Question

Using properties of proportion find x:y,given
X^2+2x/2x+4=y^2+3y/3y+9
Icse 2020
Plz solve correctly

Answer 1

Step-by-step explanation:

• Given

        $\mathbf{\frac{x^{2}+2x}{2x+4}=\frac{y^{2}+3y}{3y+9}}$

       adding 1 on both side, we get

       $\mathbf{\frac{x^{2}+2x}{2x+4}+1=\frac{y^{2}+3y}{3y+9}+1}$

       On L.H.S and R.H.S

       $\mathbf{\frac{x^{2}+2x+2x+4}{2x+4}=\frac{y^{2}+3y+3y+9}{3y+9}}$  

       $\mathbf{\frac{x^{2}+4x+4}{2x+4}=\frac{y^{2}+6y+9}{3y+9}}$          ...1)

• Numerator and denominator of equation 1) can be written as

        $\mathbf{\frac{(x+2)^{2}}{2(x+2)}=\frac{(y+3)^{2}}{3(y+3)}}$

       On cancel out term on L.H.S and R.H.S

       $\mathbf{\frac{(x+2)}{2}=\frac{(y+3)}{3}}$      ...2)

• Cross multiplication of equation 2), we get

        $\mathbf{3(x+2)=2(y+3)}$

        $\mathbf{3x+6=2y+6}$

        $\mathbf{3x=2y}$

        So

        $\mathbf{\frac{x}{y}=\frac{2}{3}=}$ This is answer