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Mathematics

Solve the following equation and check your answer:

x+52+x3=20\dfrac{x + 5}{2} + \dfrac{x}{3} = 20

Linear Eqns One Variable

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Answer

We have:

=x+52+x3=203(x+5)+2x6=203x+15+2x=20×65x+15=1205x=12015[Transposing +15 to RHS]5x=105x=1055\phantom{=} \dfrac{x + 5}{2} + \dfrac{x}{3} = 20 \\[1em] \Rightarrow \dfrac{3(x + 5) + 2x}{6} = 20 \\[1em] \Rightarrow 3x + 15 + 2x = 20 \times 6 \\[1em] \Rightarrow 5x + 15 = 120 \\[1em] \Rightarrow 5x = 120 - 15 \quad \text{[Transposing +15 to RHS]} \\[1em] \Rightarrow 5x = 105 \\[1em] \Rightarrow x = \dfrac{105}{5} \\[1em]

∴ x = 21

Check:

LHS = x+52+x3\dfrac{x + 5}{2} + \dfrac{x}{3}

= 21+52+213\dfrac{21 + 5}{2} + \dfrac{21}{3}

= 13 + 7

= 20

RHS = 20

Hence, LHS = RHS.

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