Q16. $3\text{x} + 5(\text{y} + 2) = 1$ and $3\text{x} + 8\text{y} = 0$
Show Answer
$\Rightarrow 3\text{x} + 5(\text{y} + 2) = 1$
$\Rightarrow 3\text{x} + 5\text{y} + 10 = 1$
$\Rightarrow 3\text{x} + 5\text{y} = 1 - 10$
$\Rightarrow 3\text{x} + 5\text{y} = - \ 9$......eq(i)
$\Rightarrow 3\text{x} + 8\text{y} = 0$......eq(ii)
Subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 3\text{x} & + & 5\text{y} & = & - \ 9\\ & 3\text{x} & + & 8\text{y} & = & 0\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 3\text{y} & = & - \ 9\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{9}{3} = 3$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 8(3) = 0$
$\Rightarrow 3\text{x} + 24 = 0$
$\Rightarrow 3\text{x} = - \ 24$
$\Rightarrow \text{x} = - \ \dfrac{24}{3} = - \ 8$
$\therefore$ x = $- \ 8$ and y = 3 Ans
$\Rightarrow 3\text{x} + 5\text{y} + 10 = 1$
$\Rightarrow 3\text{x} + 5\text{y} = 1 - 10$
$\Rightarrow 3\text{x} + 5\text{y} = - \ 9$......eq(i)
$\Rightarrow 3\text{x} + 8\text{y} = 0$......eq(ii)
Subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 3\text{x} & + & 5\text{y} & = & - \ 9\\ & 3\text{x} & + & 8\text{y} & = & 0\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 3\text{y} & = & - \ 9\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{9}{3} = 3$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 8(3) = 0$
$\Rightarrow 3\text{x} + 24 = 0$
$\Rightarrow 3\text{x} = - \ 24$
$\Rightarrow \text{x} = - \ \dfrac{24}{3} = - \ 8$
$\therefore$ x = $- \ 8$ and y = 3 Ans
Q17. $\dfrac{\text{x}}{3} = \dfrac{\text{y}}{2}$ and $\dfrac{2\text{x}}{3} - \dfrac{\text{y}}{2} = 2$
Show Answer
$\Rightarrow \dfrac{\text{x}}{3} = \dfrac{\text{y}}{2} \Rightarrow 2\text{x} = 3\text{y}$
$\Rightarrow 2\text{x} - 3\text{y} = 0$......eq(i)
$\Rightarrow \dfrac{2\text{x}}{3} - \dfrac{\text{y}}{2} = 2$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{2\text{x}}{3} - (6)\dfrac{\text{y}}{2} = (6)2$
$\Rightarrow 4\text{x} - 3\text{y} = 12$......eq(ii)
Multiply eq(i) by 2 and subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 4\text{x} & - & 6\text{y} & = & 0\\ & 4\text{x} & - & 3\text{y} & = & 12\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & - \ 3\text{y} & = & - \ 12\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{12}{3} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow 2\text{x} - 3(4) = 0$
$\Rightarrow 2\text{x} - 12 = 0$
$\Rightarrow 2\text{x} = 12$
$\Rightarrow \text{x} = \dfrac{12}{2} = 6$
$\therefore$ x = 6 and y = 4 Ans
$\Rightarrow 2\text{x} - 3\text{y} = 0$......eq(i)
$\Rightarrow \dfrac{2\text{x}}{3} - \dfrac{\text{y}}{2} = 2$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{2\text{x}}{3} - (6)\dfrac{\text{y}}{2} = (6)2$
$\Rightarrow 4\text{x} - 3\text{y} = 12$......eq(ii)
Multiply eq(i) by 2 and subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 4\text{x} & - & 6\text{y} & = & 0\\ & 4\text{x} & - & 3\text{y} & = & 12\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & - \ 3\text{y} & = & - \ 12\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{12}{3} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow 2\text{x} - 3(4) = 0$
$\Rightarrow 2\text{x} - 12 = 0$
$\Rightarrow 2\text{x} = 12$
$\Rightarrow \text{x} = \dfrac{12}{2} = 6$
$\therefore$ x = 6 and y = 4 Ans
Q18. $\dfrac{\text{x}}{2} - \dfrac{\text{y}}{3} = 2$ and $\dfrac{\text{x}}{5} + \dfrac{\text{y}}{3} = 15$
Show Answer
$\Rightarrow \dfrac{\text{x}}{2} - \dfrac{\text{y}}{3} = 2$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{x}}{2} - (6)\dfrac{\text{y}}{3} = (6)2$
$\Rightarrow 3\text{x} - 2\text{y} = 12$......eq(i)
$\Rightarrow \dfrac{\text{x}}{5} + \dfrac{\text{y}}{3} = 15$
Simplify by taking L.C.M. = 15
$\Rightarrow (15)\dfrac{\text{x}}{5} + (15)\dfrac{\text{y}}{3} = (15)15$
$\Rightarrow 3\text{x} + 5\text{y} = 225$......eq(ii)
