Question

Solve the following equation stepwise.

Give the accurate answer stepwise.​

Answer 1

Let's solve your equation step-by-step

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

Step 1: Simplify the equation.

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

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

Step 2: Combine like terms.

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

⇒ $(\dfrac{1}{2}x + \dfrac{1}{3}x- \dfrac{1}{4}x ) + (\dfrac{-1}{2} + \dfrac{2}{3} + \dfrac{3}{4} )= 1$

⇒ $(\dfrac{1\times 6 }{2\times 6 }x + \dfrac{1\times 4 }{3\times 4 }x- \dfrac{1\times 3 }{4\times 3 }x ) + (\dfrac{-1\times 6 }{2\times 6 } + \dfrac{2\times 4 }{3\times 4 } + \dfrac{3\times 3 }{4\times 3 } )= 1$

⇒ $(\dfrac{6 }{12}x + \dfrac{4 }{12 }x- \dfrac{3 }{12 }x ) + (\dfrac{-6 }{12 } + \dfrac{8}{12 } + \dfrac{9 }{12} )= 1$

⇒ $(\dfrac{7}{12}x ) + (\dfrac{11}{12 } )= 1$

⇒ $\dfrac{7}{12}x + \dfrac{11}{12 } = 1$

Step 3: Subtract ¹¹/₁₂ from both sides of the equation.

⇒ $\dfrac{7}{12}x + \dfrac{11}{12 } = 1$

⇒ $\dfrac{7}{12}x + \dfrac{11}{12 } - \dfrac{11}{12} = 1-\dfrac{11}{12}$

⇒ $\dfrac{7}{12}x = \dfrac{12}{12} -\dfrac{11}{12}$

⇒ $\dfrac{7}{12}x = \dfrac{1}{12}$

Step 4: Multiply ¹²/₇ from both sides of the equation.

⇒ $\dfrac{7}{12}x = \dfrac{1}{12}$

⇒ $\dfrac{12}{7}\times \dfrac{7}{12}x = \dfrac{12}{7}\times \dfrac{1}{12}$

⇒ $x = \dfrac{1}{7}$

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Verification for value of 'x'

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

⇒ $\dfrac{\frac{1}{7} -1}{2} + \dfrac{\frac{1}{7} +2}{3} - \dfrac{\frac{1}{7} -3}{4} =1$

⇒ $\dfrac{\frac{1}{7} -\frac{7}{7} }{2} + \dfrac{\frac{1}{7} +\frac{14}{7} }{3} - \dfrac{\frac{1}{7} -\frac{21}{7} }{4} =1$

⇒ $\dfrac{-\frac{6}{7} }{2} +\dfrac{\frac{15}{7} }{3} - \dfrac{-\frac{20}{7} }{4} =1$

⇒ $\dfrac{-6}{7} \times \dfrac{1}{2} + \dfrac{15}{7}\times \dfrac{1}{3} + \dfrac{20}{7} \times \dfrac{1}{4} = 1$

⇒ $\dfrac{-3}{7}+\dfrac{5}{7} +\dfrac{5}{7} = 1$

⇒ $\dfrac{7}{7} =1$

⇒ $1 = 1$

∴ LHS = RHS

$\bf \therefore x = \frac{1}{7} \ in \ the \ equation \rightarrow \dfrac{x-1}{2} + \dfrac{x+2}{3} - \dfrac{x-3}{4} =1$

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Answer 2

Answer:

Question:-

$= > \frac{x - 1}{2} + \frac{x + 2}{3} - \frac{x - 3}{4 } = 1</u></p><p></p><p><u>[tex]= > \frac{x - 1}{2} + \frac{x + 2}{3} - \frac{x - 3}{4 } = 1$

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