Solve the following equation and check your answer:
3x−53=x−33x - \dfrac{5}{3} = x - 33x−35=x−3
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We have:
=3x−53=x−3⇒3x−x=53−3[Transposing −53 to RHS and +x to LHS]⇒2x=5−93⇒2x=−43⇒x=−43×2⇒x=−23×1\phantom{=} 3x - \dfrac{5}{3} = x - 3 \\[1em] \Rightarrow 3x - x = \dfrac{5}{3} - 3 \quad \text{[Transposing } -\dfrac{5}{3} \text{ to RHS and +x to LHS]} \\[1em] \Rightarrow 2x = \dfrac{5 - 9}{3} \\[1em] \Rightarrow 2x = -\dfrac{4}{3} \\[1em] \Rightarrow x = -\dfrac{4}{3 \times 2} \\[1em] \Rightarrow x = -\dfrac{2}{3 \times 1}=3x−35=x−3⇒3x−x=35−3[Transposing −35 to RHS and +x to LHS]⇒2x=35−9⇒2x=−34⇒x=−3×24⇒x=−3×12
∴ x = −23-\dfrac{2}{3}−32
Check:
LHS=3x−53LHS=3(−23)−53LHS=−2−53LHS=−113RHS=x−3RHS=−23−3RHS=−113\text{LHS} = 3x - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = 3\left(-\dfrac{2}{3}\right) - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = -2 - \dfrac{5}{3} \\[1em] \phantom{\text{LHS}} = -\dfrac{11}{3} \\[2em] \text{RHS} = x - 3 \\[1em] \phantom{\text{RHS}} = -\dfrac{2}{3} - 3 \\[1em] \phantom{\text{RHS}} = -\dfrac{11}{3}LHS=3x−35LHS=3(−32)−35LHS=−2−35LHS=−311RHS=x−3RHS=−32−3RHS=−311
Hence, LHS = RHS.
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2(y−52)=0.32\Big(y -\dfrac{5}{2}\Big) = 0.32(y−25)=0.3
58x−6=9\dfrac{5}{8}x - 6 = 985x−6=9
0.6x - 1.9 = 0.2x + 0.5
2(3y - 2) - 4(2y - 5) = 9
