Solve (question no. 2-22) for x :
3a−15=a5+5253a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}3a−51=5a+552
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3a−15=a5+525⇒3a−15=a5+275⇒3a−a5=275+15⇒15a5−a5=275+15⇒15a−a5=27+15⇒14a5=285⇒a=28×55×14⇒a=14070⇒a=23a - \dfrac{1}{5} = \dfrac{a}{5} + 5\dfrac{2}{5}\\[1em] ⇒ 3a - \dfrac{1}{5} = \dfrac{a}{5} + \dfrac{27}{5}\\[1em] ⇒ 3a - \dfrac{a}{5} = \dfrac{27}{5} + \dfrac{1}{5}\\[1em] ⇒ \dfrac{15a}{5} - \dfrac{a}{5} = \dfrac{27}{5} + \dfrac{1}{5}\\[1em] ⇒ \dfrac{15a - a}{5} = \dfrac{27 + 1}{5} \\[1em] ⇒ \dfrac{14a}{5} = \dfrac{28}{5} \\[1em] ⇒ a = \dfrac{28 \times 5}{5 \times 14}\\[1em] ⇒ a = \dfrac{140}{70}\\[1em] ⇒ a = 23a−51=5a+552⇒3a−51=5a+527⇒3a−5a=527+51⇒515a−5a=527+51⇒515a−a=527+1⇒514a=528⇒a=5×1428×5⇒a=70140⇒a=2
Hence, the value of a is 2.
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3(x + 1) = 12 + 4(x - 1)
3x4−14(x−20)=x4+32\dfrac{3x}{4} - \dfrac{1}{4}(x - 20) = \dfrac{x}{4} + 3243x−41(x−20)=4x+32
x3−212=4x9−2x3\dfrac{x}{3} - 2\dfrac{1}{2} = \dfrac{4x}{9} - \dfrac{2x}{3}3x−221=94x−32x
4(y+2)5=7+5y13\dfrac{4(y + 2)}{5} = 7 + \dfrac{5y}{13}54(y+2)=7+135y
