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This is the second of our many quizzes on Solving equations with variables on both sides. Click on this to access more such competitive quizzes on Algebra.

Solving equations with variables on both sides Quiz 2

mathematics

 Solve $\dfrac{{x - 3}}{3} + \dfrac{{x + 4}}{4} = x - 1 + \dfrac{{2x + 3}}{6}$ for the value of $x$.

 Simplify the expression on both sides of the equation and then transfer all terms with $x$ to the LHS. 

  $  \dfrac{{x - 3}}{3} + \dfrac{{x + 4}}{4} = x - 1 + \dfrac{{2x + 3}}{6} \\   \Rightarrow \dfrac{x}{3} - 1 + \dfrac{x}{4} + 1 = x - 1 + \dfrac{x}{3} + \dfrac{1}{2} \\   \Rightarrow \dfrac{x}{3} + \dfrac{x}{4} - x - \dfrac{x}{3} =  - 1 + \dfrac{1}{2} \\   \Rightarrow x - \dfrac{x}{4} = 1 - \dfrac{1}{2} \\   \Rightarrow \dfrac{3}{4}x = \dfrac{1}{2} \Rightarrow x = \dfrac{2}{3} \\  $

 The value of $y$ which satisfies the equation $\dfrac{{3 - 2y}}{5} + 7 = \dfrac{{2\left( {y + 3} \right)}}{7} - 5$ is

 Transfer the terms with $y$ to the LHS and then simplify. 

  $  \dfrac{{3 - 2y}}{5} + 7 = \dfrac{{2\left( {y + 3} \right)}}{7} - 5 \\   \Rightarrow \dfrac{{3 - 2y}}{5} + 7 = \dfrac{{2y + 6}}{7} - 5 \\   \Rightarrow \dfrac{{3 - 2y}}{5} - \dfrac{{2y + 6}}{7} =  - 5 - 7 \\   \Rightarrow \dfrac{{21 - 14y - 10y - 30}}{{35}} =  - 12 \\   \Rightarrow \dfrac{{ - 24y - 9}}{{35}} =  - 12 \\   \Rightarrow 24y + 9 = 12 \times 35 = 420 \\   \Rightarrow y = \dfrac{{411}}{{24}} \\  $

 Solve $\dfrac{{ - 3x + 5}}{2} = \dfrac{{x\left( {2 - \dfrac{2}{x}} \right)}}{3} + 5$ for $x$ and mark the correct answer. 

$  \dfrac{{ - 3x + 5}}{2} = \dfrac{{x\left( {2 - \dfrac{2}{x}} \right)}}{3} + 5 \\   \Rightarrow \dfrac{{ - 3x + 5}}{2} = \dfrac{{2x - 2}}{3} + 5 \\   \Rightarrow \dfrac{{ - 3x + 5}}{2} - \dfrac{{2x - 2}}{3} = 5 \\   \Rightarrow \dfrac{{ - 9x + 15 - 4x + 4}}{6} = 5 \\   \Rightarrow \dfrac{{ - 13x + 19}}{6} = 5 \\   \Rightarrow  - 13x = 30 - 19 = 11 \\   \Rightarrow x =  - \dfrac{{11}}{{13}} \\  $

 The age of a father is $5 years$ more than $5$ times his son’s age. In $5$ years time, father’s age will be $5 years$ more than $3$ times his son’s age. Then the present age of the father and the son is

 Assume an age for the son or the father and correspondingly form an equation with the given conditions. The solution of this equation will give the required result.

 Let the present age of the son is $x$. Then the father's present age is $5x + 5$. $ \\ $ After $5$ years, son’s age will be $x + 5$. And father’s age will be $3\left( {x + 5} \right) + 5$. $ \\ $ Also the father’s age after $5$ years will be $5x + 5 + 5$. $ \\ $ Thus, $ 3\left( {x + 5} \right) + 5 = 5x + 5 + 5$ $\\  \Rightarrow 3x + 15 + 5 = 5x + 10 \\   \Rightarrow 5x - 3x = 20 - 10 \\   \Rightarrow 2x = 10 \Rightarrow x = 5 \\  $ Thus, the son's present age is $5 years$ and father’s present age is $30 years$.

 The numerator of a number is more than its denominator by $5$. If the numerator becomes $4$ times and the denominator becomes it’s three times, then the difference between the new number and the original number is $\dfrac{1}{8}$. The original number is

 Assume the denominator or numerator and correspondingly form an equation with the given conditions. The solution of this equation will give the required result.

 Let the denominator of the original number is $x$. $ \\ $ Then the original number is $\dfrac{{x + 5}}{x}$. $ \\ $ The new number obtained will be $\dfrac{{4\left( {x + 5} \right)}}{{3x}}$. $ \\ $ Given that, the new number is twice the original number. Therefore, $ \\  \dfrac{{4\left( {x + 5} \right)}}{{3x}} = \left( {\dfrac{{x + 5}}{x}} \right) + \dfrac{1}{8} \\   \Rightarrow \dfrac{{4x + 20}}{{3x}} = \dfrac{{8x + 40 + x}}{{8x}} \\ \Rightarrow 32x + 160 = 27x + 120 \\   \Rightarrow 32x - 27x = 120 - 160 \\   \Rightarrow 5x =  - 40 \\   \Rightarrow x =  - 8 \\  $ Thus, the original number is $\dfrac{{x + 5}}{x} = \dfrac{{ - 8 + 5}}{{ - 8}} = \dfrac{3}{8}$.