In both equations the coefficients of x being the same, subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 3\text{x} & - & 2\text{y} & = & 12\\ & 3\text{x} & + & 5\text{y} & = & 225\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 7\text{y} & = & - \ 213\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{213}{7} = 30\dfrac{3}{7}$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} - 2(\dfrac{213}{7}) = 12$
$\Rightarrow 3\text{x} - \dfrac{426}{7} = 12$
Simplify by taking L.C.M. = 7
$\Rightarrow (7)3\text{x} - 426 = (7)12$
$\Rightarrow 21\text{x} - 426 = 84$
$\Rightarrow 21\text{x} = 84 + 426$
$\Rightarrow 21\text{x} = 510$
$\Rightarrow \text{x} = \dfrac{510}{21} = \dfrac{170}{7} = 24\dfrac{2}{7}$
$\therefore$ x = $24\dfrac{2}{7}$ and y = $30\dfrac{3}{7}$ Ans
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{x}}{2} - (6)\dfrac{\text{y}}{3} = (6)2$
$\Rightarrow 3\text{x} - 2\text{y} = 12$......eq(i)
$\Rightarrow \dfrac{\text{x}}{5} + \dfrac{\text{y}}{3} = 15$
Simplify by taking L.C.M. = 15
$\Rightarrow (15)\dfrac{\text{x}}{5} + (15)\dfrac{\text{y}}{3} = (15)15$
$\Rightarrow 3\text{x} + 5\text{y} = 225$......eq(ii)
In both equations the coefficients of x being the same, subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 3\text{x} & - & 2\text{y} & = & 12\\ & 3\text{x} & + & 5\text{y} & = & 225\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 7\text{y} & = & - \ 213\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{213}{7} = 30\dfrac{3}{7}$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} - 2(\dfrac{213}{7}) = 12$
$\Rightarrow 3\text{x} - \dfrac{426}{7} = 12$
Simplify by taking L.C.M. = 7
$\Rightarrow (7)3\text{x} - 426 = (7)12$
$\Rightarrow 21\text{x} - 426 = 84$
$\Rightarrow 21\text{x} = 84 + 426$
$\Rightarrow 21\text{x} = 510$
$\Rightarrow \text{x} = \dfrac{510}{21} = \dfrac{170}{7} = 24\dfrac{2}{7}$
$\therefore$ x = $24\dfrac{2}{7}$ and y = $30\dfrac{3}{7}$ Ans
Q19. $\dfrac{\text{x}}{3} + \dfrac{\text{x} + \text{y}}{6} = 3$ and $\dfrac{\text{y}}{3} - \dfrac{\text{x} - \text{y}}{2} = 6$
Show Answer
$\Rightarrow \dfrac{\text{x}}{3} + \dfrac{\text{x} + \text{y}}{6} = 3$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{x}}{3} + (6)\dfrac{\text{x} + \text{y}}{6} = (6)3$
$\Rightarrow 2\text{x} + \text{x} + \text{y} = 18$
$\Rightarrow 3\text{x} + \text{y} = 18$......eq(i)
$\Rightarrow \dfrac{\text{y}}{3} - \dfrac{\text{x} - \text{y}}{2} = 6$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{y}}{3} - (6)\dfrac{\text{x} - \text{y}}{2} = (6)6$
$\Rightarrow 2\text{y} - 3(\text{x} - \text{y}) = 36$
$\Rightarrow 2\text{y} - 3\text{x} + 3\text{y} = 36$
$\Rightarrow - \ 3\text{x} + 5\text{y} = 36$......eq(ii)
In both equations the coefficients of x being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 3\text{x} & + & \text{y} & = & 18\\ - \ & 3\text{x} & + & 5\text{y} & = & 36\\ \hline\\ & & & 6\text{y} & = & 54\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{54}{6} = 9$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 9 = 18$
$\Rightarrow 3\text{x} = 18 - 9$
$\Rightarrow 3\text{x} = 9$
$\Rightarrow \text{x} = \dfrac{9}{3} = 3$
$\therefore$ x = 3 and y = 9 Ans
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{x}}{3} + (6)\dfrac{\text{x} + \text{y}}{6} = (6)3$
$\Rightarrow 2\text{x} + \text{x} + \text{y} = 18$
$\Rightarrow 3\text{x} + \text{y} = 18$......eq(i)
$\Rightarrow \dfrac{\text{y}}{3} - \dfrac{\text{x} - \text{y}}{2} = 6$
Simplify by taking L.C.M. = 6
$\Rightarrow (6)\dfrac{\text{y}}{3} - (6)\dfrac{\text{x} - \text{y}}{2} = (6)6$
$\Rightarrow 2\text{y} - 3(\text{x} - \text{y}) = 36$
$\Rightarrow 2\text{y} - 3\text{x} + 3\text{y} = 36$
$\Rightarrow - \ 3\text{x} + 5\text{y} = 36$......eq(ii)
