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To determine the equation that satisfies all given pairs of [tex]\((a, b)\)[/tex] values, we look for a linear equation of the form [tex]\( b = ma + c \)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(c\)[/tex] is the y-intercept.
Given the points:
- [tex]\((0, -10)\)[/tex]
- [tex]\((1, -7)\)[/tex]
- [tex]\((2, -4)\)[/tex]
Let's determine the slope [tex]\(m\)[/tex]:
1. Calculate the change in [tex]\(b\)[/tex] values:
- From [tex]\(b = -10\)[/tex] to [tex]\(b = -7\)[/tex]: difference [tex]\(= -7 - (-10) = 3\)[/tex]
- From [tex]\(b = -7\)[/tex] to [tex]\(b = -4\)[/tex]: difference [tex]\(= -4 - (-7) = 3\)[/tex]
2. Calculate the change in [tex]\(a\)[/tex] values:
- From [tex]\(a = 0\)[/tex] to [tex]\(a = 1\)[/tex]: difference [tex]\(= 1 - 0 = 1\)[/tex]
- From [tex]\(a = 1\)[/tex] to [tex]\(a = 2\)[/tex]: difference [tex]\(= 2 - 1 = 1\)[/tex]
3. Determine the slope [tex]\(m\)[/tex]:
- Slope [tex]\(m = \frac{\Delta b}{\Delta a} = \frac{3}{1} = 3\)[/tex]
Next, we find the y-intercept [tex]\(c\)[/tex]:
1. Use the point [tex]\((0, -10)\)[/tex]:
[tex]\[ b = ma + c \rightarrow -10 = 3 \cdot 0 + c \][/tex]
[tex]\[ c = -10 \][/tex]
Thus, the equation of the line is:
[tex]\[ b = 3a - 10 \][/tex]
Hence, the correct equation is [tex]\( \boxed{3a - 10} \)[/tex].
Given the points:
- [tex]\((0, -10)\)[/tex]
- [tex]\((1, -7)\)[/tex]
- [tex]\((2, -4)\)[/tex]
Let's determine the slope [tex]\(m\)[/tex]:
1. Calculate the change in [tex]\(b\)[/tex] values:
- From [tex]\(b = -10\)[/tex] to [tex]\(b = -7\)[/tex]: difference [tex]\(= -7 - (-10) = 3\)[/tex]
- From [tex]\(b = -7\)[/tex] to [tex]\(b = -4\)[/tex]: difference [tex]\(= -4 - (-7) = 3\)[/tex]
2. Calculate the change in [tex]\(a\)[/tex] values:
- From [tex]\(a = 0\)[/tex] to [tex]\(a = 1\)[/tex]: difference [tex]\(= 1 - 0 = 1\)[/tex]
- From [tex]\(a = 1\)[/tex] to [tex]\(a = 2\)[/tex]: difference [tex]\(= 2 - 1 = 1\)[/tex]
3. Determine the slope [tex]\(m\)[/tex]:
- Slope [tex]\(m = \frac{\Delta b}{\Delta a} = \frac{3}{1} = 3\)[/tex]
Next, we find the y-intercept [tex]\(c\)[/tex]:
1. Use the point [tex]\((0, -10)\)[/tex]:
[tex]\[ b = ma + c \rightarrow -10 = 3 \cdot 0 + c \][/tex]
[tex]\[ c = -10 \][/tex]
Thus, the equation of the line is:
[tex]\[ b = 3a - 10 \][/tex]
Hence, the correct equation is [tex]\( \boxed{3a - 10} \)[/tex].
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