Solution:

 \(2x - 6y = 3\)
Substituting \(x = - 5\) in the given equation,
  \(2 \times - 5 - 6y = 3\)
∴ \(- 10 - 6y = 3\)
∴ \(- 10 - 3 = 6y\)
∴ \(- 13 = 6y\)
i.e. \(6y = - 13\)

∴ \(\displaystyle y = - \frac {13}{6}\) ... (i)

Also,
 \(2x - 6y = 3\)
Substituting \(y = 0\) in the given equation,
  \(2x - 6 \times 0 = 3\)
∴ \(2x - 0 = 3\)
∴ \(2x = 3\)

∴ \(\displaystyle x = \frac {3}{2}\) ... (ii)

The completed table is given below:

Solution:

 \(\begin{eqnarray} \left| \begin{array}{rrr} 4 & 3 \\ 2 & 7 \\ \end{array} \right| \end{eqnarray}\)

= \(4 \times 7 - 3 \times 2\)
= \(28 - 6\)
= \(22\)

Solution:

 \(\begin{eqnarray} \left| \begin{array}{rrr} 5 & - 2 \\ - 3 & 1 \\ \end{array} \right| \end{eqnarray}\)

= \(5 \times 1 - (- 2) \times (- 3)\)
= \(5 - 6\)
= \(- 1\)

Solution:

 \(\begin{eqnarray} \left| \begin{array}{rrr} 3 & - 1 \\ 1 & 4 \\ \end{array} \right| \end{eqnarray}\)

= \(3 \times 4 - (- 1) \times 1\)
= \(12 + 1\)
= \(13\)

Solution:

First let’s express the given equations in the standard form:
 \(6x - 3y = - 10\)
and \(3x + 5y = 8\)

Comparing with the standard form:
\(a_{1} = 6, b_{1} =\:-\:3, c_{1} = \:-\:10,\\a_{2} = 3, b_{2} = 5, c_{2} = 8\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D = \left| \begin{array}{rrr} 6 & - 3 \\ 3 & 5 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D = 6 \times 5\:-\:(\:-\:3) \times 3\)
∴ \(D = 30 + 9\)
∴ \(D = 39\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{x} = \left| \begin{array}{rrr} - 10 & - 3 \\ 8 & 5 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{x} = \:-\:10 \times 5\:-\:(\:-\:3) \times 8\)
∴ \(D_{x} = \:-\:50 + 24\)
∴ \(D_{x} = \:-\:26\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{y} = \left| \begin{array}{rrr} 6 & - 10 \\ 3 & 8 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{y} = 6 \times 8\:-\:(\:-\:10) \times 3\)
∴ \(D_{y} = 48 + 30\)
∴ \(D_{y} = 78\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{- 2}{\:-\:26}}{\cancelto{3}{39}}\)

∴ \(\displaystyle x = - \frac {2}{3}\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{2}{78}}{\cancelto{1}{39}}\)

∴ \(\displaystyle y = 2\) ... (v)

∴ The solution is \(\displaystyle (x, y) = \left(- \frac {2}{3},\ 2\right)\)

Solution:

Comparing with the standard form:
\(a_{1} = 4, b_{1} =\:-\:2, c_{1} = \:-\:4,\\a_{2} = 4, b_{2} = 3, c_{2} = 16\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D = \left| \begin{array}{rrr} 4 & - 2 \\ 4 & 3 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D = 4 \times 3\:-\:(\:-\:2) \times 4\)
∴ \(D = 12 + 8\)
∴ \(D = 20\) ... (i)

Also, \(D_{m} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{m} = \left| \begin{array}{rrr} - 4 & - 2 \\ 16 & 3 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{m} = \:-\:4 \times 3\:-\:(\:-\:2) \times 16\)
∴ \(D_{m} = \:-\:12 + 32\)
∴ \(D_{m} = 20\) ... (ii)

And, \(D_{n} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{n} = \left| \begin{array}{rrr} 4 & - 4 \\ 4 & 16 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{n} = 4 \times 16\:-\:(\:-\:4) \times 4\)
∴ \(D_{n} = 64 + 16\)
∴ \(D_{n} = 80\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle m = \frac{D_{m}}{D}\)

∴ \(\displaystyle m = \frac{\cancelto{1}{20}}{\cancelto{1}{20}}\)

∴ \(m = 1\) ... (iv)

