Question

solve the following equation and verify your answer 3k+5/4k-3=4/9​

Answer 1

ANSWER:

• Value of k = -57/11

GIVEN:

$\implies \: \dfrac{3k + 5}{4k - 3} = \dfrac{4}{9}$

TO FIND:

• Value of 'k'.

SOLUTION:

$\implies \: \dfrac{3k + 5}{4k - 3} = \dfrac{4}{9} \\ \\ \implies \: 9(3k + 5) = 4(4k - 3) \\ \implies \: 27k + 45 = 16k - 12 \\ \implies \: 27k - 16k = - 45 - 12 \\ \implies \: 11k = - 57 \\ \implies \: k \: = \frac{ - 57}{11}$

Putting k = -57/11 in LHS:

$= \dfrac{3( \dfrac{ - 57}{11}) + 5 }{4( \dfrac{ - 57}{11} ) - 3} \\ \\ = \dfrac{ \dfrac{ - 171}{11} + 5}{ \dfrac{ - 228}{11} - 3 } \\ \\ = \frac{ \dfrac{ - 116}{11} }{ \dfrac{ - 261}{11} } \\ \\ = \frac{116}{261} \\ \\ = \dfrac{116 \div 29}{261 \div 29} \\ \\ = \dfrac{4}{9}$

LHS = RHS

Answer 2

$\huge{\longrightarrow{\text{ANSWER }}}{\longleftarrow}$

Given :-

$\large{\tt{ \dfrac{ 3k + 5 }{ 4k - 3 } = \dfrac{4}{9} }}$

Required to find :-

• Value of k

• Verification of the answer

Concept used :-

Method used in solving linear equations

Solution :-

Given :-

$\large{\tt{ \dfrac{ 3k + 5 }{ 4k - 3 } = \dfrac{4}{9} }}$

we need to find the value of k and verification of the answer

so

$\large{\tt{ \dfrac{ 3k + 5 }{ 4k - 3 } = \dfrac{4}{9} }}$

Cross multiplication on both sides

$\longrightarrow{\tt{9 ( 3k + 5 ) = 4 ( 4k - 3 ) }}$

$\longrightarrow{\tt{ 27k + 45 = 16k - 12 }}$

$\longrightarrow{\tt{ 27k - 16k = - 12 - 45 }}$

$\longrightarrow{\tt{ 11k = - 57 }}$

$\longrightarrow{\tt{ k = \dfrac{ -57 }{11 }}}$

Hence

Value of k is - 57/11

Verification :-

Since we had find the value of k let's find whether it is correct or wrong

So,

$\large{\tt{ \dfrac{ 3k + 5 }{ 4k - 3 } = \dfrac{4}{9} }}$

Here We need to substitute the value of k in L.H.S. part and the end resulting value should be equal to R.H.S. part .

If LHS = RHS then the value of k is correct

If , LHS ≠ RHS then the value of k is incorrect

$\large{\longrightarrow{\tt{ \dfrac{ 3 ( \dfrac{ - 57 }{11 }) + 5 }{ 4 ( \dfrac{ - 57 }{11 } ) - 3 } = \dfrac{ 4 }{ 9 }}}}$

$\large{\longrightarrow{\tt{ \dfrac{ \dfrac{3}{1} \times \dfrac{-57}{11} + 5 }{ \dfrac{4}{1} \times \dfrac{ - 57 }{11 } - 3 } = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ \dfrac{ - 171 }{11} + 5 }{ \dfrac{ - 228 }{11 } - 3 } = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ \dfrac{ -171 }{11} + \dfrac{5 \times 11 }{1 \times 11 }}{ \dfrac{ - 228 }{11 } - \dfrac{3 \times 11 }{ 1 \times 11 }} = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ \dfrac{ -171 + 55}{11}}{ \dfrac{ -228 - 33}{11}} = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ \dfrac{ -116 }{11 }}{ \dfrac{ -261 }{11 }} = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ -116 }{11 } \div \dfrac{ -261 }{ 11 } = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ -116 }{11 } \times \dfrac{ 11 }{ -261 } = \dfrac{4}{9}}}}$

11 , - ( minus sign ) gets cancelled on both numerator and denominator

$\large{\longrightarrow{\tt{ \dfrac{ 116 }{ 261 } = \dfrac{4}{9}}}}$

$\large{\longrightarrow{\tt{ \dfrac{ 29 ( 4 ) }{ 29 ( 9 ) } = \dfrac{4}{9}}}}$

This is because

29 x 4 = 116

29 x 9 = 261

So , 29 gets cancelled in both numerator and denominator

$\large{\implies{\red{\tt{ \dfrac{4}{9} = \dfrac{4}{9}}}}}$

Therefore,

LHS = RHS

Hence value of " k " is correct .