Q1. $\text{x} + \text{y} = 11$ and $\text{x} - \text{y} = -3$
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$\Rightarrow$ Taking one equation: $\text{x} + \text{y} = 11$
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 11 - \text{y}$
Now substituting the value of x in second equation:
$\Rightarrow 11 - \text{y} - \text{y} = -3$
$\Rightarrow - \ 2\text{y} = - \ 3 - 11$
$\Rightarrow - \ 2\text{y} = - \ 14$
$\Rightarrow \text{y} = \dfrac{14}{2} = 7$
Now substitute the value of y in any one equation:
$\Rightarrow \text{x} + 7 = 11$
$\Rightarrow \text{x} = 11 - 7 = 4$
$\therefore$ x = 4 and y = 7 Ans
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 11 - \text{y}$
Now substituting the value of x in second equation:
$\Rightarrow 11 - \text{y} - \text{y} = -3$
$\Rightarrow - \ 2\text{y} = - \ 3 - 11$
$\Rightarrow - \ 2\text{y} = - \ 14$
$\Rightarrow \text{y} = \dfrac{14}{2} = 7$
Now substitute the value of y in any one equation:
$\Rightarrow \text{x} + 7 = 11$
$\Rightarrow \text{x} = 11 - 7 = 4$
$\therefore$ x = 4 and y = 7 Ans
Q2. $\text{x} + 5\text{y} = 18$ and $3\text{x} + 2\text{y} = 41$
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$\Rightarrow$ Taking one equation: $\text{x} + 5\text{y} = 18$
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 18 - 5\text{y}$
Now substituting the value of x in second equation:
$\Rightarrow 3(18 - 5\text{y}) + 2\text{y} = 41$
$\Rightarrow 54 - 15\text{y}) + 2\text{y} = 41$
$\Rightarrow - / 13\text{y} = 41 - 54$
$\Rightarrow - / 13\text{y} = - / 13$
$\Rightarrow \text{y} = \dfrac{13}{13} = 1$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 2(1) = 41$
$\Rightarrow 3\text{x} = 41 - 2$
$\Rightarrow 3\text{x} = 39$
$\Rightarrow \text{x} = \dfrac{39}{3} = 13$
$\therefore$ x = 13 and y = 1 Ans
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 18 - 5\text{y}$
Now substituting the value of x in second equation:
$\Rightarrow 3(18 - 5\text{y}) + 2\text{y} = 41$
$\Rightarrow 54 - 15\text{y}) + 2\text{y} = 41$
$\Rightarrow - / 13\text{y} = 41 - 54$
$\Rightarrow - / 13\text{y} = - / 13$
$\Rightarrow \text{y} = \dfrac{13}{13} = 1$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 2(1) = 41$
$\Rightarrow 3\text{x} = 41 - 2$
$\Rightarrow 3\text{x} = 39$
$\Rightarrow \text{x} = \dfrac{39}{3} = 13$
$\therefore$ x = 13 and y = 1 Ans
Q3. $\text{x} + \text{y} = 0$ and $\text{y} - \text{x} = 6$
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$\Rightarrow$ Taking one equation: $\text{x} + \text{y} = 0$
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 0 - \text{y}$
Now substituting the value of x in second equation:
$\Rightarrow \text{y} - (0 - \text{y}) = 6$
$\Rightarrow \text{y} - 0 + \text{y} = 6$
$\Rightarrow 2\text{y} = 6 + 0$
$\Rightarrow \text{y} = \dfrac{6}{2} = 3$
Now substitute the value of y in any one equation:
$\Rightarrow \text{y} - \text{x} = 6$
$\Rightarrow 3 - \text{x} = 6$