In both equations the coefficients of x being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 3\text{x} & + & \text{y} & = & 18\\ - \ & 3\text{x} & + & 5\text{y} & = & 36\\ \hline\\ & & & 6\text{y} & = & 54\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{54}{6} = 9$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 9 = 18$
$\Rightarrow 3\text{x} = 18 - 9$
$\Rightarrow 3\text{x} = 9$
$\Rightarrow \text{x} = \dfrac{9}{3} = 3$
$\therefore$ x = 3 and y = 9 Ans
Q20. $\dfrac{\text{a}}{4} - \dfrac{\text{b}}{3} = 0$ and $\dfrac{3\text{a} + 8}{5} = \dfrac{2\text{b} - 1}{2}$
Show Answer
$\Rightarrow \dfrac{\text{a}}{4} - \dfrac{\text{b}}{3} = 0$
Simplify by taking L.C.M. = 12
$\Rightarrow (12)\dfrac{\text{a}}{4} - (12)\dfrac{\text{b}}{3} = (12)0$
$\Rightarrow 3\text{a} - 4\text{b} = 0$......eq(i)
$\Rightarrow \dfrac{3\text{a} + 8}{5} = \dfrac{2\text{b} - 1}{2}$
$\Rightarrow 2(3\text{a} + 8) = 5(2\text{b} - 1)$
$\Rightarrow 6\text{a} + 16 = 10\text{b} - 5$
$\Rightarrow 6\text{a} - 10\text{b} = - \ 5 - 16$
$\Rightarrow 6\text{a} - 10\text{b} = - \ 21$......eq(ii)
Now multiply eq(i) by 2 and subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 6\text{a} & - & 8\text{b} & = & 0\\ & 6\text{a} & - & 10\text{b} & = & - \ 21\\ (-) & & (+) & & & (+) \\\\ \hline\\ & & & 2\text{b} & = & 21\\ \end{array} $$ $\Rightarrow \text{b} = \dfrac{21}{2} = 10.5$
Now substitute the value of b in any one equation:
$\Rightarrow 3\text{a} - 4(\dfrac{21}{2}) = 0$
$\Rightarrow 3\text{a} - 2(21) = 0$
$\Rightarrow 3\text{a} - 42 = 0$
$\Rightarrow 3\text{a} = 0 + 42$
$\Rightarrow 3\text{a} = 42$
$\Rightarrow \text{a} = \dfrac{42}{3} = 14$
$\therefore$ a = 14 and b = 10.5 Ans
Simplify by taking L.C.M. = 12
$\Rightarrow (12)\dfrac{\text{a}}{4} - (12)\dfrac{\text{b}}{3} = (12)0$
$\Rightarrow 3\text{a} - 4\text{b} = 0$......eq(i)
$\Rightarrow \dfrac{3\text{a} + 8}{5} = \dfrac{2\text{b} - 1}{2}$
$\Rightarrow 2(3\text{a} + 8) = 5(2\text{b} - 1)$
$\Rightarrow 6\text{a} + 16 = 10\text{b} - 5$
$\Rightarrow 6\text{a} - 10\text{b} = - \ 5 - 16$
$\Rightarrow 6\text{a} - 10\text{b} = - \ 21$......eq(ii)
Now multiply eq(i) by 2 and subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 6\text{a} & - & 8\text{b} & = & 0\\ & 6\text{a} & - & 10\text{b} & = & - \ 21\\ (-) & & (+) & & & (+) \\\\ \hline\\ & & & 2\text{b} & = & 21\\ \end{array} $$ $\Rightarrow \text{b} = \dfrac{21}{2} = 10.5$
Now substitute the value of b in any one equation:
$\Rightarrow 3\text{a} - 4(\dfrac{21}{2}) = 0$
$\Rightarrow 3\text{a} - 2(21) = 0$
$\Rightarrow 3\text{a} - 42 = 0$
$\Rightarrow 3\text{a} = 0 + 42$
$\Rightarrow 3\text{a} = 42$
$\Rightarrow \text{a} = \dfrac{42}{3} = 14$
$\therefore$ a = 14 and b = 10.5 Ans
Q21. $\dfrac{\text{x} - 1}{2} + \dfrac{\text{y} + 1}{5} = 4\dfrac{1}{5}$ and $\dfrac{\text{x} + \text{y}}{3} = \text{y} - 1$
Show Answer
$\Rightarrow \dfrac{\text{x} - 1}{2} + \dfrac{\text{y} + 1}{5} = \dfrac{21}{5}$
Simplify by taking L.C.M. = 10
$\Rightarrow (10)\dfrac{\text{x} - 1}{2} + (10)\dfrac{\text{y} + 1}{5} = (10)\dfrac{21}{5}$
$\Rightarrow 5(\text{x} - 1) + 2(\text{y} + 1) = 2(21)$
$\Rightarrow 5\text{x} - 5 + 2\text{y} + 2 = 42$
$\Rightarrow 5\text{x} + 2\text{y} - 3 = 42$
$\Rightarrow 5\text{x} + 2\text{y} = 42 + 3$
$\Rightarrow 5\text{x} + 2\text{y} = 45$......eq(i)
$\Rightarrow \dfrac{\text{x} + \text{y}}{3} = \text{y} - 1$
$\Rightarrow \text{x} + \text{y} = 3(\text{y} - 1)$
$\Rightarrow \text{x} + \text{y} = 3\text{y} - 3$
$\Rightarrow \text{x} + \text{y} - 3\text{y} = - \ 3$
$\Rightarrow \text{x} - 2\text{y} = - \ 3$......eq(ii)
In both equations the coefficients of y being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 5\text{x} & + & 2\text{y} & = & 45\\ & \text{x} & - & 2\text{y} & = & - \ 3\\ \hline\\ & 6\text{x} & & & = & 42\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{42}{6} = 7$
Now substitute the value of x in any one equation:
$\Rightarrow 7 - 2\text{y} = - \ 3$
$\Rightarrow - \ 2\text{y} = - \ 3 - 7$
$\Rightarrow - \ 2\text{y} = - \ 10$
$\Rightarrow \text{y} = \dfrac{10}{2} = 5$
$\therefore$ x = 7 and y = 5 Ans
Simplify by taking L.C.M. = 10
$\Rightarrow (10)\dfrac{\text{x} - 1}{2} + (10)\dfrac{\text{y} + 1}{5} = (10)\dfrac{21}{5}$
$\Rightarrow 5(\text{x} - 1) + 2(\text{y} + 1) = 2(21)$
$\Rightarrow 5\text{x} - 5 + 2\text{y} + 2 = 42$
$\Rightarrow 5\text{x} + 2\text{y} - 3 = 42$
$\Rightarrow 5\text{x} + 2\text{y} = 42 + 3$
$\Rightarrow 5\text{x} + 2\text{y} = 45$......eq(i)
$\Rightarrow \dfrac{\text{x} + \text{y}}{3} = \text{y} - 1$
$\Rightarrow \text{x} + \text{y} = 3(\text{y} - 1)$
$\Rightarrow \text{x} + \text{y} = 3\text{y} - 3$
$\Rightarrow \text{x} + \text{y} - 3\text{y} = - \ 3$
$\Rightarrow \text{x} - 2\text{y} = - \ 3$......eq(ii)
In both equations the coefficients of y being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 5\text{x} & + & 2\text{y} & = & 45\\ & \text{x} & - & 2\text{y} & = & - \ 3\\ \hline\\ & 6\text{x} & & & = & 42\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{42}{6} = 7$
Now substitute the value of x in any one equation:
$\Rightarrow 7 - 2\text{y} = - \ 3$
$\Rightarrow - \ 2\text{y} = - \ 3 - 7$
$\Rightarrow - \ 2\text{y} = - \ 10$
$\Rightarrow \text{y} = \dfrac{10}{2} = 5$
$\therefore$ x = 7 and y = 5 Ans
Q22. $\dfrac{1}{\text{x}} + \dfrac{1}{\text{y}} = 5$ and $\dfrac{1}{\text{x}} - \dfrac{1}{\text{y}} = 1$
Show Answer
$\Rightarrow \dfrac{1}{\text{x}} + \dfrac{1}{\text{y}} = 5$......eq(i)
$\Rightarrow \dfrac{1}{\text{x}} - \dfrac{1}{\text{y}} = 1$......eq(ii)
In both equations the coefficients of y being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & \dfrac{1}{\text{x}} & + & \dfrac{1}{\text{y}} & = & 5\\\\ & \dfrac{1}{\text{x}} & - & \dfrac{1}{\text{y}} & = & 1\\\\ \hline\\ & \dfrac{2}{\text{x}} & & & = & 6\\\\ \end{array} $$ $\Rightarrow \dfrac{2}{\text{x}} = 6$
$\Rightarrow 2 = 6\text{x}$
$\Rightarrow \dfrac{2}{6} = \text{x}$
$\Rightarrow \text{x} = \dfrac{1}{3}$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{1}{\dfrac{1}{3}} + \dfrac{1}{\text{y}} = 5$
$\Rightarrow 3 + \dfrac{1}{\text{y}} = 5$
$\Rightarrow \dfrac{1}{\text{y}} = 5 - 3$
$\Rightarrow \dfrac{1}{\text{y}} = 2$
$\Rightarrow \text{y} = \dfrac{1}{2}$
$\therefore$ x = $\dfrac{1}{3}$ and y = $\dfrac{1}{2}$ Ans
$\Rightarrow \dfrac{1}{\text{x}} - \dfrac{1}{\text{y}} = 1$......eq(ii)
In both equations the coefficients of y being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & \dfrac{1}{\text{x}} & + & \dfrac{1}{\text{y}} & = & 5\\\\ & \dfrac{1}{\text{x}} & - & \dfrac{1}{\text{y}} & = & 1\\\\ \hline\\ & \dfrac{2}{\text{x}} & & & = & 6\\\\ \end{array} $$ $\Rightarrow \dfrac{2}{\text{x}} = 6$
$\Rightarrow 2 = 6\text{x}$
$\Rightarrow \dfrac{2}{6} = \text{x}$
$\Rightarrow \text{x} = \dfrac{1}{3}$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{1}{\dfrac{1}{3}} + \dfrac{1}{\text{y}} = 5$
$\Rightarrow 3 + \dfrac{1}{\text{y}} = 5$
$\Rightarrow \dfrac{1}{\text{y}} = 5 - 3$
$\Rightarrow \dfrac{1}{\text{y}} = 2$
$\Rightarrow \text{y} = \dfrac{1}{2}$
$\therefore$ x = $\dfrac{1}{3}$ and y = $\dfrac{1}{2}$ Ans
Q23. $\dfrac{3}{\text{a}} + \dfrac{4}{\text{b}} = 2$ and $\dfrac{9}{\text{a}} - \dfrac{4}{\text{b}} = 2$
Show Answer
$\Rightarrow \dfrac{3}{\text{a}} + \dfrac{4}{\text{b}} = 2$......eq(i)
$\Rightarrow \dfrac{9}{\text{a}} - \dfrac{4}{\text{b}} = 2$......eq(ii)
In both equations the coefficients of b being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & \dfrac{3}{\text{a}} & + & \dfrac{4}{\text{b}} & = & 2\\\\ & \dfrac{9}{\text{a}} & - & \dfrac{4}{\text{b}} & = & 2\\\\ \hline\\ & \dfrac{12}{\text{a}} & & & = & 4\\\\ \end{array} $$ $\Rightarrow \dfrac{12}{\text{a}} = 4$
$\Rightarrow 12 = 4\text{a}$
$\Rightarrow \dfrac{12}{4} = \text{a}$
