Question

Solve the following pair of linear equations by the substitution method. 
(i) x + y = 14 
    x – y = 4   

(ii) s – t = 3 
    \(\frac{s}{3} + \frac{t}{2}\) =6 

(iii) 3x – y = 3 
      9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3 
     0.4x + 0.5y = 2.3 

(v)\(\sqrt2x\) + \(\sqrt3y\)=0
    \(\sqrt3x\) - \(\sqrt8y\) = 0

(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
    \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

Answer 1

(i) System of equations:
1. \(x + y = 14\)
2. \(x − y = 4\)
From equation (1), \(x = 14 − y\) (3).
Substitute (3) into (2): \((14-y) -y =4\).
This simplifies to \(14 - 2y = 4\), so \(10 = 2y\), which gives \(y = 5\).
Substitute \(y = 5\) into equation (3): \(x = 14 - 5 = 9\).
Solution: \(x = 9, y = 5\).

(ii) System of equations:
1. \(s-t =3\)
2. \(\frac{s}{3} + \frac{t}{2} = 6\)
From equation (1), \(s = t + 3\) (3).
Substitute (3) into (2): \(\frac{t+3}{3} + \frac{t}{2} = 6\).
Multiply by 6 to clear denominators: \(2(t+3) + 3t = 36\).
This simplifies to \(2t + 6 + 3t = 36\), so \(5t = 30\), which gives \(t = 6\).
Substitute \(t = 6\) into equation (3): \(s = 6 + 3 = 9\).
Solution: \(s = 9, t = 6\).

(iii) System of equations:
1. \(3x − y = 3\)
2. \(9x − 3y = 9\)
From equation (1), \(y = 3x − 3\) (3).
Substitute (3) into (2): \(9x - 3(3x - 3) = 9\).
This simplifies to \(9x - 9x + 9 = 9\), resulting in \(9 = 9\).
This identity indicates that the system has infinite solutions. The relationship between the variables is given by \(y = 3x - 3\).
An example solution is \(x = 1, y = 0\).

(iv) System of equations:
1. \(0.2x + 0.3y = 1.3\)
2. \(0.4x + 0.5y = 2.3\)
From equation (1), \(x = \frac{1.3 - 0.3y}{0.2}\) (3).
Substitute (3) into (2): \(0.4(\frac{1.3 - 0.3y}{0.2}) + 0.5y = 2.3\).
This simplifies to \(2(1.3 - 0.3y) + 0.5y = 2.3\), so \(2.6 - 0.6y + 0.5y = 2.3\).
Rearranging gives \(0.3 = 0.1y\), so \(y = 3\).
Substitute \(y = 3\) into equation (3): \(x = \frac{1.3 - 0.3 \times 3}{0.2} = \frac{1.3 - 0.9}{0.2} = \frac{0.4}{0.2} = 2\).
Solution: \(x = 2, y = 3\).

(v) System of equations:
1. \(\sqrt{2}x + \sqrt{3}y = 0\)
2. \(\sqrt{3}x - \sqrt{8}y = 0\)
From equation (1), \(x = -\frac{\sqrt{3}y}{\sqrt{2}}\) (3).
Substitute (3) into (2): \(\sqrt{3}(-\frac{\sqrt{3}y}{\sqrt{2}}) - \sqrt{8}y = 0\).
This simplifies to \(-\frac{3y}{\sqrt{2}} - 2\sqrt{2}y = 0\).
Factoring out \(y\): \(y(-\frac{3}{\sqrt{2}} - 2\sqrt{2}) = 0\).
This implies \(y = 0\).
Substitute \(y = 0\) into equation (3): \(x = -\frac{\sqrt{3}(0)}{\sqrt{2}} = 0\).
Solution: \(x = 0, y = 0\).

(vi) System of equations:
1. \(\frac{3}{2}x - \frac{5}{3}y = -2\)
2. \(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)
From equation (1), multiply by 6 to get \(9x - 10y = -12\). Then, \(x = \frac{-12 + 10y}{9}\) (3).
Substitute (3) into (2): \(\frac{1}{3}(\frac{-12 + 10y}{9}) + \frac{y}{2} = \frac{13}{6}\).
This simplifies to \(\frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6}\).
Multiply by the least common multiple of the denominators (54): \(2(-12 + 10y) + 27y = 9 \times 13\).
So, \(-24 + 20y + 27y = 117\).
This gives \(47y = 141\), so \(y = 3\).
Substitute \(y = 3\) into equation (3): \(x = \frac{-12 + 10 \times 3}{9} = \frac{-12 + 30}{9} = \frac{18}{9} = 2\).
Solution: \(x = 2, y = 3\).