Also, \(\displaystyle n = \frac{D_{n}}{D}\)

∴ \(\displaystyle n = \frac{\cancelto{4}{80}}{\cancelto{1}{20}}\)

∴ \(n = 4\) ... (v)

∴ The solution is \(\displaystyle (m, n) = (1, 4)\)

Solution:

 \(\displaystyle 3x - 2y = \frac {5}{2}\) ... (i)

and \(\displaystyle \frac {1}{3}x + 3y = - \frac {4}{3}\) ... (ii)

Multiplying equation (i) by 2,

 \(\displaystyle 3x \times 2 - 2y \times 2 = \frac {5}{\cancel {2}} \times \cancel {2}\)

∴ \(6x - 4y = 5\) ... (iii)

Multiplying equation (ii) by 3,

 \(\displaystyle \frac {1}{\cancel{3}}x \times \cancel{3} + 3y \times 3 = - \frac {4}{\cancel{3}} \times \cancel{3}\)

∴ \(x + 9y = - 4\) ... (iv)

Comparing with the standard form:
\(a_{1} = 6, b_{1} =\:-\:4, c_{1} = 5,\\a_{2} = 1, b_{2} = 9, c_{2} = -4\)

 \(\displaystyle D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D = \left| \begin{array}{rrr} 6 & - 4 \\ - 4 & 9 \\ \end{array} \right| \end{eqnarray}\)

∴ \(\displaystyle D = 6 \times 9\:-\:(\:-\:4) \times 1\)
∴ \(\displaystyle D = 54 + 4\)
∴ \(\displaystyle D = 58\) ... (v)

Also, \(\displaystyle D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{x} = \left| \begin{array}{rrr} 5 & - 4 \\ - 4 & 9 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{x} = 5 \times 9\:-\:(\:-\:4) \times (-4)\)
∴ \(D_{x} = 45 - 16\)
∴ \(D_{x} = 29\) ... (vi)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{y} = \left| \begin{array}{rrr} 6 & 5 \\ 1 & - 4 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{y} = 6 \times (-4)\:-\:(5) \times 1\)
∴ \(D_{y} = -24 - 5\)
∴ \(D_{y} = - 29\) ... (vii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{1}{29}}{\cancelto{2}{58}}\)

∴ \(\displaystyle x = \frac{1}{2}\) ... (viii)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{- 1}{- 29}}{\cancelto{2}{58}}\)

∴ \(\displaystyle y = -\frac{1}{2}\) ... (ix)

∴ The solution is \(\displaystyle (x, y) = \left(\frac{1}{2}, -\frac{1}{2}\right)\)

Solution:

Let’s express the given equations in the standard form:
 \(\displaystyle 7x + 3y = 15\) ... (i)
and \(\displaystyle - 5x + 12y = 39\) ... (ii)

Comparing with the standard form:
\(a_{1} = 7, b_{1} = 3, c_{1} = 15,\\a_{2} = - 5, b_{2} = 12, c_{2} = 39\)

 \(\displaystyle D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D = \left| \begin{array}{rrr} 7 & 3 \\ - 5 & 12 \\ \end{array} \right| \end{eqnarray}\)

∴ \(\displaystyle D = 7 \times 12\:-\:3 \times (- 5)\)
∴ \(\displaystyle D = 84 + 15\)
∴ \(\displaystyle D = 99\) ... (iii)

Also, \(\displaystyle D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{x} = \left| \begin{array}{rrr} 15 & 3 \\ 39 & 12 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{x} = 15 \times 12\:-\:3 \times 39\)
∴ \(D_{x} = 180 - 117\)
∴ \(D_{x} = 63\) ... (iv)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{y} = \left| \begin{array}{rrr} 7 & 15 \\ - 5 & 39 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{y} = 7 \times 39\:-\:(15) \times (- 5)\)
∴ \(D_{y} = 273 + 75\)
∴ \(D_{y} = 348\) ... (v)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto {7}{63}}{\cancelto {11}{99}}\)

∴ \(\displaystyle x = \frac{7}{11}\) ... (vi)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto {116}{348}}{\cancelto {33}{99}}\)

∴ \(\displaystyle y = \frac{116}{33}\) ... (vii)

∴ The solution is \(\displaystyle (x, y) = \left(\frac{7}{11}, \frac{116}{33}\right)\)

Solution:

First, let’s prepare two simultaneous equations from the given relations:

