Express the following linear equations in the form \(ax + by + c = 0\) and indicate the values of \(a, b, c\) in each case:
(i) \(2x + 3y = 9.35\)
(ii) \(x - \dfrac{y}{5} - 10 = 0\)
(iii) \(-2x + 3y = 6\)
(iv) \(x = 3y\)
(v) \(2x = -5y\)
(vi) \(3x + 2 = 0\)
(vii) \(y - 2 = 0\)
(viii) \(5 = 2x\)
(i) \(2x + 3y - 9.35 = 0;\ a = 2,\ b = 3,\ c = -9.35\)
(ii) \(x - \dfrac{y}{5} - 10 = 0;\ a = 1,\ b = -\dfrac{1}{5},\ c = -10\)
(iii) \(-2x + 3y - 6 = 0;\ a = -2,\ b = 3,\ c = -6\)
(iv) \(x - 3y + 0 = 0;\ a = 1,\ b = -3,\ c = 0\)
(v) \(2x + 5y + 0 = 0;\ a = 2,\ b = 5,\ c = 0\)
(vi) \(3x + 0y + 2 = 0;\ a = 3,\ b = 0,\ c = 2\)
(vii) \(0x + 1y - 2 = 0;\ a = 0,\ b = 1,\ c = -2\)
(viii) \(-2x + 0y + 5 = 0;\ a = -2,\ b = 0,\ c = 5\)
Rewrite each equation so it matches \(ax + by + c = 0\). Identify \(a\) with the coefficient of \(x\), \(b\) with the coefficient of \(y\), and \(c\) as the constant term.
(i) Move 9.35 to the left to get \(2x + 3y - 9.35 = 0\). Then \(a = 2\), \(b = 3\), \(c = -9.35\).
(ii) The equation already reads \(x - \dfrac{y}{5} - 10 = 0\). So \(a = 1\), \(b = -\dfrac{1}{5}\), \(c = -10\).
(iii) Shift 6 left: \(-2x + 3y - 6 = 0\). Here \(a = -2\), \(b = 3\), \(c = -6\).
(iv) Write \(x - 3y = 0\) as \(x - 3y + 0 = 0\). Thus \(a = 1\), \(b = -3\), \(c = 0\).
(v) Bring terms together: \(2x + 5y = 0\) becomes \(2x + 5y + 0 = 0\). So \(a = 2\), \(b = 5\), \(c = 0\).
(vi) \(3x + 2 = 0\) already fits: \(a = 3\), \(b = 0\), \(c = 2\).
(vii) \(y - 2 = 0\) matches \(0x + 1y - 2 = 0\). So \(a = 0\), \(b = 1\), \(c = -2\).
(viii) \(5 = 2x\) rearranges to \(-2x + 5 = 0\) or \(-2x + 0y + 5 = 0\). Here \(a = -2\), \(b = 0\), \(c = 5\).