Equate the straight line through (-1; 3) and (2; 4).

y = kx + b, then substituting the data solved by the system of equations, we find k and b
3 = -k + b
4 = 2k + b. “B = 4-2k
3 = -k + 4-2k
3 = -3k + 4
-3k = -1 |: (- 1)
3k = 1
K = 1/3 “b = 4-2 * 1/3 = 2/3
We get the equation of the straight line:
y = 1 / 3x + 2/3



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