 \(\displaystyle \bbox[yellow, 5pt, border: 2px dotted red]{\frac {x + y - 8}{2} = \frac {x + 2y - 14}{3}} = \frac {3x - y}{4}\)

∴\(\displaystyle \frac {x + y - 8}{2} = \frac {x + 2y - 14}{3}\)

By cross multiplication,
 \(3(x + y - 8) = 2(x + 2y - 14)\)
∴ \(3x + 3y - 24 = 2x + 4y - 28\)
∴ \(3x + 3y - 2x - 4y = - 28 + 24\)
∴ \(x - y = - 4\) ... (i)

Also, \(\displaystyle \frac {x + y - 8}{2} = \bbox[yellow, 5pt, border: 2px dotted red]{\frac {x + 2y - 14}{3} = \frac {3x - y}{4}}\)

∴ \(\displaystyle \frac {x + 2y - 14}{3} = \frac {3x - y}{4}\)

By cross multiplication,
 \(4(x + 2y - 14) = 3(3x - y)\)
∴ \(4x + 8y - 56 = 9x - 3y\)
∴ \(4x + 8y - 9x + 3y = 56\)
∴ \(- 5x + 11y = 56\) ... (ii)

Now, comparing with the standard form:
\(a_{1} = 1, b_{1} = - 1, c_{1} = - 4,\\a_{2} = - 5, b_{2} = 11, c_{2} = 56\)

 \(\displaystyle D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D = \left| \begin{array}{rrr} 1 & - 1 \\ - 5 & 11 \\ \end{array} \right| \end{eqnarray}\)

∴ \(\displaystyle D = 1 \times 11\:-\:(- 1) \times (- 5)\)
∴ \(\displaystyle D = 11 - 5\)
∴ \(\displaystyle D = 6\) ... (iii)

Also, \(\displaystyle D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{x} = \left| \begin{array}{rrr} - 4 & - 1 \\ 56 & 11 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{x} = (- 4) \times 11\:-\:(- 1) \times 56\)
∴ \(D_{x} = - 44 + 56\)
∴ \(D_{x} = 12\) ... (iv)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(\begin{eqnarray} D_{x} = \left| \begin{array}{rrr} 1 & - 4 \\ - 5 & 56 \\ \end{array} \right| \end{eqnarray}\)

∴ \(D_{y} = 1 \times 56\:-\:(- 4) \times (- 5)\)
∴ \(D_{y} = 56 - 20\)
∴ \(D_{y} = 36\) ... (v)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto {2}{12}}{\cancelto {1}{6}}\)

∴ \(\displaystyle x = 2\) ... (vi)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto {6}{36}}{\cancelto {1}{6}}\)

∴ \(y = 6\) ... (vii)

∴ The solution is \((x, y) = (2, 6)\)

Solution:

 \(\displaystyle \frac {2}{x} + \frac {2}{3y} = \frac {1}{6}\) ... (i)

and \(\displaystyle \frac {3}{x} + \frac {2}{y} = 0\) ... (ii)

Let \(\displaystyle \frac {1}{x} = a\) and \(\displaystyle \frac {1}{y} = b\),

∴ The given equations:

 \(\displaystyle 2a + \frac {2b}{3} = \frac {1}{6}\) ... (iii)

and \(3a + 2b = 0\) ... (iv)

Multiplying equation (iii) by 6,

 \(\displaystyle 2a \times 6 + \frac {2b}{\cancelto {1}{3}} \times \cancelto {2}{6} = \frac {1}{\cancel{6}} \times \cancel{6}\)

∴ \(12a + 4b = 1\) ... (v)

Multiplying equation (iv) by 2,

 \(3a \times 2 + 2b \times 2 = 0 \times 2\)

∴ \(6a + 4b = 0\) ... (vi)

Subtracting (vi) from (v),

∴ \(\displaystyle a = \frac {1}{6}\) ... (vii)

Substituting the value of \(a\) in (vi),

 \(6a + 4b = 0\) ... (vi)

∴ \(\displaystyle \cancel {6} \times \frac {1}{\cancel {6}} + 4b = 0\)

∴ \(1 + 4b = 0\)

∴ \(4b = - 1\)

∴ \(\displaystyle b = - \frac {1}{4}\) ... (viii)

Replacing the value of \(a\),

 \(\displaystyle \frac {1}{x} = a\)

