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 : $$a=\frac{1}{\sqrt{x}}andb=\frac{1}{\sqrt{y}}=-1$$ 

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)

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

$$or4a+6b-4a+9b=4+1$$

$$or15b=5$$

$$orb=\frac{5}{15}$$

$$orb=\frac{1}{3}$$

Now, substitute b in equation (1)

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

$$or2a+1=2$$

$$or2a=1$$

$$\therefore a=\frac{1}{2}$$

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)