(1) \(~~ 3x-2y=2 \), \(9x^2+4y^2=2 \)
Solution:
Let, \(3x-2y=2 ⇒ x = \frac{2+2y}{3} \) ...(1)
and \(9x^2+4y^2=2 \) ...(2)
⇒ \( 9[\frac{2+2y}{3}]^2+4y^2 = 2 \) [using (1)]
⇒ \(9[\frac{4+4y^2+8y}{9}]+4y^2 = 2 \)
⇒ \( 4+4y^2+8y+4y^2=2 \)
⇒ \( 8y^2+8y+2 = 0 \)
⇒ \( 4y^2+4y+1 = 0 \)
⇒ \( (2y+1)^2 = 0 \)
⇒ \( y = -\frac{1}{2} \) or \( -\frac{1}{2} \)
When \( y = -\frac{1}{2} \)
(1)⇒ \(x = \frac{2+2(-\frac{1}{2})}{3} = \frac{2-1}{3} = \frac{1}{3} \)
Since both the value of \(y \) are equal, the required solutions are \( (\frac{1}{3}, -\frac{1}{2}) \) and \( (\frac{1}{3}, -\frac{1}{2}) \)
(8) \(~~ x+y+\sqrt{xy}=28 \), \(x^2+y^2+xy=336 \)
Solution:
Let, \(x+y+\sqrt{xy}=28 \) ...(1)
and \(x^2+y^2+xy=336 \) ...(2)
⇒ \( (x+y)^2-2xy+xy = 336 \)
⇒ \((x+y)^2-xy = 336 \)
⇒ \((x+y)^2-(\sqrt{xy})^2 = 336 \)
⇒ \( (x+y+\sqrt{xy})(x+y-\sqrt{xy}) = 336 \)
⇒ \( 28(x+y-\sqrt{xy}) = 336 \) [using (1)]
⇒ \( x+y - \sqrt{xy}=12 \) ...(3)
(1)+(3) ⇒ \(x+y+\sqrt{xy}+x+y-\sqrt{xy} = 28+12 \)
⇒ \( 2(x+y) = 40 \)
⇒ \(x+y = 20 \)
⇒ \(x = 20-y \) ...(4)
(2) ⇒ \( (20-y)^2+y^2+(20-y) = 336 \) [using (4)]
⇒ \( 400-40y+y^2+y^2-y^2+20y = 336 \)
⇒ \(y^2-16y-4y+64 = 0 \)
⇒ \(y(y-16)-4(y-16) = 0 \)
⇒ \( y = 4 \) or \(16 \)
When \(y = 4 \), (4)⇒ \(x = 20-4 = 16 \)
When \(y = 16 \), (4)⇒ \(x = 20-16 = 4 \)
⸫ The required solutions are \( (4, 16) \) and \( (16, 4) \)
(10) \(~~ \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = \frac{10}{3} \), \( x+y=10 \)
Solution:
Let, \( \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = \frac{10}{3} \) ...(1)
and \( x+y=10 \) ...(2)
⇒ \( x = 10-y \) ...(3)
(1)⇒ \( \frac{x+y}{\sqrt{xy}} = \frac{10}{3} \)
⇒ \( \frac{10}{\sqrt{xy}} = \frac{10}{3} \) [using (2)]
⇒ \( \sqrt{xy} = 3 \)
⇒ \( \sqrt{(10-y)y} = 3 \)
⇒ \( (\sqrt{10y - y^2})^2 = 3^3 \)
⇒ \( 10y - y^2 = 9 \)
⇒ \( y^2-10y+9 = 0 \)
⇒ \( y^2-9y-y+9 = 0 \)
⇒ \( y(y-9)-1(y-9) = 0 \)
⇒ \((y-1)(y-9) = 0 \)
⇒ \( y = 1 \) or \( 9 \)
When \(y = 1 \), (3)⇒ \(x = 10-1 = 9 \)
When \(y = 9 \), (3)⇒ \(x = 10-9 = 1 \)
⸫ The required solutions are \( (4, 16) \) and \( (16, 4) \)
(12) \(~~ x+y = p+q \), \( \frac{p}{q}+\frac{q}{y} = 2 \)
Solution:
Let, \(x+y=p+q \) ...(1)
⇒ \( x-p = q-y \)
⇒ \( p-x = y-q \) ...(2)
and \( \frac{p}{q}+\frac{q}{y} = 2 \)
⇒ \( \frac{p}{q}-1+\frac{q}{y}-1 = 0 \) ...(3)
⇒ \( \frac{p-x}{x}+\frac{q-y}{y} = 0 \)
⇒ \( \frac{y-q}{x}+\frac{y-q}{y} = 0 \) [using (2)]
⇒ \( (y-q)(\frac{1}{x}-\frac{1}{y}) = 0 \)