$\Rightarrow - \ \text{x} = 6 - 3 = 3$
$\Rightarrow \text{x} = - \ 3$
$\therefore$ x = $- \ 3$ and y = 3 Ans
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = 0 - \text{y}$
Now substituting the value of x in second equation:
$\Rightarrow \text{y} - (0 - \text{y}) = 6$
$\Rightarrow \text{y} - 0 + \text{y} = 6$
$\Rightarrow 2\text{y} = 6 + 0$
$\Rightarrow \text{y} = \dfrac{6}{2} = 3$
Now substitute the value of y in any one equation:
$\Rightarrow \text{y} - \text{x} = 6$
$\Rightarrow 3 - \text{x} = 6$
$\Rightarrow - \ \text{x} = 6 - 3 = 3$
$\Rightarrow \text{x} = - \ 3$
$\therefore$ x = $- \ 3$ and y = 3 Ans
Q4. $\text{x} - 4\text{y} = - \ 8$ and $\text{x} - 2\text{y} = 0$
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$\Rightarrow$ Taking one equation: $\text{x} - 4\text{y} = -8$
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = - \ 8 + 4\text{y}$
Now substituting the value of x in second equation:
$\Rightarrow - \ 8 + 4\text{y} - 2\text{y} = 0$
$\Rightarrow 2\text{y} = 0 + 8$
$\Rightarrow \text{y} = \dfrac{8}{2} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow \text{x} - 4(4) = - \ 8$
$\Rightarrow \text{x} - 16 = - \ 8$
$\Rightarrow \text{x} = 16 - 8 = 8$
$\therefore$ x = 8 and y = 4 Ans
$\Rightarrow$ Solving for x:
$\Rightarrow \text{x} = - \ 8 + 4\text{y}$
Now substituting the value of x in second equation:
$\Rightarrow - \ 8 + 4\text{y} - 2\text{y} = 0$
$\Rightarrow 2\text{y} = 0 + 8$
$\Rightarrow \text{y} = \dfrac{8}{2} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow \text{x} - 4(4) = - \ 8$
$\Rightarrow \text{x} - 16 = - \ 8$
$\Rightarrow \text{x} = 16 - 8 = 8$
$\therefore$ x = 8 and y = 4 Ans
Q5. $4\text{a} - \text{b} = 10$ and $2\text{a} + 3\text{b} = 12$
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$\Rightarrow$ Taking one equation: $4\text{a} - \text{b} = 10$
$\Rightarrow$ Solving for a:
$\Rightarrow \text{a} = \dfrac{10 + \text{b}}{4}$
Now substituting the value of a in second equation:
$\Rightarrow 2(\dfrac{10 + \text{b}}{4}) + 3\text{b} = 12$
Simplify by taking L.C.M. = 4
$\Rightarrow 2(10 + \text{b}) + (4)3\text{b} = (4)12$
$\Rightarrow 20 + 2\text{b} + 12\text{b} = 48$
$\Rightarrow 14\text{b} = 48 - 20$
$\Rightarrow \text{b} = \dfrac{28}{14} = 2$
Now substitute the value of b in any one equation:
$\Rightarrow 4\text{a} - 2 = 10$
$\Rightarrow 4\text{a} = 10 + 2$
$\Rightarrow 4\text{a} = 12$
$\Rightarrow \text{a} = \dfrac{12}{4} = 3$
$\therefore$ a = 3 and b = 2 Ans
$\Rightarrow$ Solving for a:
$\Rightarrow \text{a} = \dfrac{10 + \text{b}}{4}$
Now substituting the value of a in second equation:
$\Rightarrow 2(\dfrac{10 + \text{b}}{4}) + 3\text{b} = 12$
Simplify by taking L.C.M. = 4
$\Rightarrow 2(10 + \text{b}) + (4)3\text{b} = (4)12$
$\Rightarrow 20 + 2\text{b} + 12\text{b} = 48$
$\Rightarrow 14\text{b} = 48 - 20$
$\Rightarrow \text{b} = \dfrac{28}{14} = 2$
Now substitute the value of b in any one equation:
$\Rightarrow 4\text{a} - 2 = 10$
$\Rightarrow 4\text{a} = 10 + 2$
$\Rightarrow 4\text{a} = 12$
$\Rightarrow \text{a} = \dfrac{12}{4} = 3$
$\therefore$ a = 3 and b = 2 Ans
Q6. $2\text{a} + 3\text{b} = 6$ and $3\text{a} + 5\text{b} = 15$
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$\Rightarrow$ Taking one equation: $2\text{a} + 3\text{b} = 6$
$\Rightarrow$ Solving for a:
$\Rightarrow \text{a} = \dfrac{6 - 3\text{b}}{2}$
Now substituting the value of a in second equation:
$\Rightarrow 3(\dfrac{6 - 3\text{b}}{2}) + 5\text{b} = 15$
Simplify by taking L.C.M. = 2
$\Rightarrow 3(6 - 3\text{b}) + (2)5\text{b} = (2)15$
$\Rightarrow 18 - 9\text{b} + 10\text{b} = 30$
$\Rightarrow \text{b} = 30 - 18$
$\Rightarrow \text{b} = 12$
Now substitute the value of b in any one equation:
$\Rightarrow 2\text{a} + 3(12) = 6$
$\Rightarrow 2\text{a} + 36 = 6$
$\Rightarrow 2\text{a} = 6 - 36$
$\Rightarrow 2\text{a} = - \ 30$
$\Rightarrow \text{a} = - \ \dfrac{30}{2} = - \ 15$
$\therefore$ a = $- \ 15$ and b = 12 Ans
$\Rightarrow$ Solving for a:
$\Rightarrow \text{a} = \dfrac{6 - 3\text{b}}{2}$
Now substituting the value of a in second equation:
$\Rightarrow 3(\dfrac{6 - 3\text{b}}{2}) + 5\text{b} = 15$
Simplify by taking L.C.M. = 2
$\Rightarrow 3(6 - 3\text{b}) + (2)5\text{b} = (2)15$
$\Rightarrow 18 - 9\text{b} + 10\text{b} = 30$
$\Rightarrow \text{b} = 30 - 18$
$\Rightarrow \text{b} = 12$
Now substitute the value of b in any one equation:
$\Rightarrow 2\text{a} + 3(12) = 6$
$\Rightarrow 2\text{a} + 36 = 6$
$\Rightarrow 2\text{a} = 6 - 36$
$\Rightarrow 2\text{a} = - \ 30$
$\Rightarrow \text{a} = - \ \dfrac{30}{2} = - \ 15$
$\therefore$ a = $- \ 15$ and b = 12 Ans
Q7. $2\text{x} - \text{y} = 9$ and $3\text{x} - 7\text{y} = 19$
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$\Rightarrow 2\text{x} - \text{y} = 9$......eq(i)
$\Rightarrow 3\text{x} - 7\text{y} = 19$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 2, then subtract eq(ii) by eq(i):
$$ \begin{array}{c c c c c c} & 6\text{x} & - & 3\text{y} & = & 27\\ & 6\text{x} & - & 14\text{y} & = & 38\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & 11\text{y} & = & - \ 11\\ \end{array} $$ $\Rightarrow \text{y} = - \ \dfrac{11}{11} = - \ 1$
Now substitute the value of y in any one equation:
$\Rightarrow 2\text{x} - (- \ 1) = 9$
$\Rightarrow 2\text{x} + 1 = 9$
$\Rightarrow 2\text{x} = 9 - 1$
$\Rightarrow 2\text{x} = 8$
$\Rightarrow \text{x} = \dfrac{8}{2} = 4$
$\therefore$ x = 4 and y = $- \ 1$ Ans
$\Rightarrow 3\text{x} - 7\text{y} = 19$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 2, then subtract eq(ii) by eq(i):
$$ \begin{array}{c c c c c c} & 6\text{x} & - & 3\text{y} & = & 27\\ & 6\text{x} & - & 14\text{y} & = & 38\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & 11\text{y} & = & - \ 11\\ \end{array} $$ $\Rightarrow \text{y} = - \ \dfrac{11}{11} = - \ 1$
Now substitute the value of y in any one equation:
