(i). \(x + y = 3.3\) and \(\dfrac{0.6}{3x - 2y} = -1\), \(3x - 2y \ne 0\).
(ii). Solve: \(\dfrac{x}{3} + \dfrac{y}{4} = 4\) and \(\dfrac{5x}{6} - \dfrac{y}{8} = 4\).
(iii). Solve: \(4x + \dfrac{6}{y} = 15\) and \(6x - \dfrac{8}{y} = 14\), \(y \ne 0\).
(iv). Solve: \(\dfrac{1}{2x} - \dfrac{1}{y} = -1\) and \(\dfrac{1}{x} + \dfrac{1}{2y} = 8\), \(x,y \ne 0\).
(v). Solve: \(43x + 67y = -24\) and \(67x + 43y = 24\).
(vi). Solve: \(\dfrac{x}{a} + \dfrac{y}{b} = a + b\) and \(\dfrac{x}{a^2} + \dfrac{y}{b^2} = 2\), \(a,b \ne 0\).
(vii). Solve: \(\dfrac{2xy}{x + y} = \dfrac{3}{2}\) and \(\dfrac{xy}{2x - y} = -\dfrac{3}{10}\), with \(x + y \ne 0\) and \(2x - y \ne 0\).
(i). \(x = 1.2\), \(y = 2.1\).
(ii). \(x = 6\), \(y = 8\).
(iii). \(x = 3\), \(y = 2\).
(iv). \(x = \dfrac{1}{6}\), \(y = \dfrac{1}{4}\).
(v). \(x = 1\), \(y = -1\).
(vi). \(x = a^2\), \(y = b^2\).
(vii). \(x = \dfrac{1}{2}\), \(y = -\dfrac{3}{2}\).
We are given two equations:
1) \(x + y = 3.3\)
2) \(\dfrac{0.6}{3x - 2y} = -1\)
From (2):
\(\dfrac{0.6}{3x - 2y} = -1\)
This means: \(3x - 2y = -0.6\).
So the system is:
\(x + y = 3.3\)
\(3x - 2y = -0.6\)
Multiply the first by 2:
\(2x + 2y = 6.6\)
Now add with the second:
\(2x + 2y + 3x - 2y = 6.6 - 0.6\)
\(5x = 6.0\)
So, \(x = 1.2\).
Put in \(x + y = 3.3\):
\(1.2 + y = 3.3\)
So, \(y = 2.1\).
Equations:
\(\dfrac{x}{3} + \dfrac{y}{4} = 4\)
\(\dfrac{5x}{6} - \dfrac{y}{8} = 4\)
Clear fractions by multiplying:
First ×12: \(4x + 3y = 48\)
Second ×24: \(20x - 3y = 96\)
Add them:
\(4x + 20x + 3y - 3y = 48 + 96\)
\(24x = 144\)
So, \(x = 6\).
Put in \(4x + 3y = 48\):
\(24 + 3y = 48\)
\(3y = 24\)
\(y = 8\).
Equations:
\(4x + \dfrac{6}{y} = 15\)
\(6x - \dfrac{8}{y} = 14\)
Let \(t = \dfrac{1}{y}\).
Then:
\(4x + 6t = 15\)
\(6x - 8t = 14\)
Multiply first by 3:
\(12x + 18t = 45\)
Multiply second by 2:
\(12x - 16t = 28\)
Subtract:
\(34t = 17\)
\(t = \dfrac{1}{2}\)
So, \(y = 2\).
Put in first: \(4x + 6(1/2) = 15\)
\(4x + 3 = 15\)
\(4x = 12\)
\(x = 3\).
Equations:
\(\dfrac{1}{2x} - \dfrac{1}{y} = -1\)
\(\dfrac{1}{x} + \dfrac{1}{2y} = 8\)
Let \(u = \dfrac{1}{x}\), \(v = \dfrac{1}{y}\).
So equations become:
\(\dfrac{1}{2}u - v = -1\)
\(u + \dfrac{1}{2}v = 8\)
Multiply both by 2:
\(u - 2v = -2\)
\(2u + v = 16\)
From first: \(u = -2 + 2v\).
Substitute in second:
\(2(-2 + 2v) + v = 16\)
\(-4 + 4v + v = 16\)
\(5v = 20\)
\(v = 4\)
So, \(y = 1/v = 1/4\).
Then \(u = -2 + 2(4) = 6\)
So, \(x = 1/u = 1/6\).
Equations:
\(43x + 67y = -24\)
\(67x + 43y = 24\)
Add them:
\(110x + 110y = 0\)
\(x + y = 0\)
So, \(y = -x\).
Put in first:
\(43x + 67(-x) = -24\)
\(-24x = -24\)
\(x = 1\)
So, \(y = -1\).
Equations:
\(\dfrac{x}{a} + \dfrac{y}{b} = a + b\)
\(\dfrac{x}{a^2} + \dfrac{y}{b^2} = 2\)
Let \(U = \dfrac{x}{a}\), \(V = \dfrac{y}{b}\).
Then:
\(U + V = a + b\)
\(\dfrac{U}{a} + \dfrac{V}{b} = 2\)
Multiply the second by \(ab\):
\(Ub + Va = 2ab\)
But from first: \(U + V = a + b\).
Check consistency: solution is \(U = a, V = b\).
So, \(x = a^2, y = b^2\).
Equation 1:
\(\dfrac{2xy}{x + y} = \dfrac{3}{2}\)
Cross multiply:
\(4xy = 3(x + y)\)
Rearrange:
\(y(4x - 3) = 3x\)
So, \(y = \dfrac{3x}{4x - 3}\).
Equation 2:
\(\dfrac{xy}{2x - y} = -\dfrac{3}{10}\)
Substitute y:
\(\dfrac{x(3x/(4x - 3))}{2x - (3x/(4x - 3))} = -3/10\)
Simplify (step by step):
\(\dfrac{3x^2}{8x^2 - 18x + 9} = -3/10\)
Cancel 3:
\(\dfrac{x^2}{8x^2 - 18x + 9} = -1/10\)
Cross multiply:
\(10x^2 = -(8x^2 - 18x + 9)\)
\(10x^2 = -8x^2 + 18x - 9\)
\(18x^2 - 18x + 9 = 0\)
\(2x^2 - 2x + 1 = 0\) (divided by 9)
Check discriminant: \(D = (-2)^2 - 4*2*1 = 4 - 8 = -4\).
Oops, mistake in earlier simplification.
Let’s go back carefully:
From earlier, simpler substitution works:
\(\dfrac{3x}{8x - 9} = -3/10\)
Cancel 3:
\(\dfrac{x}{8x - 9} = -1/10\)
Cross multiply: \(10x = -(8x - 9)\)
\(10x = -8x + 9\)
\(18x = 9\)
\(x = 1/2\).
Find y:
\(y = \dfrac{3(1/2)}{4(1/2) - 3} = \dfrac{1.5}{2 - 3} = -3/2\).