Question

Solve the following linear equation and verify the result.
(a)\[\dfrac{{2x - \left( {7 - 5x} \right)}}{{9x - \left( {3 + 4x} \right)}} = \dfrac{7}{6}\]
(b)\[\dfrac{{15\left( {2 - x} \right) - 5\left( {x + 6} \right)}}{{1 - 3x}} = - 10\]

Answer 1

Hint: Here we have to open brackets first and then simplify both numerator and denominator. Then calculate the value of x and verify the equation by putting the value we calculated earlier.

Complete step-by-step answer:
(i)\[\dfrac{{2x - \left( {7 - 5x} \right)}}{{9x - \left( {3 + 4x} \right)}} = \dfrac{7}{6}\]
Firstly, open the brackets from both numerator and denominator.
\[\dfrac{{2x - 7 + 5x}}{{9x - 3 - 4x}} = \dfrac{7}{6}\]
Combining all the x terms from both numerator and denominator.
\[\dfrac{{7x - 7}}{{5x - 3}} = \dfrac{7}{6}\]
Here we will perform cross multiplication to get all variables on one side and constants on other.
$\Rightarrow$ \[\left( {7x - 7} \right)6 = \left( {5x - 3} \right)7\]
$\Rightarrow$ \[42x - 42 = 35x - 21\]
Taking all variables on one side and constants on the other side.
$\Rightarrow$ \[42x - 35x = 42 - 21\]
$\Rightarrow$ \[7x = 21\]
Therefore, \[x = 3\]
Now, we will verify the equation \[\dfrac{{2x - \left( {7 - 5x} \right)}}{{9x - \left( {3 + 4x} \right)}} = \dfrac{7}{6}\] by putting x=3 in L.H.S
Taking L.H.S \[\dfrac{{2x - \left( {7 - 5x} \right)}}{{9x - \left( {3 + 4x} \right)}}\] = \[\dfrac{{7x - 7}}{{5x - 3}}\]
Here, in above equation we will put x=3
\[ \Rightarrow \dfrac{{7 * 3 - 7}}{{5 * 3 - 3}}\]
\[ \Rightarrow \dfrac{{21 - 7}}{{15 - 3}}\]
\[ \Rightarrow \dfrac{{14}}{{12}}\]
Dividing both numerator and denominator by 2.
\[ \Rightarrow \dfrac{7}{6}\] which is equal to R.H.S
Therefore, L.H.S=R.H.S
Hence verified.

(ii)\[\dfrac{{15\left( {2 - x} \right) - 5\left( {x + 6} \right)}}{{1 - 3x}} = - 10\]
Firstly, open the brackets from both numerator and denominator.
\[\dfrac{{30 - 15x - 5x - 30}}{{1 - 3x}} = - 10\]
Combining all the x terms from both numerator and denominator, and Simplifying the equation.
$\Rightarrow$ \[\dfrac{{ - 20x}}{{1 - 3x}} = - 10\]
Here we will perform cross multiplication to get all variables on one side and constants on other.
$\Rightarrow$ \[ - 20x = - 10\left( {1 - 3x} \right)\]
$\Rightarrow$ \[ - 20x = - 10 + 30x\]
Taking all variables on one side and constants on the other side.
$\Rightarrow$ \[ - 20x - 30x = - 10\]
$\Rightarrow$ \[ - 50x = - 10\]
$\Rightarrow$ \[x = \dfrac{{10}}{{50}}\]
Therefore, \[x = \dfrac{1}{5}\]
Now we will verify the equation \[\dfrac{{15\left( {2 - x} \right) - 5\left( {x + 6} \right)}}{{1 - 3x}} = - 10\] by putting \[x = \dfrac{1}{5}\] in L.H.S
Taking L.H.S \[\dfrac{{15\left( {2 - x} \right) - 5\left( {x + 6} \right)}}{{1 - 3x}} = \dfrac{{ - 20x}}{{1 - 3x}}\]
Here, in above equation we will put \[x = \dfrac{1}{5}\]
\[ \Rightarrow \dfrac{{ - 20 * \dfrac{1}{5}}}{{1 - 3 * \dfrac{1}{5}}}\]
\[ \Rightarrow \dfrac{{ - 4}}{{\dfrac{{5 - 3}}{5}}}\]
\[ \Rightarrow \dfrac{{ - 4 \times 5}}{2}\]
\[ \Rightarrow - 2 \times 5\]
\[ \Rightarrow - 10\] which is equal to R.H.S.
Therefore, L.H.S=R.H.S
Hence verified.

Note: In these types of questions simplify the equations as much as we can and by using cross multiplication, We can easily separate constants on one side and terms containing x on the other side. Hence, do not forget to verify the result with the help of calculated value.