Solve (question no. 2-22) for x :
3x4−14(x−20)=x4+32\dfrac{3x}{4} - \dfrac{1}{4}(x - 20) = \dfrac{x}{4} + 3243x−41(x−20)=4x+32
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3x4−14(x−20)=x4+32⇒3x4−x4+204=x4+32⇒3x4−x4+5=x4+32⇒3x−x4+5=x4+32⇒2x4+5=x4+32⇒2x4−x4=32−5⇒2x−x4=32−5⇒x4=27⇒x=27×4⇒x=108\dfrac{3x}{4} - \dfrac{1}{4}(x - 20) = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x}{4} - \dfrac{x}{4} + \dfrac{20}{4} = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x}{4} - \dfrac{x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{3x - x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{2x}{4} + 5 = \dfrac{x}{4} + 32\\[1em] ⇒ \dfrac{2x}{4} - \dfrac{x}{4} = 32 - 5\\[1em] ⇒ \dfrac{2x - x}{4} = 32 - 5\\[1em] ⇒ \dfrac{x}{4} = 27\\[1em] ⇒ x = 27 \times 4\\[1em] ⇒ x = 10843x−41(x−20)=4x+32⇒43x−4x+420=4x+32⇒43x−4x+5=4x+32⇒43x−x+5=4x+32⇒42x+5=4x+32⇒42x−4x=32−5⇒42x−x=32−5⇒4x=27⇒x=27×4⇒x=108
Hence, the value of x is 108.
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5(8x + 3) = 9(4x + 7)
3(x + 1) = 12 + 4(x - 1)
3a−15=a5+5253a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}3a−51=5a+552
x3−212=4x9−2x3\dfrac{x}{3} - 2\dfrac{1}{2} = \dfrac{4x}{9} - \dfrac{2x}{3}3x−221=94x−32x