$\Rightarrow 2\text{x} - (- \ 1) = 9$
$\Rightarrow 2\text{x} + 1 = 9$
$\Rightarrow 2\text{x} = 9 - 1$
$\Rightarrow 2\text{x} = 8$
$\Rightarrow \text{x} = \dfrac{8}{2} = 4$
$\therefore$ x = 4 and y = $- \ 1$ Ans
Q8. $8\text{x} = 5\text{y}$ and $13\text{x} = 8\text{y} + 1$
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$\Rightarrow 8\text{x} = 5\text{y} \Rightarrow 8\text{x} - 5\text{y} = 0$......eq(i)
$\Rightarrow 13\text{x} = 8\text{y} + 1 \Rightarrow 13\text{x} - 8\text{y} = 1$......eq(ii)
Multiply eq(i) by 8 and eq(ii) by 5, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 64\text{x} & - & 40\text{y} & = & 0\\ & 65\text{x} & - & 40\text{y} & = & 5\\ (-) & & (+) & & & (-) \\\\ \hline\\ & - \ \text{x} & & & = & - \ 5\\ \end{array} $$ $\Rightarrow \text{x} = 5$
Now substitute the value of x in any one equation:
$\Rightarrow 8(5) - 5\text{y} = 0$
$\Rightarrow 40 - 5\text{y} = 0$
$\Rightarrow - \ 5\text{y} = 0 - 40$
$\Rightarrow - \ 5\text{y} = - \ 40$
$\Rightarrow \text{y} = \dfrac{40}{5} = 8$
$\therefore$ x = 5 and y = 8 Ans
$\Rightarrow 13\text{x} = 8\text{y} + 1 \Rightarrow 13\text{x} - 8\text{y} = 1$......eq(ii)
Multiply eq(i) by 8 and eq(ii) by 5, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 64\text{x} & - & 40\text{y} & = & 0\\ & 65\text{x} & - & 40\text{y} & = & 5\\ (-) & & (+) & & & (-) \\\\ \hline\\ & - \ \text{x} & & & = & - \ 5\\ \end{array} $$ $\Rightarrow \text{x} = 5$
Now substitute the value of x in any one equation:
$\Rightarrow 8(5) - 5\text{y} = 0$
$\Rightarrow 40 - 5\text{y} = 0$
$\Rightarrow - \ 5\text{y} = 0 - 40$
$\Rightarrow - \ 5\text{y} = - \ 40$
$\Rightarrow \text{y} = \dfrac{40}{5} = 8$
$\therefore$ x = 5 and y = 8 Ans
Q9. $\text{x} + 2\text{y} = 11$ and $2\text{x} - \text{y} = 2$
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$\Rightarrow \text{x} + 2\text{y} = 11$......eq(i)
$\Rightarrow 2\text{x} - \text{y} = 2$......eq(ii)
Multiply eq(ii) by 2, then add eq(i) from eq(ii):
$$ \begin{array}{c c c c c c} & \text{x} & + & 2\text{y} & = & 11\\ & 4\text{x} & - & 2\text{y} & = & 4\\ \hline\\ & 5\text{x} & & & = & 15\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{15}{5} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow 3 + 2\text{y} = 11$
$\Rightarrow 2\text{y} = 11 - 3$
$\Rightarrow 2\text{y} = 8$
$\Rightarrow \text{y} = \dfrac{8}{2} = 4$
$\therefore$ x = 3 and y = 4 Ans
$\Rightarrow 2\text{x} - \text{y} = 2$......eq(ii)
Multiply eq(ii) by 2, then add eq(i) from eq(ii):
$$ \begin{array}{c c c c c c} & \text{x} & + & 2\text{y} & = & 11\\ & 4\text{x} & - & 2\text{y} & = & 4\\ \hline\\ & 5\text{x} & & & = & 15\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{15}{5} = 3$
Now substitute the value of x in any one equation:
$\Rightarrow 3 + 2\text{y} = 11$
$\Rightarrow 2\text{y} = 11 - 3$
$\Rightarrow 2\text{y} = 8$
$\Rightarrow \text{y} = \dfrac{8}{2} = 4$
