Class X Session 2023-24 Question 28-optional

 MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041)

Question 28

\[Solve:  \; \frac{2}{\sqrt{x}} +  \frac{3}{\sqrt{y}} = 2; \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1, x,y>0 \]

Explanation:  

Let's consider : \[a =  \; \frac{1}{\sqrt{x}} \; and \; b = \frac{1}{\sqrt{y}} = -1\] 

Now equation becomes 

2a+3b = 2 -----------------(1)

4a - 9b = -1 -------------------------(2)

now solve both equations (1) and (2) First, multiply equation (1) by 2 to align the coefficients of a and subtract from each other 

2x(2a+3b) - (4a -9b ) = 4 -(-1)

\[2\times (2a+3b) - (4a-9b) = 4-(-1)\]

\[ or \; 4a+6b -4a + 9b = 4+1\]

\[ or \; 15b = 5\]

\[ or \; b = \frac{5}{15}\]

\[ or \; b = \frac{1}{3}\]

Now, substitute b in equation (1)

\[\Rightarrow 2a + 3\times \frac{1}{3} = 2\]

\[\ or\; 2a + 1 = 2\]

\[\ or\; 2a = 1\]

\[\therefore a = \frac{1}{2}\]

Now we have a and b so we can find the values of x and y 

\[\ a = \frac{1}{\sqrt{x}} \Rightarrow \frac{1}{2} = \frac{1}{\sqrt{x}} \]

\[\ \sqrt{x} = 2 \]

\[\therefore x = 4 \]

\[\ b = \frac{1}{\sqrt{y}} \Rightarrow \frac{1}{3} = \frac{1}{\sqrt{y}}\]

\[\ \sqrt{y} = 3 \]

\[\therefore y = 9 \]

So the solution to the system of equations is (x,y)= (4,9)