∴ \(\displaystyle \frac {1}{x} = \frac {1}{6}\)

By invertendo,
 \(x = 6\) ... (ix)

Replacing the value of \(b\),

 \(\displaystyle \frac {1}{y} = b\)

∴ \(\displaystyle \frac {1}{y} = - \frac {1}{4}\)

By invertendo,
 \(y = - 4\) ... (x)

∴ The solution is \((x, y) = (6, - 4)\)

Solution:

 \(\displaystyle \frac {7}{2x + 1} + \frac {13}{y + 2} = 27\) ... (i)

and \(\displaystyle \frac {13}{2x + 1} + \frac {7}{y + 2} = 33\) ... (ii)

Let \(\displaystyle \frac {1}{2x + 1} = a\) and \(\displaystyle \frac {1}{y + 2} = b\),

∴ The given equations:
  \(7a + 13b = 27\) ... (iii)
and \(13a + 7b = 33\) ... (iv)

Adding (iii) and (iv),

Dividing both sides by 20,
and \(a + b = 3\) ... (v)

Subtracting (iv) from (iii),

Dividing both sides by − 6,
and \(a − b = 1\) ... (vi)

Adding (v) and (vi),

∴ \(\displaystyle a = \frac {\cancelto {2}{4}}{\cancelto {1}{2}}\)

∴ \(a = 2\) ... (vii)

Substituting the value of \(a\) in equation (v),
 \(a + b = 3\) ... (v)
∴ \(2 + b = 3\)
∴ \(b = 3 − 2\)
∴ \(b = 1\) ... (viii)

Replacing the value of \(a\),

 \(\displaystyle \frac {1}{2x + 1} = a\)

∴ \(\displaystyle \frac {1}{2x + 1} = 2\)

∴ \(1 = 2(2x + 1)\)
∴ \(1 = 4x + 2\)
∴ \(1 - 2 = 4x\)
∴ \(- 1 = 4x\)
∴ \(4x = - 1\)

∴ \(\displaystyle x = - \frac {1}{4}\) ... (ix)

Replacing the value of \(b\),

 \(\displaystyle \frac {1}{y + 2} = b\)

∴ \(\displaystyle \frac {1}{y + 2} = 1\)

∴ \(1 = 1(y + 2)\)
∴ \(1 = y + 2\)
∴ \(1 - 2 = y\)
∴ \(- 1 = y\)
i.e. \(y = - 1\) ... (x)

∴ The solution is \(\displaystyle (x, y) = \left(- \frac {1}{4}, - 1\right)\)

Solution:

 \(\displaystyle \frac {148}{x} + \frac {231}{y} = \frac {527}{xy}\) ... (i)

and \(\displaystyle \frac {231}{x} + \frac {148}{y} = \frac {610}{xy}\) ... (ii)

Multiplying equation (i) by \(xy\),

 \(\displaystyle \frac {148}{\cancel{x}} \times \cancel{x}y + \frac {231}{\cancel{y}} \times x\cancel{y} = \frac {527}{\cancel{xy}} \times \cancel{xy}\)

∴ \(148y + 231x = 527\) ... (iii)

Multiplying equation (ii) by \(xy\),

 \(\displaystyle \frac {231}{\cancel{x}} \times \cancel{x}y + \frac {148}{\cancel{y}} \times x\cancel{y} = \frac {610}{\cancel{xy}} \times \cancel{xy}\)

∴ \(231y + 148x = 610\) ... (iv)

Adding (iii) and (iv),

Dividing both sides by 379,
and \(y + x = 3\) ... (v)

Subtracting (iv) from (iii),

Dividing both sides by − 83,
and \(y − x = 1\) ... (vi)

Adding (v) and (vi),

∴ \(\displaystyle y = \frac {\cancelto {2}{4}}{\cancelto {1}{2}}\)

∴ \(y = 2\) ... (vii)

Substituting the value of \(y\) in equation (iii),
 \(y + x = 3\) ... (iii)
∴ \(2 + x = 3\)
∴ \(x = 3 − 2\)
∴ \(x = 1\) ... (viii)

∴ The solution is \((x, y) = (1, 2)\)

Solution:

 \(\displaystyle \frac {7x - 2y}{xy} = 5\) ... (i)

∴ \(\displaystyle \frac {7\cancel{x}}{\cancel{x}y} - \frac {2\cancel{y}}{x\cancel{y}} = 5\)

