else the slope wouldn't be real. c is the x-intercept.
Point-Slope form
y-y1=m(x-x1) m is the slope and (x1,y1) is any point on the line.
Intercept form
Two-point form
Parametric form
Normal form
I'm not going to go over those but those are other forms. If you would like me to go over them, just tell me.
Let's say a problem told you to solve for d in this linear equation:
q+x/d=px/2+a
You must isolate d and get it by itself. The first step is to get rid of the fractions by multiplying the whole thing by 2d (this will cancel out the two denomanators on the fractions).
The result is: q(2d)+2x=px(2)+a(2d) = 2dq+2x=pxd+2da
Next step, get all the parts with a d in it on one side. The result would be:
2dq-pxd-a2d=-2x
Now, you must factor out the d. This would look like:
d(2q-px-a2)=-2x
Finally, get rid of the excess "stuff" on the side of the d by dividing both sides by it. The final result should be:
d=(-2x)/(2q-px-a2)
Right there, is solving for d.
Any questions? Anything I missed that you would like me to go over? I hope my lesson was beneficial to you. I know, it may be difficult at first if you still don't get it but you must practice. If you would like further help or would like problems to try on your own, feel free to ask me.
Your Mathematics Teacher
Evey