$\Rightarrow \text{a} = 3$
Now substitute the value of a in any one equation:
$\Rightarrow \dfrac{9}{3} - \dfrac{4}{\text{b}} = 2$
$\Rightarrow 3 - \dfrac{4}{\text{b}} = 2$
$\Rightarrow - \ \dfrac{4}{\text{b}} = 2 - 3$
$\Rightarrow - \ \dfrac{4}{\text{b}} = - \ 1$
$\Rightarrow - \ 4 = - \ \text{b}$
$\Rightarrow \text{b} = 4$
$\therefore$ a = 3 and b = 4 Ans
$\Rightarrow \dfrac{9}{\text{a}} - \dfrac{4}{\text{b}} = 2$......eq(ii)
In both equations the coefficients of b being the same, add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & \dfrac{3}{\text{a}} & + & \dfrac{4}{\text{b}} & = & 2\\\\ & \dfrac{9}{\text{a}} & - & \dfrac{4}{\text{b}} & = & 2\\\\ \hline\\ & \dfrac{12}{\text{a}} & & & = & 4\\\\ \end{array} $$ $\Rightarrow \dfrac{12}{\text{a}} = 4$
$\Rightarrow 12 = 4\text{a}$
$\Rightarrow \dfrac{12}{4} = \text{a}$
$\Rightarrow \text{a} = 3$
Now substitute the value of a in any one equation:
$\Rightarrow \dfrac{9}{3} - \dfrac{4}{\text{b}} = 2$
$\Rightarrow 3 - \dfrac{4}{\text{b}} = 2$
$\Rightarrow - \ \dfrac{4}{\text{b}} = 2 - 3$
$\Rightarrow - \ \dfrac{4}{\text{b}} = - \ 1$
$\Rightarrow - \ 4 = - \ \text{b}$
$\Rightarrow \text{b} = 4$
$\therefore$ a = 3 and b = 4 Ans
Q24. $\dfrac{8}{\text{x}} - \dfrac{9}{\text{y}} = 1$ and $\dfrac{10}{\text{x}} + \dfrac{6}{\text{y}} = 7$
Show Answer
$\Rightarrow \dfrac{8}{\text{x}} - \dfrac{9}{\text{y}} = 1$......eq(i)
$\Rightarrow \dfrac{10}{\text{x}} + \dfrac{6}{\text{y}} = 7$......eq(ii)
Multiply eq(i) by 6 and eq(ii) by 9, and add eq(i) and eq(ii) :
$$ \begin{array}{c c c c c c} & \dfrac{48}{\text{x}} & - & \dfrac{54}{\text{y}} & = & 6\\\\ & \dfrac{90}{\text{x}} & + & \dfrac{54}{\text{y}} & = & 63\\\\ \hline\\ & \dfrac{138}{\text{x}} & & & = & 69\\\\ \end{array} $$ $\Rightarrow \dfrac{138}{\text{x}} = 69$
$\Rightarrow 138 = 69\text{x}$
$\Rightarrow \dfrac{138}{69} = \text{x}$
$\Rightarrow \text{x} = 2$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{10}{2} + \dfrac{6}{\text{y}} = 7$
$\Rightarrow 5 + \dfrac{6}{\text{y}} = 7$
$\Rightarrow \dfrac{6}{\text{y}} = 7 - 5$
$\Rightarrow \dfrac{6}{\text{y}} = 2$
$\Rightarrow 6 = 2\text{y}$
$\Rightarrow \text{y} = \dfrac{6}{2} = 3$
$\therefore$ x = 2 and y = 3 Ans
$\Rightarrow \dfrac{10}{\text{x}} + \dfrac{6}{\text{y}} = 7$......eq(ii)
Multiply eq(i) by 6 and eq(ii) by 9, and add eq(i) and eq(ii) :
$$ \begin{array}{c c c c c c} & \dfrac{48}{\text{x}} & - & \dfrac{54}{\text{y}} & = & 6\\\\ & \dfrac{90}{\text{x}} & + & \dfrac{54}{\text{y}} & = & 63\\\\ \hline\\ & \dfrac{138}{\text{x}} & & & = & 69\\\\ \end{array} $$ $\Rightarrow \dfrac{138}{\text{x}} = 69$
$\Rightarrow 138 = 69\text{x}$
$\Rightarrow \dfrac{138}{69} = \text{x}$
$\Rightarrow \text{x} = 2$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{10}{2} + \dfrac{6}{\text{y}} = 7$
$\Rightarrow 5 + \dfrac{6}{\text{y}} = 7$
$\Rightarrow \dfrac{6}{\text{y}} = 7 - 5$
$\Rightarrow \dfrac{6}{\text{y}} = 2$
$\Rightarrow 6 = 2\text{y}$
$\Rightarrow \text{y} = \dfrac{6}{2} = 3$
$\therefore$ x = 2 and y = 3 Ans
Q25. $\dfrac{6}{\text{x}} - \dfrac{2}{\text{y}} = 1$ and $\dfrac{9}{\text{x}} - \dfrac{6}{\text{y}} = 0$
Show Answer
$\Rightarrow \dfrac{6}{\text{x}} - \dfrac{2}{\text{y}} = 1$......eq(i)
$\Rightarrow \dfrac{9}{\text{x}} - \dfrac{6}{\text{y}} = 0$......eq(ii)
Multiply eq(i) by 3 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & \dfrac{18}{\text{x}} & - & \dfrac{6}{\text{y}} & = & 3\\\\ & \dfrac{9}{\text{x}} & - & \dfrac{6}{\text{y}} & = & 0\\\\ (-) & & (+) & & & (-)\\\\ \hline\\ & \dfrac{9}{\text{x}} & & & = & 3\\\\ \end{array} $$ $\Rightarrow \dfrac{9}{\text{x}} = 3$
$\Rightarrow 9 = 3\text{x}$
$\Rightarrow \dfrac{9}{3} = \text{x}$
$\Rightarrow \text{x} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{9}{3} - \dfrac{6}{\text{y}} = 0$
$\Rightarrow 3 - \dfrac{6}{\text{y}} = 0$