$\therefore$ x = 3 and y = 4 Ans
Q10. $3\text{x} - 7\text{y} = 35$ and $2\text{x} + 5\text{y} = 4$
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$\Rightarrow 3\text{x} - 7\text{y} = 35$......eq(i)
$\Rightarrow 2\text{x} + 5\text{y} = 4$......eq(ii)
Multiply eq(i) by 5 and eq(ii) by 7, then add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 15\text{x} & - & 35\text{y} & = & 175\\ & 14\text{x} & + & 35\text{y} & = & 28\\ \hline\\ & 29\text{x} & & & = & 203\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{203}{29} = 7$
Now substitute the value of x in any one equation:
$\Rightarrow 3(7) - 7\text{y} = 35$
$\Rightarrow 21 - 7\text{y} = 35$
$\Rightarrow - \ 7\text{y} = 35 - 21$
$\Rightarrow - \ 7\text{y} = 14$
$\Rightarrow - \ \text{y} = \dfrac{14}{7} = 2$
$\Rightarrow \text{y} = - \ 2$
$\therefore$ x = 7 and y = $- \ 2$ Ans
$\Rightarrow 2\text{x} + 5\text{y} = 4$......eq(ii)
Multiply eq(i) by 5 and eq(ii) by 7, then add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 15\text{x} & - & 35\text{y} & = & 175\\ & 14\text{x} & + & 35\text{y} & = & 28\\ \hline\\ & 29\text{x} & & & = & 203\\ \end{array} $$ $\Rightarrow \text{x} = \dfrac{203}{29} = 7$
Now substitute the value of x in any one equation:
$\Rightarrow 3(7) - 7\text{y} = 35$
$\Rightarrow 21 - 7\text{y} = 35$
$\Rightarrow - \ 7\text{y} = 35 - 21$
$\Rightarrow - \ 7\text{y} = 14$
$\Rightarrow - \ \text{y} = \dfrac{14}{7} = 2$
$\Rightarrow \text{y} = - \ 2$
$\therefore$ x = 7 and y = $- \ 2$ Ans
Q11. $4\text{x} - 3\text{y} = 8$ and $3\text{x} - 4\text{y} = - \ 1$
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$\Rightarrow 4\text{x} - 3\text{y} = 8$......eq(i)
$\Rightarrow 3\text{x} - 4\text{y} = - \ 1$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 4, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 12\text{x} & - & 9\text{y} & = & 24\\ & 12\text{x} & - & 16\text{y} & = & - \ 4\\ (-) & & (+) & & & (+) \\\\ \hline\\ & & & 7\text{y} & = & 28\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{28}{7} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow 4\text{x} - 3(4) = 8$
$\Rightarrow 4\text{x} - 12 = 8$
$\Rightarrow 4\text{x} = 8 + 12$
$\Rightarrow 4\text{x} = 20$
$\Rightarrow \text{x} = \dfrac{20}{4} = 5$
$\therefore$ x = 5 and y = 4 Ans
$\Rightarrow 3\text{x} - 4\text{y} = - \ 1$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 4, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 12\text{x} & - & 9\text{y} & = & 24\\ & 12\text{x} & - & 16\text{y} & = & - \ 4\\ (-) & & (+) & & & (+) \\\\ \hline\\ & & & 7\text{y} & = & 28\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{28}{7} = 4$
Now substitute the value of y in any one equation:
$\Rightarrow 4\text{x} - 3(4) = 8$
$\Rightarrow 4\text{x} - 12 = 8$
$\Rightarrow 4\text{x} = 8 + 12$
$\Rightarrow 4\text{x} = 20$
$\Rightarrow \text{x} = \dfrac{20}{4} = 5$
$\therefore$ x = 5 and y = 4 Ans
