Write two equations whose graphs each pass through (3,4)
y=x+1
y=10-2x
or how about the trivial ones
x=3
y=4
To find two equations whose graphs pass through the point (3,4), we need to consider that there are infinitely many possible solutions. However, I will show you two examples.
Equation 1:
We can start by using the slope-intercept form of a linear equation, y = mx + b. Since the point (3,4) lies on the graph, we substitute x = 3 and y = 4 into the equation. Let's find the value of the slope (m) first.
m = (y2 - y1) / (x2 - x1)
= (4 - y1) / (3 - x1)
For simplicity, let's choose (x1, y1) as (0,0). Now we can calculate m:
m = (4 - 0) / (3 - 0)
= 4 / 3
Substituting the slope (m) and the point (3,4) into the equation, we get:
4 = (4/3) * 3 + b
4 = 4 + b
b = 0
Therefore, one equation that passes through the point (3,4) is y = (4/3)x.
Equation 2:
Let's consider a quadratic equation for our second example. The general form of a quadratic equation is y = ax^2 + bx + c. We can substitute the values of x = 3 and y = 4 into this equation to find the values of a, b, and c.
4 = a(3)^2 + b * 3 + c
4 = 9a + 3b + c
Since there are infinitely many solutions for a, b, and c, we need to introduce additional constraints to determine specific values. For instance, let's say we want the graph to also pass through the point (1,10). Substituting this point into the equation, we get:
10 = a(1)^2 + b * 1 + c
10 = a + b + c
Now we have a system of two equations:
4 = 9a + 3b + c
10 = a + b + c
To solve this system of equations, we can use various methods such as substitution or elimination. Solving this system will yield specific values for a, b, and c, giving us a quadratic equation that passes through both (3,4) and (1,10).