∴ \(\displaystyle \frac {7}{y} - \frac {2}{x} = 5\) ... (ii)

and \(\displaystyle \frac {8x + 7y}{xy} = 15\) ... (iii)

∴ \(\displaystyle \frac {8\cancel{x}}{\cancel{x}y} + \frac {7\cancel{y}}{x\cancel{y}} = 15\)

∴ \(\displaystyle \frac {8}{y} + \frac {7}{x} = 15\) ... (iv)

Substituting \(\displaystyle \frac {1}{y} = a\) and \(\displaystyle \frac {1}{x} = b\)

The given equations:
 \(7a - 2b = 5\) ... (v)
and \(8a + 7b = 15\) ... (vi)

Multiplying equation (v) by 7,
 \(49a - 14b = 35\) ... (vii)

Multiplying equation (vi) by 2,
 \(16a + 14b = 30\) ... (viii)

Adding (vii) and (viii),

∴ \(\displaystyle a = \frac {\cancel{65}}{\cancel{65}}\)

∴ \(a = 1\) ... (ix)

Substituting the value of \(a\) in (vi),
 \(8a + 7b = 15\) ... (vi)
∴ \(8 \times 1 + 7b = 15\)
∴ \(8 + 7b = 15\)
∴ \(7b = 15 - 8\)
∴ \(7b = 7\)

∴ \(\displaystyle b = \frac {\cancel{7}}{\cancel{7}}\)

∴ \(b = 1\) ... (x)

Replacing the value of \(a\),

 \(\displaystyle \frac {1}{y} = a\)

∴ \(\displaystyle \frac {1}{y} = 1\)

∴ \(1 = y\)

i.e. \(y = 1\) ... (xi)

Replacing the value of \(b\),

 \(\displaystyle \frac {1}{x} = b\)

∴ \(\displaystyle \frac {1}{x} = 1\)

∴ \(1 = x\)

i.e. \(x = 1\) ... (xii)

∴ The solution is \((x, y) = (1, 1)\)

Solution:

 \(\displaystyle \frac {1}{2(3x + 4y)} + \frac {1}{5(2x - 3y)} = \frac {1}{4}\) ... (i)

 \(\displaystyle \frac {5}{(3x + 4y)} - \frac {2}{(2x - 3y)} = - \frac {3}{2}\) ... (ii)

Substituting \(\displaystyle \frac {1}{3x + 4y} = a\) and \(\displaystyle \frac {1}{2x - 3y} = b\)

The given equations:

 \(\displaystyle \frac {a}{2} + \frac {b}{5} = \frac {1}{4}\) ... (iii)

and \(\displaystyle 5a - 2b = - \frac {3}{2}\) ... (iv)

Multiplying equation (iii) by 20,

 \(\displaystyle \frac {a}{\cancelto {1}{2}} \times \cancelto {10}{20} + \frac {b}{\cancelto {1}{5}} \times \cancelto {4}{20} = \frac {1}{\cancelto {1}{4}} \times \cancelto {5}{20} \)

 \(10a + 4b = 5\) ... (v)

Multiplying equation (iv) by 2,

 \(\displaystyle 5a \times {2} - 2b \times {2} = - \frac {3}{\cancel {2}} \times {\cancel {2}}\)

 \(10a - 4b = - 3\) ... (vi)

Adding (v) and (vi),

∴ \(\displaystyle a = \frac {\cancelto {1}{2}}{\cancelto {10}{20}}\)

∴ \(\displaystyle a = \frac {1}{10}\) ... (vii)

Substituting the value of \(a\) in (v),
 \(10a + 4b = 5\) ... (v)

∴ \(\displaystyle \cancel {10} \times \frac {1}{\cancel {10}} + 4b = 5\)

∴ \(1 + 4b = 5\)
∴ \(4b = 5 - 1\)
∴ \(4b = 4\)

∴ \(\displaystyle b = \frac {\cancel{4}}{\cancel{4}}\)

∴ \(b = 1\) ... (viii)

Replacing the value of \(a\),

 \(\displaystyle \frac {1}{3x + 4y} = a\)

∴ \(\displaystyle \frac {1}{3x + 4y} = \frac {1}{10}\)

By Invertendo,
 \(3x + 4y = 10\) ... (ix)

Replacing the value of \(b\),

 \(\displaystyle \frac {1}{2x - 3y} = b\)