Q12. $8\text{a} - 7\text{b} = 1$ and $4\text{a} = 3\text{b} + 5$
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$\Rightarrow 8\text{a} - 7\text{b} = 1$......eq(i)
$\Rightarrow 4\text{a} = 3\text{b} + 5 \Rightarrow 4\text{a} - 3\text{b} = 5$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 7, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 24\text{a} & - & 21\text{b} & = & 3\\ & 28\text{a} & - & 21\text{b} & = & 35\\ (-) & & (+) & & & (-) \\\\ \hline\\ & - \ 4\text{a} & & & = & - \ 32\\ \end{array} $$ $\Rightarrow \text{a} = \dfrac{32}{4} = 8$
Now substitute the value of a in any one equation:
$\Rightarrow 8(8) - 7\text{b} = 1$
$\Rightarrow 64 - 7\text{b} = 1$
$\Rightarrow - \ 7\text{b} = 1 - 64$
$\Rightarrow - \ 7\text{b} = - \ 63$
$\Rightarrow \text{b} = \dfrac{63}{7} = 9$
$\therefore$ a = 8 and b = 9 Ans
$\Rightarrow 4\text{a} = 3\text{b} + 5 \Rightarrow 4\text{a} - 3\text{b} = 5$......eq(ii)
Multiply eq(i) by 3 and eq(ii) by 7, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 24\text{a} & - & 21\text{b} & = & 3\\ & 28\text{a} & - & 21\text{b} & = & 35\\ (-) & & (+) & & & (-) \\\\ \hline\\ & - \ 4\text{a} & & & = & - \ 32\\ \end{array} $$ $\Rightarrow \text{a} = \dfrac{32}{4} = 8$
Now substitute the value of a in any one equation:
$\Rightarrow 8(8) - 7\text{b} = 1$
$\Rightarrow 64 - 7\text{b} = 1$
$\Rightarrow - \ 7\text{b} = 1 - 64$
$\Rightarrow - \ 7\text{b} = - \ 63$
$\Rightarrow \text{b} = \dfrac{63}{7} = 9$
$\therefore$ a = 8 and b = 9 Ans
Q13. $5\text{x} - 6\text{y} = 8$ and $7\text{y} - 15\text{x} = 9$
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$\Rightarrow 5\text{x} - 6\text{y} = 8$......eq(i)
$\Rightarrow 7\text{y} - 15\text{x} = 9 \Rightarrow - \ 15\text{x} + 7\text{y} = 9$......eq(ii)
Multiply eq(i) by 3, then add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 15\text{x} & - & 18\text{y} & = & 24\\ & - \ 15\text{x} & + & 7\text{y} & = & 9\\ \hline\\ & & & - \ 11\text{y} & = & 33\\ \end{array} $$ $\Rightarrow - \ \text{y} = \dfrac{33}{11} = 3$
$\Rightarrow \text{y} = - \ 3$
Now substitute the value of y in any one equation:
$\Rightarrow 5\text{x} - 6(- \ 3) = 8$
$\Rightarrow 5\text{x} + 18 = 8$
$\Rightarrow 5\text{x} = 8 - 18$
$\Rightarrow 5\text{x} = - \ 10$
$\Rightarrow \text{x} = - \ \dfrac{10}{5} = - \ 2$
$\Rightarrow \text{x} = - \ 2$
$\therefore$ x = $- \ 2$ and y = $- \ 3$ Ans
$\Rightarrow 7\text{y} - 15\text{x} = 9 \Rightarrow - \ 15\text{x} + 7\text{y} = 9$......eq(ii)
Multiply eq(i) by 3, then add eq(i) and eq(ii):
$$ \begin{array}{c c c c c c} & 15\text{x} & - & 18\text{y} & = & 24\\ & - \ 15\text{x} & + & 7\text{y} & = & 9\\ \hline\\ & & & - \ 11\text{y} & = & 33\\ \end{array} $$ $\Rightarrow - \ \text{y} = \dfrac{33}{11} = 3$
$\Rightarrow \text{y} = - \ 3$
Now substitute the value of y in any one equation:
$\Rightarrow 5\text{x} - 6(- \ 3) = 8$
$\Rightarrow 5\text{x} + 18 = 8$
$\Rightarrow 5\text{x} = 8 - 18$
