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Write the linear equation for a line with a slope of 3 that goes through the point (2,9).
Given this information you can find the equation of the line. The slope is the "m" in the equation for a line, y=mx + b. Then all you have to do is find the value of "b". To do this you plug the point (2,9) into the equation. x = 2 and y = 9. The equation at this point would look like 9 = (3)2 + b. By multiplying you get 9 = 6 + b.
Subtract 6 from both sides to get "b" by itself. In this case, b = 3.
Then plug that back into the original equation, y = 3x + b, to get the equation of the line, y = 3x + 3.
| Y=3x+3 | Here is the equation |
| Y=3x+3
-3 -3 |
Subtract 3 from both sides of the equation |
| -3=3x | This is the result of subtracting 3 from both sides |
| -3=3x
3 |
Divide both sides by 3 |
| -1=x | The value of x has been defined |
This equation contains
a fraction, which are often nasty to deal with.
Although it comes from
a college book it is not very difficult to solve.
| q = 3/2q - 4 | Here is the equation |
| 2(q) = 2(3/2q - 4) | Multiply both sides by 2 , the denominator of the fracion, to get rid of the fraction |
| 2q = 3q - 8 | This is the result from multiplying both sides by 2 |
| 2q = 3q - 8
-3q -3q |
Subtract 3q from both sides to isolate the variable |
| -q = -8 | This is the result of subtractig 3 q from both sides |
| -1(-q) = -1(-8) | Multiply both sides by -1 |
| q = 8 | The value of q has been defined |
This equation is Cailletet's
and Mathia's Law. It is an example of a literal equation.
A literal equation is
an equation where no variables have been defined with numbers.
The way to solve a literal
equation is to get the determined variable by itself on one side of the
equation.
To determine the variable
to solve for, the equation is written
1/2(d1 + d2)
= A + BT : d1
| 1/2(d1 + d2) = A + BT | Here is the equation |
| 2(1/2(d1 + d2) = 2(A + BT) | Multiply both sides by 2 to get rid of the fraction |
| d1 + d2 = 2A + 2BT | This is the result of multiplying both sides by 2 |
| d1 + d2
= 2A + 2BT
- d2 -d2 |
Subtract d2 from both sides to get d1 by itself |
| d1 = 2A + 2BT - d2 | The value of d1 has been defined |
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