The problem provides a table of $x$ and $y$ values that satisfy a linear equation of the form $y = mx + c$, where $m$ and $c$ are constants. The goal is to find the equation of the line described by the table.

AlgebraLinear EquationsSlope-intercept formCoordinate Geometry
2025/4/29

1. Problem Description

The problem provides a table of xx and yy values that satisfy a linear equation of the form y=mx+cy = mx + c, where mm and cc are constants. The goal is to find the equation of the line described by the table.

2. Solution Steps

The given equation is y=mx+cy = mx + c.
We can use two points from the table to determine the values of mm and cc. Let's use the points (0,1)(0, 1) and (2,2)(2, 2).
Using the point (0,1)(0, 1):
1=m(0)+c1 = m(0) + c
1=0+c1 = 0 + c
c=1c = 1
Now that we know c=1c = 1, the equation is y=mx+1y = mx + 1. We can use the point (2,2)(2, 2) to find mm:
2=m(2)+12 = m(2) + 1
21=2m2 - 1 = 2m
1=2m1 = 2m
m=12m = \frac{1}{2}
So the equation of the line is y=12x+1y = \frac{1}{2}x + 1. Let's check if this equation holds true for the third point (4,3)(4, 3):
y=12(4)+1y = \frac{1}{2}(4) + 1
y=2+1y = 2 + 1
y=3y = 3
Since the equation holds true for all three points, the equation of the line is y=12x+1y = \frac{1}{2}x + 1.

3. Final Answer

D. y=12x+1y = \frac{1}{2}x + 1

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