Introduction to Relationship Between Variables in the Equation of a Line
Mathematical discoveries, small or great are never born of spontaneous generation.
—Jules Henri PoincarÉ (1854–1912) French Mathematician
In this lesson, you'll learn about input/output tables: how to find the rule of a table and how to find a missing value in a table.
The variables in the equation of a line have a relationship between each other. As the value of one variable changes, so does the value of the other variable. In the equation y = 4x – 1, when x = 3, y is equal to 4(3) – 1 = 12 – 1 =11. If x is equal to 4, then y is equal to 4(4) – 1 = 16 – 1 = 15.
For any equation of a line, we can make an input/output table to show the relationship between x and y. The input/output table of y = 4x – 1 shows the value of y for different values of x:
The first column shows the value of x. We say that x is the independent variable, because it determines the value of y, and we say that y is the dependent variable, because we find its value based on the value of x. The middle column of the table shows the substitution of a number for x, and the last column shows the value of y after those calculations have been performed. Most input/output tables do not show the middle column, just the values of x and y.
The equation used for this table is listed right at the top of the table, but it usually is not given to us; we will need to figure out what the equation is based on the values in the table.
This input/output table is the kind of table we will usually see—just 4 or 5 x values with their corresponding y values. Previously, we learned that the slope of a line is the change in the yvalues between two points on a line divided by the change in the xvalues of those points. Because we were given equations in that lesson, we could find the slope of a line by putting the equation in slopeintercept form. Now that we have an input/output table, we can figure out the slope of a line ourselves!
We can find the slope of a line using any two points on the line. We will use the first two rows of this input/output table. First, find the difference between the y values: 6 – 5 = 1. Next, find the difference between the x values: 1 – 0 = 1. Now, divide the difference between the y values by the difference between the x values: 1 ÷ 1 = 1. The slope of this equation is 1.
We have the slope of the equation, and now we need the yintercept. Remember: Slopeintercept form is y = mx + b, where m is the slope of the line and b is the yintercept. We have plenty of x and y values, and we know the slope of the line is 1. We can substitute these values into the equation and solve for b. Any row from the table will do. Let's use the last row, where x is 4 and y is 9. Substitute these values into the equation y = mx + b:
 9 = 1(4) + b
 9 = 4 + b
Subtract 4 from both sides of the equation:
 9 – 4 = 4 + b – 4
 5 = b
The yintercept of the line is 5. Now that we have the slope and the yintercept, we can write the equation for this table: y = 1x + 5, or y = x + 5. Looking back at the table, we can see that each value of y is 5 greater than its x value, so our equation is correct.
Tip:If x = 0 in the input/output table, its y value is the yintercept of the equation, because the yintercept of an equation is the value of y when x = 0. In the previous example, the first row of the table shows that y = 5 when x = 0, and 5 is the yintercept of the equation. 
We can find the equation used to make the following table in the same way.

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