$\Rightarrow - \ \dfrac{6}{\text{y}} = - \ 3$
$\Rightarrow - \ 6 = - \ 3\text{y}$
$\Rightarrow \text{y} = \dfrac{6}{3} = 2$
$\therefore$ x = 3 and y = 2 Ans
$\Rightarrow \dfrac{9}{\text{x}} - \dfrac{6}{\text{y}} = 0$......eq(ii)
Multiply eq(i) by 3 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & \dfrac{18}{\text{x}} & - & \dfrac{6}{\text{y}} & = & 3\\\\ & \dfrac{9}{\text{x}} & - & \dfrac{6}{\text{y}} & = & 0\\\\ (-) & & (+) & & & (-)\\\\ \hline\\ & \dfrac{9}{\text{x}} & & & = & 3\\\\ \end{array} $$ $\Rightarrow \dfrac{9}{\text{x}} = 3$
$\Rightarrow 9 = 3\text{x}$
$\Rightarrow \dfrac{9}{3} = \text{x}$
$\Rightarrow \text{x} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow \dfrac{9}{3} - \dfrac{6}{\text{y}} = 0$
$\Rightarrow 3 - \dfrac{6}{\text{y}} = 0$
$\Rightarrow - \ \dfrac{6}{\text{y}} = - \ 3$
$\Rightarrow - \ 6 = - \ 3\text{y}$
$\Rightarrow \text{y} = \dfrac{6}{3} = 2$
$\therefore$ x = 3 and y = 2 Ans
Q26. $3\text{x} + \dfrac{1}{\text{y}} = 13$ and $\dfrac{2}{\text{y}} - \text{x} = 5$
Show Answer
$\Rightarrow 3\text{x} + \dfrac{1}{\text{y}} = 13$......eq(i)
$\Rightarrow \dfrac{2}{\text{y}} - \text{x} = 5$
$\Rightarrow - \ \text{x} + \dfrac{2}{\text{y}} = 5$......eq(ii)
Multiply eq(i) by 2 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & 6\text{x} & + & \dfrac{2}{\text{y}} & = & 26\\\\ - & \text{x} & + & \dfrac{2}{\text{y}} & = & 5\\\\ (+) & & (-) & & & (-)\\\\ \hline\\ & 7\text{x} & & & = & 21\\\\ \end{array} $$ $\Rightarrow 7\text{x} = 21$
$\Rightarrow \text{x} = \dfrac{21}{7} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow - \ 3 + \dfrac{2}{\text{y}} = 5$
$\Rightarrow \dfrac{2}{\text{y}} = 5 + 3$
$\Rightarrow \dfrac{2}{\text{y}} = 8$
$\Rightarrow 2 = 8\text{y}$
$\Rightarrow \text{y} = \dfrac{2}{8} = \dfrac{1}{4}$
$\therefore$ x = 3 and y = $\dfrac{1}{4}$ Ans
$\Rightarrow \dfrac{2}{\text{y}} - \text{x} = 5$
$\Rightarrow - \ \text{x} + \dfrac{2}{\text{y}} = 5$......eq(ii)
Multiply eq(i) by 2 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & 6\text{x} & + & \dfrac{2}{\text{y}} & = & 26\\\\ - & \text{x} & + & \dfrac{2}{\text{y}} & = & 5\\\\ (+) & & (-) & & & (-)\\\\ \hline\\ & 7\text{x} & & & = & 21\\\\ \end{array} $$ $\Rightarrow 7\text{x} = 21$
$\Rightarrow \text{x} = \dfrac{21}{7} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow - \ 3 + \dfrac{2}{\text{y}} = 5$
$\Rightarrow \dfrac{2}{\text{y}} = 5 + 3$
$\Rightarrow \dfrac{2}{\text{y}} = 8$
$\Rightarrow 2 = 8\text{y}$
$\Rightarrow \text{y} = \dfrac{2}{8} = \dfrac{1}{4}$
$\therefore$ x = 3 and y = $\dfrac{1}{4}$ Ans
Q27. $4\text{x} + \dfrac{3}{\text{y}} = 1$ and $3\text{x} - \dfrac{2}{\text{y}} = 5$
Show Answer
$\Rightarrow 4\text{x} + \dfrac{3}{\text{y}} = 1$......eq(i)
$\Rightarrow 3\text{x} - \dfrac{2}{\text{y}} = 5$......eq(ii)
Multiply eq(i) by 2 and eq(ii) by 3, and add eq(i) and eq(i) :
$$ \begin{array}{c c c c c c} & 8\text{x} & + & \dfrac{6}{\text{y}} & = & 2\\\\ & 9\text{x} & - & \dfrac{6}{\text{y}} & = & 15\\\\ \hline\\ & 17\text{x} & & & = & 17\\\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{17}{17} = 1$
Now substitute the value of x in any one equation:
$\Rightarrow 4(1) + \dfrac{3}{\text{y}} = 1$
$\Rightarrow 4 + \dfrac{3}{\text{y}} = 1$
$\Rightarrow \dfrac{3}{\text{y}} = 1 - 4$
$\Rightarrow \dfrac{3}{\text{y}} = - \ 3$
$\Rightarrow 3 = - \ 3\text{y}$
$\Rightarrow - \ \text{y} = \dfrac{3}{3}$
$\Rightarrow \text{y} = - \ 1$
$\therefore$ x = 1 and y = $- \ 1$ Ans
$\Rightarrow 3\text{x} - \dfrac{2}{\text{y}} = 5$......eq(ii)
Multiply eq(i) by 2 and eq(ii) by 3, and add eq(i) and eq(i) :
$$ \begin{array}{c c c c c c} & 8\text{x} & + & \dfrac{6}{\text{y}} & = & 2\\\\ & 9\text{x} & - & \dfrac{6}{\text{y}} & = & 15\\\\ \hline\\ & 17\text{x} & & & = & 17\\\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{17}{17} = 1$