∴ \(\displaystyle \frac {1}{2x - 3y} = 1\)

By Invertendo,
 \(2x - 3y = 1\) ... (x)

Multiplying equation (ix) by 3,
 \(9x + 12y = 30\) ... (xi)

Multiplying equation (x) by 4,
 \(8x - 12y = 4\) ... (xii)

Adding (xi) and (xii),

∴ \(\displaystyle x = \frac {\cancelto {2}{34}}{\cancelto {1}{17}}\)

∴ \(x = 2\) ... (xiii)

Substituting the value of \(x\) in equation (ix),
 \(3x + 4y = 10\) ... (ix)
∴ \(3 \times 2 + 4y = 10\)
∴ \(6 + 4y = 10\)
∴ \(4y = 10 - 6\)
∴ \(4y = 4\)

∴ \(\displaystyle y = \frac {\cancelto {1}{4}}{\cancelto {1}{4}}\)

∴ \(y = 1\) ... (xiv)

∴ The solution is \((x, y) = (2, 1)\)

Solution:

Let the digit in unit’s place be \(x\) and that in the ten’s place be \(y\).

∴ The number = \(\bbox[white, 5pt, border: 2px solid red]{10}y + x\).

The number obtained by interchanging the digits is \(\bbox[white, 5pt, border: 2px solid red]{10}x + y\).

According to first condition:
two digit number + the number obtained by interchanging the digits = 143

∴ \(\bbox[white, 5pt, border: 2px solid red]{10y + x} + \bbox[white, 5pt, border: 2px solid red]{10x + y}\) = 143

∴ \(\bbox[white, 5pt, border: 2px solid red]{11x} + \bbox[white, 5pt, border: 2px solid red]{11y}\) = 143

∴ \(11(x + y)= 143\)

Dividing both sides by 11,

 \(x + y = \bbox[white, 5pt, border: 2px solid red]{13}\) ... (I)

From the second condition,
the digit in unit’s place = digit in the ten’s place + 3

∴ \(x = \bbox[white, 5pt, border: 2px solid red]{y} + 3\)

∴ \(x - y = 3\) ... (II)

Adding equations (I) and (II),

∴ \(\displaystyle x = \frac {\cancelto {8}{16}}{\cancelto {1}{2}}\)

∴ \(\bbox[white, 5pt, border: 2px solid red]{x = 8}\)

Putting this value of \(x\) in equation (I),
 \(x + y = 13\) ... (I)
∴ \(8 + \bbox[white, 5pt, border: 2px solid red]{y} = 13\)
∴ \(y = 13 - 8\)
∴ \(y = \bbox[white, 5pt, border: 2px solid red]{5}\)

Now, the original number is \(10y + x\)
= \(10 \times {5} + 8\)
= \(\bbox[white, 5pt, border: 2px solid red]{50} + 8\)
= \(58\)

Solution:

Let, the rate of 1 kg tea be ₹ \(x\) and that of 1 kg sugar be ₹ \(y\).
[Here, we are assuming that the rates at both the places are same.]

From the first condition:
 \(1.5x + 5y + 50 = 700\)
∴ \(1.5x + 5y = 700 - 50\)
∴ \(1.5x + 5y = 650\) ... (i)

From the second condition:
 \(2x + 7y = 880\) ... (ii)

Multiplying equation (i) by 4,
 \(6x + 20y = 2600\) ... (iii)

Multiplying equation (ii) by 3,
 \(6x + 21y = 2640\) ... (iv)

Subtracting (iv) from (iii),

i.e. \(y = 40\) ... (v)

Substituting the value of \(y\) in equation (ii),
 \(2x + 7y = 880\) ... (ii)
∴ \(2x + 7 \times {40} = 880\)
∴ \(2x + 280 = 880\)
∴ \(2x = 880 - 280\)
∴ \(2x = 600\)

∴ \(\displaystyle x = \frac {\cancelto {300}{600}}{\cancelto {1}{2}}\)

∴ \(\displaystyle x = 300\) ... (vi)

∴ The rate of sugar is ₹ 40 per kg and the rate of tea is ₹ 300 per kg.

Solution:

[As the blanks given in the question are ambiguous, we will rephrase the question and solve it. Then you can fill in the blanks as told in your school.]