$\Rightarrow 5\text{x} = - \ 10$
$\Rightarrow \text{x} = - \ \dfrac{10}{5} = - \ 2$
$\Rightarrow \text{x} = - \ 2$
$\therefore$ x = $- \ 2$ and y = $- \ 3$ Ans
Q14. $3\text{x} + 2\text{y} = -1$ and $6\text{y} = 5(1 - \text{x})$
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$\Rightarrow 3\text{x} + 2\text{y} = -1$......eq(i)
$\Rightarrow 6\text{y} = 5(1 - \text{x})$
$\Rightarrow 6\text{y} = 5 - 5\text{x}$
$\Rightarrow 5\text{x} + 6\text{y} = 5$......eq(ii)
Multiply eq(i) by 5 and eq(ii) by 3, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 15\text{x} & + & 10\text{y} & = & - \ 5\\ & 15\text{x} & + & 18\text{y} & = & 15\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 8\text{y} & = & - \ 20\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{20}{8} = \dfrac{5}{2} = 2.5$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 2(2.5) = -1$
$\Rightarrow 3\text{x} + 5 = -1$
$\Rightarrow 3\text{x} = - \ 1 - 5$
$\Rightarrow 3\text{x} = - \ 6$
$\Rightarrow \text{x} = - \ \dfrac{6}{3} = - \ 2$
$\therefore$ x = $- \ 2$ and y = 2.5 Ans
$\Rightarrow 6\text{y} = 5(1 - \text{x})$
$\Rightarrow 6\text{y} = 5 - 5\text{x}$
$\Rightarrow 5\text{x} + 6\text{y} = 5$......eq(ii)
Multiply eq(i) by 5 and eq(ii) by 3, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 15\text{x} & + & 10\text{y} & = & - \ 5\\ & 15\text{x} & + & 18\text{y} & = & 15\\ (-) & & (-) & & & (-) \\\\ \hline\\ & & & - \ 8\text{y} & = & - \ 20\\ \end{array} $$ $\Rightarrow \text{y} = \dfrac{20}{8} = \dfrac{5}{2} = 2.5$
Now substitute the value of y in any one equation:
$\Rightarrow 3\text{x} + 2(2.5) = -1$
$\Rightarrow 3\text{x} + 5 = -1$
$\Rightarrow 3\text{x} = - \ 1 - 5$
$\Rightarrow 3\text{x} = - \ 6$
$\Rightarrow \text{x} = - \ \dfrac{6}{3} = - \ 2$
$\therefore$ x = $- \ 2$ and y = 2.5 Ans
Q15. $\text{a} = \text{b} + 2$ and $2\text{a} - \text{b} = 7$
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$\Rightarrow \text{a} = \text{b} + 2 \Rightarrow \text{a} - \text{b} = 2$......eq(i)
$\Rightarrow 2\text{a} - \text{b} = 7$......eq(ii)
Multiply eq(i) by 2, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 2\text{a} & - & 2\text{b} & = & 4\\ & 2\text{a} & - & \text{b} & = & 7\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & - \ \text{b} & = & - \ 3\\ \end{array} $$ $\Rightarrow \text{b} = 3$
Now substitute the value of b in any one equation:
$\Rightarrow \text{a} = 3 + 2$
$\Rightarrow \text{a} = 5$
$\therefore$ a = 5 and b = 3 Ans
$\Rightarrow 2\text{a} - \text{b} = 7$......eq(ii)
Multiply eq(i) by 2, then subtract eq(ii) from eq(i):
$$ \begin{array}{c c c c c c} & 2\text{a} & - & 2\text{b} & = & 4\\ & 2\text{a} & - & \text{b} & = & 7\\ (-) & & (+) & & & (-) \\\\ \hline\\ & & & - \ \text{b} & = & - \ 3\\ \end{array} $$ $\Rightarrow \text{b} = 3$
Now substitute the value of b in any one equation:
$\Rightarrow \text{a} = 3 + 2$
$\Rightarrow \text{a} = 5$
$\therefore$ a = 5 and b = 3 Ans