Either \( y = q \)
or \( \frac{1}{x}-\frac{1}{y} \)
⇒ \( \frac{1}{x} = \frac{1}{y} \)
⇒ \( x = y \)
When \( y = q \) , (1)⇒ \( x+q = p+q ⇒ x = p \)
When \( y = x \) , (1)⇒ \(x+x = p+q ⇒ x = \frac{p+q}{2} \) \[\ ⸫~~ y = \frac{p+q}{2} ~~[⸪~~ x = y] \] ⸫ The required solutions are \( (p, q) \) and \( (\frac{p+q}{2}, \frac{p+q}{2}) \)
(14) \(~~ \frac{a}{x}+\frac{b}{y} = 2 \), \(\frac{a^2}{x^2}+\frac{b^2}{y^2} = 2 \)
Solution:
Let, \(\frac{a}{x}+\frac{b}{y} = 2 \) ...(1)
and \(\frac{a^2}{x^2}+\frac{b^2}{y^2} = 2 \)
⇒ \( (\frac{a}{x}+\frac{b}{y})^2 -\frac{2ab}{xy} = 2 \)
⇒ \( (2)^2 -\frac{2ab}{xy} = 2 \) [using (1)]
⇒ \( 2 = \frac{2ab}{xy} \)
⇒ \( \frac{ab}{xy} = 1 \)
⇒ \( \frac{b}{y} = \frac{x}{a} \) ...(2)
(1)⇒ \( \frac{a}{x}+\frac{x}{a} = 2 \) [using (2)]
⇒ \( \frac{a^2+x^2}{ax} = 2 \)
⇒ \( a^2+x^2 = 2ax \)
⇒ \( a^2+x^2-2ax = 0 \)
⇒ \( (x-a)^2 = 0 \)
⇒ \( (x-a)(x-a) = 0 \)
⇒ \( x = a \) or \( a \)
When \( x = a \) , (2)⇒ \( \frac{b}{y}= \frac{a}{a} ⇒ \frac{b}{y}=1 ⇒ y = b \)
Since both the value of \(x \) are equal, the required solutions are \( (a, b) \) and \((a, b) \)
(16) \(~~ 8.2^{xy}=4^y \), \(27.9^x.3^{xy} = 1 \)
Solution:
Let, \( 8.2^{xy}=4^y \)
⇒ \( 2^3.2^{xy} = (2^2)^y \)
⇒ \( 2^{3+xy} = 2^{2y} \)
⇒ \( 3+xy = 2y \)
⇒ \( xy = 2y-3 \) ...(1)
and \(27.9^x.3^{xy} = 1 \)
⇒ \( 3^3.(3^{2x}.3^{xy} = 1 \)
⇒ \( 3^{3+2x+xy} = 3^0 \)
⇒ \( 3+2x+xy = 0 \)
⇒ \( 3+2x+2y-3 = 0 \) [using (1)]
⇒ \( 2(x+y) = 0 \)
⇒ \( x = -y \) ...(2)
(1)⇒ \( (-y) y = 2y-3 \)
⇒ \( -y^2-2y+3 = 0 \)
⇒ \( y^2+2y-3 = 0 \)
⇒ \( y^2+3y-y-3 = 0 \)
⇒ \( y(y+3)-1(y+3) = 0 \)
⇒ \( (y+3)(y-1) = 0 \)
⇒ \( y = -3 \) or \( 1 \)
When \( y = -3 \) , (2)⇒ \( x = -(-3) ⇒ x = 3 \)
When \( y = 1 \) , (2)⇒ \( x = -(1) ⇒ x = -1 \)
⸫ The required solutions are \( (-1, 3) \) and \(1, -3) \)
20. Find the coordinates of the points at which x-axis and y-axis intersect the curves represented by the following equations:
(i) \(2x^2-3xy+y^2+x-2y-3=0\)
Solution:
\(2x^2-3xy+y^2+x-2y-3=0\) ...(1)
At x-axis, \(y=0\), then
(1) ⇒ \(2x^2+x-3=0\)
⇒ \(2x^2+3x-2x-3=0\)
⇒ \(x(2x+3)-1(2x+3)=0\)
⇒ \((2x+3)(x-1)=0\)
⇒ \(x=-\frac{3}{2}\) or \(x=1\)
⸫ The curve intersects x-axis at (\(-\frac{3}{2}, 0\)) and (\(1,0\))
At y-axis, \(x=0\), then
(1) ⇒ \(y^2-2y-3=0\)
⇒ \(y^2-3y+y-3=0\)
⇒ \(y(y-3)+1(y-3)=0\)
⇒ \((y-3)(y+1)=0\)
⇒ \(x=3\) or \(x=-1\)
⸫ The curve intersects y-axis at (\(0, 3\)) and (\(0, -1\))