Question: Anand gave some ₹ 100 and some ₹ 50 notes to Anushka. The total amount was ₹ 2500. If Anand would have given her the amount by interchanging number of notes, Anushka would have received ₹ 500 less than the previous amount Find the number of ₹ 100 and ₹ 50 notes given to Anushka.

Answer:

Suppose Anushka was given \(x\) notes of ₹ 100, and \(y\) notes of ₹ 50.

From the first condition,
 \(100x + 50y = 2500\) ... (i)

From the second condition,
 \(50x + 100y = 2000\) ... (ii)

Adding (i) and (ii),

Dividing both sides by 150,
 \(x + y = 30\) ... (iii)

Subtracting (ii) from (i),

Dividing both sides by 50,
 \(x - y = 10\) ... (iv)

Adding (iii) and (iv),

∴ \(\displaystyle x = \frac {\cancelto {20}{40}}{\cancelto {1}{2}}\)

∴ \(x = 20\) ... (v)

Substituting the value of \(x\) in equation (iii),
 \(x + y = 30\) ... (iii)
∴ \(20 + y = 30\)
∴ \(y = 30 - 20\)
∴ \(y = 10\) ... (vi)

∴ 20 notes of ₹ 100 and 10 notes of ₹ 50 were given to Anushka.

Solution:

Let the present age of Manish be \(x\) years and the present age of Savita be \(y\) years.

According to the first condition:
 \(x + y = 31\) ... (i)

According to the second condition:
 \(x - 3 = 4(y - 3)\)
∴ \(x - 3 = 4y - 12\)
∴ \(x - 4y = - 12 + 3\)
∴ \(x - 4y = - 9\) ... (ii)

Subtracting (ii) from (i),

∴ \(\displaystyle y = \frac {\cancelto {8}{40}}{\cancelto {1}{5}}\)

∴ \(y = 8\) ... (iii)

Substituting the value of \(y\) in equation (i),
 \(x + y = 31\) ... (i)
∴ \(x + 8 = 31\)
∴ \(x = 31 - 8\)
∴ \(x = 23\) ... (iv)

∴ The present age of Manish is 23 years and the present age of Savita is 8 years.

Solution:

Let the salary of the skilled worker be ₹ \(x\) and that of the unskilled worker be ₹ \(y\).

From the first condition,

 \(\displaystyle \frac {x}{y} = \frac {5}{3}\)

By cross multiplication
 \(3x = 5y\)
∴ \(3x - 5y = 0\) ... (i)

From the second condition,

 \(x + y = 720\) ... (ii)

Multiplying equation (ii) by 5,
 \(5x + 5y = 3600\) ... (iii)

Adding (i) and (iii),

∴ \(\displaystyle x = \frac {\cancelto {450}{3600}}{\cancelto {1}{8}}\)

∴ \(x = 450\) ... (iv)

Substituting the value of \(x\) in equation (ii),
 \(x + y = 720\) ... (ii)
∴ \(450 + y = 720\)
∴ \(y = 720 - 450\)
∴ \(y = 270\) ... (v)

∴ The salary of the skilled worker is ₹ 450/- and that of the unskilled worker is ₹ 270/-.

Solution:

Let, the speed of the bike starting from point ‘A’ be \(x\) km/hr and the speed of the bike starting from point ‘B’ be \(y\) km/hr.

Here, \(x \gt y\)

From the first condition,

 \(\displaystyle x \times {\frac {20}{60}} + y \times {\frac {20}{60}} = 30\)

Multiplying both sides by 3,

 \(\displaystyle x \times {\frac {\cancel {20}}{\cancel {60}}} \times {\cancel {3}} + y \times {\frac {\bcancel {20}}{\bcancel {60}}} \times {\bcancel{3}} = 30 \times {3}\)

∴ \(x + y = 90\) ... (i)

From the second condition,
 \(3x - 3y = 30\) ... (ii)

Multiplying equation (i) by 3,
 \(3x + 3y = 270\) ... (iii)

Adding (ii) and (iii),

∴ \(\displaystyle x = \frac {\cancelto {50}{300}}{\cancelto {1}{6}}\)

∴ \(x = 50\) km/hr ... (iv)

Substituting the value of \(x\) in equation (i),
 \(x + y = 90\) ... (i)
∴ \(50 + y = 90\)
∴ \(y = 90 - 50\)
∴ \(y = 40\) km/hr ... (v)

∴ Hamid’s speed is 50 km/hr and Joseph’s speed is 40 km/hr.