Now substitute the value of x in any one equation:
$\Rightarrow 4(1) + \dfrac{3}{\text{y}} = 1$
$\Rightarrow 4 + \dfrac{3}{\text{y}} = 1$
$\Rightarrow \dfrac{3}{\text{y}} = 1 - 4$
$\Rightarrow \dfrac{3}{\text{y}} = - \ 3$
$\Rightarrow 3 = - \ 3\text{y}$
$\Rightarrow - \ \text{y} = \dfrac{3}{3}$
$\Rightarrow \text{y} = - \ 1$
$\therefore$ x = 1 and y = $- \ 1$ Ans
Q28. $\text{y} - \dfrac{3}{\text{x}} = 8$ and $2\text{y} + \dfrac{7}{\text{x}} = 3$
Show Answer
$\Rightarrow \text{y} - \dfrac{3}{\text{x}} = 8$......eq(i)
$\Rightarrow 2\text{y} + \dfrac{7}{\text{x}} = 3$......eq(ii)
Multiply eq(i) by 2 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & 2\text{y} & - & \dfrac{6}{\text{x}} & = & 16\\\\ & 2\text{y} & + & \dfrac{7}{\text{x}} & = & 5\\\\ (-) & & (-) & & & (-)\\\\ \hline\\ & & & - \ \dfrac{13}{\text{x}} & = & 13\\\\ \end{array} $$ $\Rightarrow - \ 13 = 13\text{x}$
$\Rightarrow \text{x} = - \ \dfrac{13}{13} = - \ 1$
Now substitute the value of x in any one equation:
$\Rightarrow 2\text{y} + \dfrac{7}{(- \ 1)} = 3$
$\Rightarrow 2\text{y} - \ 7 = 3$
$\Rightarrow 2\text{y} = 3 + 7$
$\Rightarrow 2\text{y} = 10$
$\Rightarrow \text{y} = \dfrac{10}{2} = 5$
$\therefore$ x = $- \ 1$ and y = 5 Ans
$\Rightarrow 2\text{y} + \dfrac{7}{\text{x}} = 3$......eq(ii)
Multiply eq(i) by 2 and Subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & 2\text{y} & - & \dfrac{6}{\text{x}} & = & 16\\\\ & 2\text{y} & + & \dfrac{7}{\text{x}} & = & 5\\\\ (-) & & (-) & & & (-)\\\\ \hline\\ & & & - \ \dfrac{13}{\text{x}} & = & 13\\\\ \end{array} $$ $\Rightarrow - \ 13 = 13\text{x}$
$\Rightarrow \text{x} = - \ \dfrac{13}{13} = - \ 1$
Now substitute the value of x in any one equation:
$\Rightarrow 2\text{y} + \dfrac{7}{(- \ 1)} = 3$
$\Rightarrow 2\text{y} - \ 7 = 3$
$\Rightarrow 2\text{y} = 3 + 7$
$\Rightarrow 2\text{y} = 10$
$\Rightarrow \text{y} = \dfrac{10}{2} = 5$
$\therefore$ x = $- \ 1$ and y = 5 Ans
Q29. $\dfrac{\text{x} + 1}{\text{y} + 1} = 2$ and $\dfrac{2\text{x} + 1}{2\text{y} + 1} = \dfrac{1}{3}$
Show Answer
$\Rightarrow \dfrac{\text{x} + 1}{\text{y} + 1} = 2$
$\Rightarrow \text{x} + 1 = 2(\text{y} + 1)$
$\Rightarrow \text{x} + 1 = 2\text{y} + 2$
$\Rightarrow \text{x} - 2\text{y} = 2 - 1$
$\Rightarrow \text{x} - 2\text{y} = 1$......eq(i)
$\Rightarrow \dfrac{2\text{x} + 1}{2\text{y} + 1} = \dfrac{1}{3}$
$\Rightarrow 3(2\text{x} + 1) = 2\text{y} + 1$
$\Rightarrow 6\text{x} + 3 = 2\text{y} + 1$
$\Rightarrow 6\text{x} - 2\text{y} = 1 - 3$
$\Rightarrow 6\text{x} - 2\text{y} = - \ 2$......eq(ii)
In both equations coefficients of y being the same, subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & \text{x} & - & 2\text{y} & = & 1\\\\ & 6\text{x} & - & 2\text{y} & = & - \ 2\\\\ (-) & & (+) & & & (+)\\\\ \hline\\ & - \ 5\text{x} & & & = & 3\\\\ \end{array} $$ $\Rightarrow \text{x} = - \ \dfrac{3}{5}$
Now substitute the value of x in any one equation:
$\Rightarrow 6(- \ \dfrac{3}{5}) - 2\text{y} = - \ 2$
$\Rightarrow - \ \dfrac{18}{5} - 2\text{y} = - \ 2$
$\Rightarrow - \ 2\text{y} = \dfrac{18}{5} - 2$
$\Rightarrow - \ 2\text{y} = \dfrac{18 - 10}{5}$
$\Rightarrow - \ 2\text{y} = \dfrac{8}{5}$
$\Rightarrow - \ \text{y} = \dfrac{8}{5 \times 2}$
$\Rightarrow - \ \text{y} = \dfrac{4}{5}$
$\Rightarrow \text{y} = - \ \dfrac{4}{5}$
$\therefore$ x = $- \ \dfrac{3}{5}$ and y = $- \ \dfrac{4}{5}$ Ans
$\Rightarrow \text{x} + 1 = 2(\text{y} + 1)$
$\Rightarrow \text{x} + 1 = 2\text{y} + 2$
$\Rightarrow \text{x} - 2\text{y} = 2 - 1$
$\Rightarrow \text{x} - 2\text{y} = 1$......eq(i)
$\Rightarrow \dfrac{2\text{x} + 1}{2\text{y} + 1} = \dfrac{1}{3}$
$\Rightarrow 3(2\text{x} + 1) = 2\text{y} + 1$
$\Rightarrow 6\text{x} + 3 = 2\text{y} + 1$
$\Rightarrow 6\text{x} - 2\text{y} = 1 - 3$
$\Rightarrow 6\text{x} - 2\text{y} = - \ 2$......eq(ii)
