f(x) = -3

g(x) = 5

h(x) = -3 + 5 therefore h(x) = 2

Here we notice that we get a line where the slope is 0 and the y-intercept is 2, the sum of the y-intercepts of f(x) and g(x)

What happens when we make n=6

f(x) = 6x - 3

g(x) = 6x + 5

h(x) = (6x - 3) + (6x + 5)

This time not only can we see how the y-intercept of h(x) is equal to the sum of the y-intercepts of f(x) and g(x), but also look closely at the slope. The slope of both f(x) and g(x) is equal to 6 and the slope of h(x) is the sum of the slopes of f(x) and g(x)

Does it always work that way...and will h(x) always be a linear graph? Let's look once more with n=20

f(x) = 20x - 3

g(x) = 20x + 5

h(x) = (20x - 3) + (20x + 5)

The y-intercept of h(x) = 2, the sum of the y-intercept of f(x) and y-intercept of g(x) and the slope is 40, the sum of the slope of f(x) and the slope of g(x).

Now let's look closer at the equation, will it always be linear

Consider:

f(x) = ax + b

g(x) = cx + d

Letting h(x) = f(x) + g(x); h(x) = (ax + b) + (cx + d)

Using the communative property we can obtain

h(x) = (ax + cx) + (b + d)

h(x) = (a+c)x + (b + d)

Since addition of real numbers yield a real number we can say

m = (a + c)

n = (b + d)

Therefore, h(x) = mx + n

Therefore, h(x) is also a linear equation whose slope is equal to the sum of the slopes of f(x) and g(x), as is the y-intercept of h(x) equal to the sum of the y-intercepts of f(x) and g(x)

RETURN TO EXPLORING LINEAR EQUATIONS