Question

4x+9/ 3x+11 = 4x+3/3x+2​

Answer 1

Answer:

$\rm{Value\; of\; x\; is\;\dfrac{-5}{6}}$.

Step-by-step explanation:

$\bf{Q)}\;\rm{\dfrac{4x+9}{3x+11}=\dfrac{4x+3}{3x+2}}$

By cross-multiplication,

$\sf{\longrightarrow (4x+9)(3x+2)=(3x+11)(4x+3)}$

Multiplying the values,

$\sf{\longrightarrow4x(3x+2)+9(3x+2)=3x(4x+3)+11(4x+3)}$

$\sf{\longrightarrow12x^2+8x+27x+18=12x^2+9x+44x+33}$

$\sf{\longrightarrow12x^2+35x+18=12x^2+53x+33}$

Removing 12x² from both sides,

$\sf{\longrightarrow35x+18=53x+33}$

Transposing variables to LHS and constants to RHS,

$\sf{\longrightarrow35x-53x=33-18}$

Subtracting the numbers,

$\sf{\longrightarrow-18x=15}$

Transposing 18 from LHS to RHS and changing its sign,

$\sf{\longrightarrow-x=\dfrac{15}{18}}$

$\sf{\longrightarrow x=\dfrac{-15}{18}}$

Reducing the numbers,

$\sf{\longrightarrow x=\dfrac{-15\div3}{18\div3}}$

$\sf{\longrightarrow x=\dfrac{-5}{6}}$

Hence, value of x is -5/6.

Verification:

 $\rm{Substituting \;value \;of\; x\; to\; :\dfrac{4x+9}{3x+11}=\dfrac{4x+3}{3x+2}}$.

LHS:

$\rm{\longmapsto\dfrac{4\times\dfrac{-5}{6}+9}{3\times\dfrac{-5}{6}+11}}$

$\rm{\longmapsto\dfrac{2\times\dfrac{-5}{3}+9}{\dfrac{-5}{2}+11}}$

$\rm{\longmapsto\dfrac{\dfrac{-10}{3}+9}{\dfrac{-5}{2}+11}}$

$\rm{\longmapsto\dfrac{\dfrac{-10+27}{3}}{\dfrac{-5+22}{2}}}$

$\rm{\longmapsto\dfrac{\dfrac{17}{3}}{\dfrac{17}{2}}}$

$\rm{\longmapsto\dfrac{17\times2}{17\times3}=\dfrac{2}{3}}$

RHS:

$\rm{\longmapsto\dfrac{4\times\dfrac{-5}{6}+3}{3\times\dfrac{-5}{6}+2}}$

$\rm{\longmapsto\dfrac{2\times\dfrac{-5}{3}+3}{\dfrac{-5}{2}+2}}$

$\rm{\longmapsto\dfrac{\dfrac{-10}{3}+3}{\dfrac{-5}{2}+2}}$

$\rm{\longmapsto\dfrac{\dfrac{-10+9}{3}}{\dfrac{-5+4}{2}}}$

$\rm{\longmapsto\dfrac{\dfrac{-1}{3}}{\dfrac{-1}{2}}}$

$\rm{\longmapsto\dfrac{-1\times2}{-1\times3}=\dfrac{2}{3}}$

As,

• LHS = RHS

Hence, verified.

Answer 2

Answer:

$\\ \longrightarrow\sf {\dfrac {4x + 9}{3x + 11} = \dfrac {4x + 3}{3x + 2}}$

Taking cross multiplication ,

$\\ \longrightarrow\sf {(4x + 9)(3x + 2) = (4x + 3)(3x + 11)}$

$\\ \longrightarrow\sf {4x (3x + 2) + 9 ( 3x + 2 ) = 4x (3x + 11) + 3 ( 3x + 11)}$

$\\ \longrightarrow\sf {12 {x}^{2} + 8x + 27x + 18 = 12 {x}^{2} + 44x + 9x + 33 }$

Similar terms on both sides get cancelled

$\\ \longrightarrow\sf { 8x + 27x + 18 = 44x + 9x + 33 }$

$\\ \longrightarrow\sf { ( 8 + 27 )x + 18 = ( 44 + 9)x + 33 }$

$\\ \longrightarrow\sf { 35x + 18 = 53x + 33 }$

$\\ \longrightarrow\sf { 18 - 33 = 53x - 35x }$

$\\ \longrightarrow\sf { - 15 = 18x }$

$\\ \longrightarrow\sf { \dfrac {-15 ÷ 3}{18 ÷ 3} = x }$

$\\ \longrightarrow\sf { \dfrac {-5}{6} = x }$

Substituting value of x .

$\\ \longrightarrow\sf {\dfrac {4x + 9}{3x + 11} = \dfrac {4x + 3}{3x + 2}}$

$\\ \longrightarrow\sf {\dfrac {4\times \dfrac {-5}{6} + 9}{3\times \dfrac {-5}{6} + 11} = \dfrac {4\times \dfrac {-5}{6} + 3}{3\times \dfrac {5}{6} + 2}}$

$\\ \longrightarrow\sf {\dfrac {2\times \dfrac {-5}{3} + 9}{\dfrac {-5}{2} + 11} = \dfrac {2\times \dfrac {-5}{3} + 3}{\dfrac {-5}{2} + 2}}$

$\\ \longrightarrow\sf {\dfrac { \dfrac {-10}{3} + 9}{\dfrac {-5}{2} + 11} = \dfrac { \dfrac {-10}{3} + 3}{\dfrac {-5}{2} + 2}}$

$\\ \longrightarrow\sf {\dfrac { \dfrac {-10 + 27}{3}}{\dfrac {-5 + 22 }{2}} = \dfrac { \dfrac {-10 + 9}{3}}{\dfrac {-5 + 4}{2}}}$

$\\ \longrightarrow\sf {\dfrac { \dfrac {17}{3}}{\dfrac {17 }{2}} = \dfrac { \dfrac {- 1}{3}}{\dfrac {- 1}{2}}}$

$\\ \longrightarrow\sf {\dfrac {17}{3}\times \dfrac {2}{17} = \dfrac {- 1}{3}\times \dfrac {2}{-1}}$

$\\ \longrightarrow\sf {\dfrac {34÷17}{51÷17} = \dfrac {- 2}{-3}}$

$\\ \longrightarrow\sf {\dfrac {2}{3} = \dfrac {2}{3}}$

$\\ \longrightarrow\sf {LHS = RHS}$