In both equations coefficients of y being the same, subtract eq(ii) from eq(i) :
$$ \begin{array}{c c c c c c} & \text{x} & - & 2\text{y} & = & 1\\\\ & 6\text{x} & - & 2\text{y} & = & - \ 2\\\\ (-) & & (+) & & & (+)\\\\ \hline\\ & - \ 5\text{x} & & & = & 3\\\\ \end{array} $$ $\Rightarrow \text{x} = - \ \dfrac{3}{5}$
Now substitute the value of x in any one equation:
$\Rightarrow 6(- \ \dfrac{3}{5}) - 2\text{y} = - \ 2$
$\Rightarrow - \ \dfrac{18}{5} - 2\text{y} = - \ 2$
$\Rightarrow - \ 2\text{y} = \dfrac{18}{5} - 2$
$\Rightarrow - \ 2\text{y} = \dfrac{18 - 10}{5}$
$\Rightarrow - \ 2\text{y} = \dfrac{8}{5}$
$\Rightarrow - \ \text{y} = \dfrac{8}{5 \times 2}$
$\Rightarrow - \ \text{y} = \dfrac{4}{5}$
$\Rightarrow \text{y} = - \ \dfrac{4}{5}$
$\therefore$ x = $- \ \dfrac{3}{5}$ and y = $- \ \dfrac{4}{5}$ Ans
Q30. $\dfrac{\text{x} - 1}{3} + \dfrac{\text{y} + 2}{2} = 3$ and $\dfrac{1 - \text{x}}{6} - \dfrac{\text{y} - 4}{2} = \dfrac{1}{2}$
Show Answer
For equation (i):
$\Rightarrow \dfrac{\text{x} - 1}{3} + \dfrac{\text{y} + 2}{2} = 3$
Simplify by taking L.C.M. = 6
$\Rightarrow 6(\dfrac{\text{x} - 1}{3}) + 6(\dfrac{\text{y} + 2}{2}) = 6(3)$
$\Rightarrow 2(\text{x} - 1) + 3(\text{y} + 2) = 18$
$\Rightarrow 2\text{x} - 2 + 3\text{y} + 6 = 18$
$\Rightarrow 2\text{x} + 3\text{y} + 4 = 18$
$\Rightarrow 2\text{x} + 3\text{y} = 18 - 4$
$\Rightarrow 2\text{x} + 3\text{y} = 14$......eq(i)
For equation (ii):
$\Rightarrow \dfrac{1 - \text{x}}{6} - \dfrac{\text{y} - 4}{2} = \dfrac{1}{2}$
Simplify by taking L.C.M. = 6
$\Rightarrow 6(\dfrac{1 - \text{x}}{6}) - 6(\dfrac{\text{y} - 4}{2}) = 6(\dfrac{1}{2})$
$\Rightarrow 1 - \text{x} - 3(\text{y} - 4) = 3(1)$
$\Rightarrow 1 - \text{x} - 3\text{y} + 12 = 3$
$\Rightarrow - \ \text{x} - 3\text{y} + 13 = 3$
$\Rightarrow - \ \text{x} - 3\text{y} = 3 - 13$
$\Rightarrow - \ \text{x} - 3\text{y} = - \ 10$......eq(ii)
In both equations coefficients of y being the same, add eq(i) and eq(ii) :
$$ \begin{array}{c c c c c c} & 2\text{x} & + & 3\text{y} & = & 14\\\\ & - \ \text{x} & - & 3\text{y} & = & - \ 10\\\\ \hline\\ & \text{x} & & & = & 4\\\\ \end{array} $$ $\Rightarrow \text{x} = 4$
Now substitute the value of x in any one equation:
$\Rightarrow 2(4) + 3\text{y} = 14$
$\Rightarrow 8 + 3\text{y} = 14$
$\Rightarrow 3\text{y} = 14 - 8$
$\Rightarrow 3\text{y} = 6$
$\Rightarrow \text{y} = \dfrac{6}{3} = 2$
$\therefore$ x = 4 and y = 2 Ans
$\Rightarrow \dfrac{\text{x} - 1}{3} + \dfrac{\text{y} + 2}{2} = 3$
Simplify by taking L.C.M. = 6
$\Rightarrow 6(\dfrac{\text{x} - 1}{3}) + 6(\dfrac{\text{y} + 2}{2}) = 6(3)$
$\Rightarrow 2(\text{x} - 1) + 3(\text{y} + 2) = 18$
$\Rightarrow 2\text{x} - 2 + 3\text{y} + 6 = 18$
$\Rightarrow 2\text{x} + 3\text{y} + 4 = 18$
$\Rightarrow 2\text{x} + 3\text{y} = 18 - 4$
$\Rightarrow 2\text{x} + 3\text{y} = 14$......eq(i)
For equation (ii):
$\Rightarrow \dfrac{1 - \text{x}}{6} - \dfrac{\text{y} - 4}{2} = \dfrac{1}{2}$
Simplify by taking L.C.M. = 6
$\Rightarrow 6(\dfrac{1 - \text{x}}{6}) - 6(\dfrac{\text{y} - 4}{2}) = 6(\dfrac{1}{2})$
$\Rightarrow 1 - \text{x} - 3(\text{y} - 4) = 3(1)$
$\Rightarrow 1 - \text{x} - 3\text{y} + 12 = 3$
$\Rightarrow - \ \text{x} - 3\text{y} + 13 = 3$
$\Rightarrow - \ \text{x} - 3\text{y} = 3 - 13$
$\Rightarrow - \ \text{x} - 3\text{y} = - \ 10$......eq(ii)
In both equations coefficients of y being the same, add eq(i) and eq(ii) :
$$ \begin{array}{c c c c c c} & 2\text{x} & + & 3\text{y} & = & 14\\\\ & - \ \text{x} & - & 3\text{y} & = & - \ 10\\\\ \hline\\ & \text{x} & & & = & 4\\\\ \end{array} $$ $\Rightarrow \text{x} = 4$
Now substitute the value of x in any one equation:
$\Rightarrow 2(4) + 3\text{y} = 14$
$\Rightarrow 8 + 3\text{y} = 14$
$\Rightarrow 3\text{y} = 14 - 8$
$\Rightarrow 3\text{y} = 6$
$\Rightarrow \text{y} = \dfrac{6}{3} = 2$
$\therefore$ x = 4 and y = 2 Ans